Chapter 7 : Triangles - Rd Sharma Solutions for Class 10 Maths CBSE

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Chapter 7 - Triangles Excercise Ex. 7.1

Question 1

Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)

(iv) Two triangles are similar, if their corresponding angles are __________. (proportional, equal)

(v) Two triangles are similar, if their corresponding sides are __________. (proportional, equal)
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Solution 1

(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.

(iv) Two triangles are similar, if their corresponding angles are equal.

(v) Two triangles are similar, if their corresponding sides are proportional.
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Question 2

Write the truth value (T/F) of each of the following statements:

(i) Any two similar figures are congruent.

(ii) Any two congruent figures are similar.

(iii) Two polygons are similar, if their corresponding sides are proportional.

(iv) Two polygons are similar, if their corresponding angles are proportional.

(v) Two triangles are similar if their corresponding sides are proportional.

(vi) Two triangles are similar if their corresponding angles are proportional

Solution 2

(i) False

(ii) True

(iii) False

(iv) False

(v) True

(vi) True

Chapter 7 - Triangles Excercise Ex. 7.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

In Fig 7.35, state if PQ || EF.

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Chapter 7 - Triangles Excercise Ex. 7.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

in Fig. 7.57, AE is the AE is the bisector of the exterior $\angle$ CAD Meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Chapter 7 - Triangles Excercise Ex. 7.4

Question 1

In fig., if AB||CD, find the value of x.

Solution 1

Question 2

In fig., if AB || CD, find the value of x.

Solution 2

Question 3

In fig., AB||CD. If OA = 3x - 19, OB = x - 4, OC = x - 3 and OD = 4, find x.

Solution 3

Chapter 7 - Triangles Excercise Ex. 7.5

Question 1

Solution 1

Question 2

In Fig. 7.137, AB || QR. Find the length of PB.

Solution 2

Question 3

In Fig. 7.138, XY || BC. Find the length of XY.

Solution 3

Question 4

In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.

Solution 4

We have:

Question 5

In Fig. 7.140, $\angle$ ABC = 90^oand BD $\perp$ AC. If BD = 8 cm and AD = 4 cm, find CD.

Solution 5

Question 6

In Fig. 7.140, $\angle$ ABC = 90^o and BD $\perp$ AC> If AB = 5.7 cm , BD = 3.8 cm and CD = 5.4 cm, find BC.

Solution 6

Question 7

In fig. 7.141, DE || BC such that AE = (1/4) AC. If AB = 6 cm, find AD

Solution 7

Question 8

Solution 8

Question 9

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show that

Solution 9

Question 10

If ABC and AMP are two right triangles, right angled at B and M respectively such that MAP = BAC. Prove that

Solution 10

Question 11

A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower.

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.

Solution 17

Question 18

In ABC, AL and CM are the perpendiculars from the vertices A anf C to BC and AB respectively. If AL and CM intersect at O, prove that:

(i)

(ii)

Solution 18

Question 19

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Solution 19

Question 20

In an isosceles ABC, the base AB is produced both the ways to P and Q such that AP BQ = AC². Prove that .

Solution 20

Question 21

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2m/sec. If the lamp is 3.6m above the ground, find the length of her shadow after 4 seconds.

Solution 21

Question 22

A vertical stick of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution 22

Question 23

In fig. 7.144, ΔABC is right angled at C and DE $\perp$ AB. prove that ΔABC $\sim$ ΔADE and hence find the lengths of AE and DE.

Solution 23

Question 24

Solution 24

Question 25

In Fig. 7.144, We have AB||CD||EF, if AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.

Solution 25

Chapter 7 - Triangles Excercise Ex. 7.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

The areas of two similar traingles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?

Solution 7

Question 8

The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

Solution 8

Question 9

The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

Solution 9

Question 10

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Two isosceles triangles have equal vertical angles and their areas area in the ratio 36:25. Find the ratio of their corresponding heights.

Solution 24

Question 25

In Fig.7.180, ΔABC and ΔDBC are two triangles on the same base BC. If AD intersects BC at O,

show that

Solution 25

Since ABC and DBC are one same base,
Therefore ratio between their areas will be as ratio of their heights.
Let us draw two perpendiculars AP and DM on line BC.

In APO and DMO,
APO = DMO    (Each is90^o)
AOP = DOM         (vertically opposite angles)
OAP = ODM        (remaining angle)
Therefore APO ~ DMO   (By AAA rule)

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 7 - Triangles Excercise Ex. 7.7

Question 1

Solution 1

Question 2

The sides of a triangle are a = 7 cm, b = 24 cm and c = 25 cm. Determine whether it is a right triangle.

Solution 2

Question 3

The sides of a triangle are a = 9 cm, b = 16 cm and c = 18 cm. Determine whether it is a right triangle.

Solution 3

Question 4

The sides of a triangle are a = 1.6 cm, b = 3.8 cm and c = 4 cm. Determine whether it is a right triangle.

Solution 4

Question 5

The sides of a triangle are a = 8 cm, b = 10 cm and c = 6 cm. Determine whether it is a right triangle.

Solution 5

Question 6

A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?

Solution 6

Question 7

A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

Solution 7

Question 8

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution 8

Let CD and AB be the poles of height 11 and 6 m.
Therefore CP = 11 - 6 = 5 m
From the figure we may observe that AP = 12m
In triangle APC, by applying Pythagoras theorem

Therefore distance between their tops = 13 m.

Question 9

In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Using pythagoras theorem determine the length of AD terms of b and c shown in Fig. 7.221.

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

In Fig. 7.222, $\angle$ B<90^o and segment AD $\perp$ BC, show that

Solution 20

(i)

Question 21

$begin mathsize 12px style In space an space equilateral space increment space ABC comma space AD perpendicular BC comma space prove space that space AD squared equals 3 BD squared end style$

Solution 21

Question 22

ABD is a right triangle right angled at A and AC BD. Show that
(i)   AB² = BC . BD
(ii)   AC² = BC . DC
(iii)   AD² = BD . CD

(iv) AB²/ AC² = BD/ DC

Solution 22

Question 23

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Solution 23

Question 24

Determine whether the triangle having sides (a - 1) cm, cm and (a + 1) cm is a right angled triangle.

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

In Fig. 7.223, D is the mid-point of side BC and AE $\perp$ BC. If

Solution 27

Question 28

Solution 28

(i)

Question 29

Solution 29

Question 30

Solution 30

Question 31

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after

Solution 31

Chapter 7 - Triangles Excercise Rev. 7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

In Fig. 7.226, DE || CB. Determine AC and AE

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

In the following figure, you find two triangles. Indicate whether the traingles are similar. Give reasons in support of your answer.

Solution 13

Incomplete question (two triangles are not given in the figure).

Question 14

In the following figure, you find two triangles. Indicate whether the traingle are similar. Give reasons in support of your answer.

Solution 14

Incomplete question (two triangles are not given in the figure).

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

In Fig. 7.229, Δ $\sim$ CMD; determine MD in terms of x, y and z.

Solution 18

Question 19

Solution 19

Question 20

In Fig. 7.230, l || m

Solution 20

Question 21

In Fig. 7.231, AB || DC Prove that

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

In Fig. 7.232, each of PA , QB, RC and SD is perpendicular to l. if AB = 6 cm, BC = 9 cm, CD = 12 cm and PS = 36 cm, then determine PQ, QR and RS.

Solution 28

Question 29

In each of the figure given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segments are marked in each figure. Determine x,y,z in each case.

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

There is a staircase as shown in Fig. 7.234, connecting points A and B. Measurements of steps are marketed in the figure. Find the straight line distance between A and B.

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (See fig.)? If she pulls the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds.

Solution 46

The given information can be represented by the figure given below.

Chapter 7 - Triangles Excercise 7.131

Question 1

Sides of two similar triangles are in the ratio 4:9.

Areas of these triangles are in the ratio

(a) 2:3 (b) 4:9 (c) 81:16 (d)16:81

Solution 1

We know if sides of two similar triangles are in ratio a:b then area of these triangles are in ratio a²b²

According to question, ratio of sides= 4:9

Hence ratio of areas = 4²:9²

= 16:81

So, the correct option is (d).

Question 2

The areas of two similar triangles are in respectively 9 cm² and 16 cm². The ratio of this corresponding sides is

(a) 3:4 (b) 4:3 (c)2:3 (d) 4:5

Solution 2

$begin mathsize 11px style table attributes columnalign left end attributes row cell If straight space areas straight space of straight space two straight space similar straight space triangles straight space are space in space the straight space ratio straight space straight a colon straight b straight space end cell row cell then straight space their straight space corresponding straight space sides straight space are straight space in space the straight space ratio square root of straight a colon square root of straight b end cell row cell Hence comma given space ratio space of space areas space is space 9 colon 16 end cell row cell so space ratio space of space corresponding space sides space is space square root of 3 colon square root of 16 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 3 colon 4 end cell end table end style$

So, the correct option is (a).

Question 3

The areas of two similar triangles ∆ ABC and ∆ DEF are 144 cm² and 81 cm² respectively. If the longest side of larger ∆ABC be 36cm, then longest side of smaller triangle ∆DEF is

(a) 20 cm (b) 26 cm (c)27 cm (d) 30 cm

Solution 3

$begin mathsize 11px style table attributes columnalign left end attributes row cell ratio text of areas of triangle = end text 144 over 81 end cell row cell ratio text of their corresponding sides = end text fraction numerator square root of 144 over denominator square root of 81 end fraction end cell row cell text = end text 12 over 9 end cell row cell rightwards double arrow text end text fraction numerator side text of end text straight capital delta text ABC end text over denominator side text of end text ΔDEF end fraction equals 12 over 9 end cell row cell rightwards double arrow text end text fraction numerator 36 over denominator sides text of end text ΔDEF end fraction equals 12 over 9 end cell row cell rightwards double arrow text side of end text ΔDEF equals fraction numerator 36 straight x 9 over denominator 12 end fraction end cell row cell text = 27 cm end text end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Chapter 7 - Triangles Excercise 7.132

Question 1

∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of areas of triangles ABC and BDE is

(a) 2:1 (b) 1:2 (c) 4:1 (d)1:4

Solution 1

$begin mathsize 11px style table attributes columnalign left end attributes row cell Equilateral space triangles space are space also space similar space triangles. end cell row cell As space straight D space is space mid space point space of space BC space end cell row cell Hence space BC space equals space 2 space BD end cell row cell space space space space space rightwards double arrow straight space BC over BD equals 2 colon 1 end cell row cell ratio space of space corresponding space sides space are space 2 colon 1 end cell row cell ratio space of space areas space of space triangles space is space space 2 to the power of straight 2 colon 1 squared end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 colon 1 end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 2

If ∆ABC and ∆DEF are similar such that 2AB+DE and BC=8 cm, then EF=

(a) 16 cm (b)12 cm (c) 8 cm (d) 4 cm

Solution 2

$begin mathsize 11px style table attributes columnalign left end attributes row cell ΔABC space is space similar space to space ΔDEF end cell row Hence row cell space space space space space space AB over DE equals BC over EF space space space space space space space space space space space space space space minus negative negative box enclose 1 end cell row cell rightwards double arrow space space given space 2 AB space equals space DE space space space and space BC equals 8 space cm end cell row cell space space space space space space from space box enclose 1 end cell row cell space space space space space space 1 half equals 8 over EF end cell row cell rightwards double arrow space space EF space equals space 16 space cm end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 3

$begin mathsize 11px style table attributes columnalign left end attributes row cell If text end text ΔABC text end text and text end text ΔDEF text end text are text end text two text end text triangles text end text such text end text that end cell row cell AB over DE equals BC over EF equals CA over FD equals 2 over 5 comma text then area end text left parenthesis straight capital delta text ABC end text right parenthesis text : Area end text left parenthesis ΔDEF right parenthesis equals end cell row cell left parenthesis straight a right parenthesis 2 colon 5 text end text left parenthesis straight b right parenthesis 4 colon 25 text end text left parenthesis straight c right parenthesis 4 colon 15 text end text left parenthesis straight d right parenthesis 8 colon 125 end cell end table end style$

Solution 3

All these pairs of corresponding sides are in the same proportion so by SSS similarity criteria triangle ∆ABC are similar.

Given ratio of sides = 2.5

So, ratio of areas = 2²:5²

= 4:25

So, the correct option is (b).

Question 4

XY is drawn parallel to the base BC of a ∆ABC cutting AB at x and X and AC at Y. If AB=4BX and YC=2 cm

Then AY=

(a) 2 cm (b) 4 cm (c) 6 cm (d) 8 cm

Solution 4

$begin mathsize 11px style table attributes columnalign left end attributes row cell Given straight space AB over BX equals 4 end cell row cell XY parallel to BC space Hence end cell row cell angle AXY space equals space angle ABC space space space and end cell row cell angle AYX space equals space angle ACB end cell row cell By space AAA space similarity space criteria comma space triangles space are space similar end cell row cell Given space space AB over BX = 4 ---- box enclose 1 end cell row cell rightwards double arrow straight space AB over BX minus 1 equals 4 minus 1 end cell row cell rightwards double arrow straight space fraction numerator AB minus BX over denominator BX end fraction equals 3 end cell row cell rightwards double arrow straight space AX over BX equals 3 end cell row cell space rightwards double arrow straight space BX over AX equals 1 third ----- box enclose 2 end cell row cell multiply straight space box enclose 1 space space by space box enclose 2 end cell row cell rightwards double arrow straight space AB over BX straight space cross times straight space BX over AX straight space cross times 4 cross times 1 third end cell row cell rightwards double arrow AB over AX = 4 over 3 end cell row cell straight space rightwards double arrow Also straight space AB over AX equals AC over AY space space left parenthesis Since space XY parallel to BC right parenthesis end cell row cell rightwards double arrow straight space 4 over 3 equals fraction numerator AY plus YC over denominator AY end fraction end cell row cell rightwards double arrow 4 over 3 equals fraction numerator AY plus 2 over denominator AY end fraction end cell row cell rightwards double arrow space 4 AY space equals space 3 AY plus 6 end cell row cell rightwards double arrow straight space box enclose AY equals 6 space cm end enclose end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 5

Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is

(a) 12 m (b)14 m (c)13 m (d)11 m

Solution 5

$begin mathsize 11px style angle ACB equals angle DCE space left parenthesis common space angle right parenthesis angle ABC equals angle DEC space left parenthesis both space are space right space angles right parenthesis Since space DE parallel to AC comma space angle BAC equals angle EDC space left parenthesis corresponding space angles right parenthesis table attributes columnalign left end attributes row cell ΔABC straight space is space similar space to space ΔDEC end cell row cell by straight space AAA space similarity space criterion end cell row cell so space AB over DE straight space equals straight space BC over EC equals straight space AC over DC end cell row cell rightwards double arrow straight space 11 over 6 equals straight space fraction numerator BE space plus space EC over denominator EC end fraction end cell row cell rightwards double arrow straight space 11 over 6 equals straight space BE over EC plus 1 end cell row cell rightwards double arrow straight space 5 over 6 equals straight space BE over EC end cell row cell rightwards double arrow space EC space equals space fraction numerator BE straight space cross times space 6 over denominator 5 end fraction equals straight space fraction numerator 12 cross times 6 over denominator 5 end fraction end cell row cell rightwards double arrow space EC space equals space 14.4 space straight m end cell row cell In space ΔDEC space comma space DC to the power of straight 2 equals space DE to the power of straight 2 plus EC to the power of straight 2 end cell row cell space space space space space space space space space space space space space space space space space space space space rightwards double arrow space DC to the power of straight 2 equals 6 squared plus 14.4 squared rightwards double arrow DC equals 15.6 space straight m end cell row cell In space ΔABC space comma space AC to the power of straight 2 equals space AB to the power of straight 2 plus BC to the power of straight 2 end cell row cell space space space space space space space space space space space space space space space space space space space space rightwards double arrow space AC to the power of straight 2 equals 11 squared plus 26.4 squared rightwards double arrow AC equals 28.6 space straight m end cell row cell space space space space space space space space space space space space space space space space space AD space equals space AC space minus space DC equals 13 space straight m end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 6

In ∆ABC, D and E are points on side AB and AC respectively such that DE∥BC and AD:DB=3:1

If EA = 3.3 cm, then AC=

(a) 1.1 cm (b) 4 cm (c) 4.4 cm (d) 5. 5 cm

Solution 6

$begin mathsize 11px style table attributes columnalign left end attributes row cell AD over DB equals 3 over 1 space space space space space space space space space space space space EA equals 3.3 space cm space minus negative negative negative negative negative box enclose 1 end cell row cell ΔADE space is space similar space to space ΔABC end cell row cell so space space space AD over DB equals AE over AC space space space space minus negative negative negative negative box enclose 2 end cell row cell gives straight space AD over BD equals 3 end cell row cell rightwards double arrow straight space AD over BD plus 1 equals 3 plus 1 end cell row cell rightwards double arrow straight space fraction numerator AD plus BD over denominator BD end fraction equals 4 end cell row cell rightwards double arrow straight space AB over BD equals 4 end cell row cell rightwards double arrow straight space BD over AB equals 1 fourth space rightwards double arrow straight space 1 minus BD over AB equals 1 minus 1 fourth end cell row cell rightwards double arrow straight space AD over AB equals 3 over 4 space space space space space space space space space minus negative negative negative box enclose 3 end cell row cell from straight space box enclose 1 comma straight space box enclose 2 comma straight space box enclose 3 end cell row cell space space space space space space space 3 over 4 equals fraction numerator 3.3 over denominator AC end fraction end cell row cell rightwards double arrow space space AC space equals space 4.4 cm end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 7

In triangles ABC and DEF, ∠A =∠E=40°, AB:ED= AC : EF and ∠F = 65°, then ∠B=

(a) 35° (b) 65° (c) 75° (d) 85°

Solution 7

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB over ED equals AC over EF end cell row cell ΔABC space and space ΔEDF space are space similar space by space SAS space criterion end cell row cell so space angle straight B equals angle straight D end cell row cell given straight space angle straight D equals 180 degree minus 40 degree minus 65 degree end cell row cell space space space space space space equals space 75 degree end cell row cell so straight space box enclose angle straight B equals 75 degree end enclose end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 8

If ABC and DEF are similar triangles such that ∠A=47° and ∠E = 83°, then ∠C=

(a)50° (b)60° (c)70° (d)80°

Solution 8

$begin mathsize 11px style table attributes columnalign left end attributes row cell As straight space ΔABC straight space and straight space ΔDEF space are straight space similar colon end cell row cell So space space space angle straight A straight equals angle straight D & angle straight C equals angle straight F end cell row cell rightwards double arrow straight space angle straight D space equals space 47 degree end cell row cell so comma straight space angle straight F space equals space 180 degree minus 83 degree minus 47 degree end cell row cell = 50 degree end cell row cell so box enclose angle straight C space equals space 50 degree end enclose end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 9

If D, E, F are the mid-points of sides BC, CA and AB respectively of ∆ABC, then the ratio of areas of triangles DEF and ABC is

(a) 1:4 (b) 1:2 (c) 2:3 (d)4:5

Solution 9

$begin mathsize 11px style table attributes columnalign left end attributes row cell FE parallel to BC end cell row cell and space straight F comma space straight E space are space mid space points end cell row cell so space EF space equals space 1 half BC end cell row cell similarly space DE space equals space 1 half AB end cell row cell space space space space space space space & space space space FD space equals space 1 half AC end cell row cell So straight space ΔABC space and space ΔDEF space are space similar space and space their space sides space are space in end cell row cell ratio space 2 colon 1 end cell row cell Hence space ratio space of space areas space 2 to the power of straight 2 colon 1 squared end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space 4 colon 1 end cell row cell So space ratio space of space areas space of space ΔDEF space and space ΔABC space is space 1 colon 4 end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 10

In an equilateral triangle ABC, if AD ⏊ BC, then

(a) 2AB²=3AD² (b) 4 AB² =3AD²

Solution 10

$begin mathsize 11px style table attributes columnalign left end attributes row cell BD squared plus AD squared equals AB squared --- box enclose 1 end cell row cell DC squared plus AD squared equals AC squared --- box enclose 2 end cell row cell ΔABC space is space equilateral comma space AB space equals space BC space equals space AC space equals space straight a end cell row cell space Let space space space space space space BD space equals space straight x space then space DC space equals space BC space minus space BD end cell row cell space space space space space space = straight a minus straight x end cell row cell from straight space box enclose 1 & straight space box enclose 2 end cell row cell straight x squared plus AD squared equals straight a squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space and space space space space space space space space space space space space space space space left parenthesis straight a minus straight x right parenthesis squared plus AD squared equals straight a squared end cell row cell rightwards double arrow left parenthesis straight a minus square root of straight a squared minus AD squared end root right parenthesis squared plus AD squared equals straight a squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow left parenthesis straight a minus straight x right parenthesis squared equals straight a squared minus AD squared end cell row cell rightwards double arrow straight a squared plus straight a squared minus AD squared minus 2 straight a square root of straight a squared minus AD squared end root plus AD squared equals straight a squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow straight a minus straight x equals square root of straight a squared minus AD squared end root end cell row cell rightwards double arrow straight a squared equals 2 straight a square root of straight a squared minus AD squared end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space straight x space equals space straight a minus square root of straight a squared minus AD squared end root end cell row cell rightwards double arrow straight a to the power of 4 equals 4 straight a squared left parenthesis straight a squared minus AD squared right parenthesis end cell row cell rightwards double arrow straight a squared equals 4 left parenthesis straight a squared minus AD squared right parenthesis end cell row cell rightwards double arrow straight a to the power of 4 equals 4 straight a squared minus 4 AD squared end cell row cell rightwards double arrow 4 AD squared equals 3 straight a squared end cell row cell rightwards double arrow 4 AD squared equals 3 AB squared end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 11

$begin mathsize 11px style If space increment ABC space is space an space equilateral space triangle space such space that space AD perpendicular BC comma space space then space AD squared equals space space left parenthesis straight a right parenthesis 3 over 2 DC squared text end text left parenthesis 2 right parenthesis 2 DC squared text end text left parenthesis 3 right parenthesis 2 CD squared text end text 4 DC squared end style$

Solution 11

$Error converting from MathML to accessible text.$

Question 12

In a ∆ABC, AD is the bisector of ∠BAC. If AB = 6 cm,

AC=5 cm and BD=3 cm, then DC=

(a) 11.3 cm (b)2.5cm (c) 3.5 cm (d) None of these

Solution 12

$begin mathsize 11px style By space angle space bisector space theorem table attributes columnalign left end attributes row cell text end text AC over AB equals CD over DB end cell row cell rightwards double arrow text end text 5 over 6 equals CD over 3 end cell row cell rightwards double arrow CD equals 5 over 2 equals 2.5 cm end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 13

In a ∆ABC, AD is bisector of ∠BAC. If AB=8cm, BD=6cm, and DC=3cm, Find AC

(a) 4cm (b)6cm (c)3cm (d)8cm

Solution 13

$begin mathsize 11px style table attributes columnalign left end attributes row cell By space angle space bisector space theorem end cell row cell space space space AC over AB equals CD over BD end cell row cell rightwards double arrow AC over 8 equals 3 over 6 end cell row cell rightwards double arrow AC over 8 equals 1 half end cell row cell rightwards double arrow AC equals 4 space cm end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 14

$begin mathsize 11px style ABCD space is space straight a space trapezium space such space that space BC parallel to AD space and space AB equals 4 space cm. space If space the space diagonals space AC space and space BD space intersect space at space straight O space such space that space AO over OC equals DO over OB equals 1 half comma then text BC= end text left parenthesis straight a right parenthesis space 7 space cm space space space space straight b right parenthesis 8 space cm space space space space space space space left parenthesis straight c right parenthesis 9 space cm space space space space space space left parenthesis straight d right parenthesis 6 cm end style$

Solution 14

$Error converting from MathML to accessible text.$

Chapter 7 - Triangles Excercise 7.133

Question 1

If ABC is a right triangle, right angled at B and M, N are mid points of AB and BC respectively, then 4(AN²+CM²)=

(a) 4 AC² (b) 5 AC² (a) 4 AC² (c) 4 AC²

Solution 1

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB squared plus BN squared equals AN squared space space space space space space space space space space space space space minus negative negative negative box enclose 1 end cell row cell BC squared plus BM squared equals MC squared space space space space space space space space space space space space space minus negative negative negative box enclose 2 end cell row cell AB squared plus BC squared equals AC squared space space space space space space space space space space space space space minus negative negative negative box enclose 3 end cell row cell Since space straight M space and space straight N space are space mid minus points space of space AB space and space BC comma space MN space is space parallel space to space AC space and space ΔBMN space is space similar space to space space ΔBAC. end cell row cell rightwards double arrow space Hence space BM over BA equals BN over BC equals MN over AC end cell row cell and space BN space equals space 1 half BC space and space BM equals 1 half BA end cell row cell so straight space MN over AC equals 1 half ----- box enclose 4 end cell row cell from space adding space box enclose 1 & box enclose 2 end cell row cell AB squared plus BC squared plus BN squared plus BM squared equals AN squared plus MC squared end cell row cell rightwards double arrow space AC squared plus MN squared equals AN squared plus MC squared end cell row cell From straight space box enclose 4 space space space space space AC squared plus left parenthesis AC over 2 right parenthesis to the power of straight 2 = AN squared plus MC squared end cell row cell rightwards double arrow 4 left parenthesis AN squared plus MC squared right parenthesis equals 5 AC squared end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 2

$begin mathsize 11px style table attributes columnalign left end attributes row cell table attributes columnalign left end attributes row cell If text end text in text end text straight capital delta text end text ABC text end text and text end text straight capital delta text end text DEF comma end cell end table text end text AB over DE equals BC over FD comma then text end text straight capital delta text end text ABC tilde straight capital delta text end text DEF end cell row when row cell left parenthesis straight a right parenthesis angle straight A equals angle straight F text end text left parenthesis straight b right parenthesis angle straight A equals angle straight D text end text left parenthesis straight c right parenthesis angle straight B equals angle straight D text end text left parenthesis straight b right parenthesis angle straight B equals angle straight E end cell end table end style$

Solution 2

For triangles to be similar by SAS

∠B = ∠D

So, the correct option is (c).

Question 3

$begin mathsize 11px style table attributes columnalign left end attributes row cell If text in two trianles ABC and DEF, end text AB over DE equals BC over FE equals CA over FD end cell row then row cell left parenthesis straight a right parenthesis ΔFDE tilde ΔCAB text end text left parenthesis straight b right parenthesis ΔFDE tilde ΔABC end cell row cell left parenthesis straight c right parenthesis ΔCBA tilde ΔFDE text end text left parenthesis straight c right parenthesis ΔBCA tilde ΔFDE end cell end table end style$

Solution 3

$begin mathsize 11px style Given text end text AB over DE equals BC over FE equals CA over FD ΔCAB tilde ΔFDE So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 4

$begin mathsize 11px style table attributes columnalign left end attributes row cell ΔABC tilde ΔDEF comma text ar end text left parenthesis ΔABC right parenthesis equals 9 cm squared comma text ar end text left parenthesis ΔDEF right parenthesis equals 16 cm squared. end cell row cell If text BC=2.1 cm, then the measure of EF is end text end cell row cell left parenthesis straight a right parenthesis 2.8 cm text end text left parenthesis straight b right parenthesis 4.2 cm text end text left parenthesis straight c right parenthesis 2.5 cm text end text left parenthesis straight d right parenthesis 4.1 cm end cell end table end style$

Solution 4

$begin mathsize 11px style table attributes columnalign left end attributes row cell fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔDEF right parenthesis end fraction equals 9 over 16 end cell row cell so space as space the space triangles space are space similar end cell row cell BC over EF equals square root of fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔDEF right parenthesis end fraction end root end cell row cell rightwards double arrow straight space fraction numerator 2.1 over denominator EF end fraction equals square root of 9 over 16 end root end cell row cell rightwards double arrow straight space fraction numerator 2.1 over denominator EF end fraction equals 3 over 4 end cell row cell rightwards double arrow EF equals fraction numerator 2.1 space straight x space 4 over denominator 3 end fraction equals 2.8 cm end cell end table end style$

So, the correct option is (a).

Question 5

$begin mathsize 11px style The space length space of space the space hypotenuse space of space an space isosceles space right space triangle space whose space one space side space is space 4 square root of 2 space cm space is left parenthesis straight a right parenthesis 12 cm space space space space space left parenthesis straight b right parenthesis 8 cm space space space space left parenthesis straight c right parenthesis 8 square root of 2 cm space space space left parenthesis straight d right parenthesis 12 square root of 2 cm space space space end style$

Solution 5

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB equals AC end cell row cell and space space AB space equals space 4 square root of 2 space cm end cell row cell also space space space AB to the power of straight 2 plus AC to the power of straight 2 equals BC to the power of straight 2 end cell row cell rightwards double arrow space space space 2 AB to the power of straight 2 equals BC to the power of straight 2 end cell row cell rightwards double arrow space space BC space equals space square root of 2 straight space AB end cell row cell space space space space space space space space space space space equals space square root of 2 space straight x space 4 square root of 2 end cell row cell space space space space space space space space space space space equals space 8 space cm end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 6

A man goes 24 m due west and then 7 m due north.

How far is he from the starting point?

(a)31m (b)17m (c)25m (d)26m

Solution 6

$begin mathsize 11px style table attributes columnalign left end attributes row cell He space is space AC space straight m space for end cell row cell and space space AC to the power of straight 2 equals AB squared plus BC squared end cell row cell space space space space space space space space space space space space space space equals space 24 to the power of straight 2 plus 7 squared end cell row cell space space space space space space space space space space space space space space equals space 576 space plus space 49 end cell row cell space space space space space space space space space space space space space space equals space 625 end cell row cell space space space space space space space space AC space space equals space square root of 625 end cell row cell space space space space space space space space AC space space equals space 25 straight m end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 7

$begin mathsize 11px style table attributes columnalign left end attributes row cell ΔABC tilde ΔDEF. text If BC=3cm, EF=4 cm and ar end text left parenthesis ΔABC right parenthesis equals 54 cm squared end cell row cell then text ar end text left parenthesis ΔDEF right parenthesis equals end cell row cell left parenthesis straight a right parenthesis 108 cm squared text end text left parenthesis straight b right parenthesis 96 cm squared text end text left parenthesis straight c right parenthesis 48 cm squared text end text left parenthesis straight d right parenthesis 100 cm squared end cell end table end style$

Solution 7

$begin mathsize 11px style table attributes columnalign left end attributes row cell Triangles space are space similar. end cell row cell rightwards double arrow space fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔDEF right parenthesis end fraction equals left parenthesis BC over EF right parenthesis squared end cell row cell rightwards double arrow straight space fraction numerator 54 over denominator ar left parenthesis ΔDEF right parenthesis end fraction equals left parenthesis 3 over 4 right parenthesis squared end cell row cell rightwards double arrow straight space ar left parenthesis ΔDEF right parenthesis equals left parenthesis fraction numerator 54 straight space straight x straight space 16 over denominator 9 end fraction right parenthesis end cell row cell space space space space space space space space space space space space space space space space space space space space equals space 96 space cm to the power of straight 2 end cell end table end style$

So, the correct option is (b).

Question 8

$begin mathsize 11px style table attributes columnalign left end attributes row cell ΔABC tilde ΔPQR text such that ar end text left parenthesis ΔABC right parenthesis equals 4 ar left parenthesis ΔPQR right parenthesis. text If BC=12 cm end text end cell row cell text then QR= end text end cell row cell left parenthesis straight a right parenthesis 9 cm text end text left parenthesis straight b right parenthesis 10 cm text end text left parenthesis straight c right parenthesis 6 cm text end text left parenthesis straight d right parenthesis 8 cm end cell end table end style$

Solution 8

$begin mathsize 11px style table attributes columnalign left end attributes row cell fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔPQR right parenthesis end fraction equals left parenthesis BC over QR right parenthesis squared ---- box enclose 1 end cell row cell Given ar left parenthesis ΔABC right parenthesis equals 4 straight space ar left parenthesis ΔPQR right parenthesis end cell row cell rightwards double arrow fraction numerator ar left parenthesis ΔABC right parenthesis over denominator ar left parenthesis ΔPQR right parenthesis end fraction equals 4 ---- box enclose 2 end cell row cell and BC equals space 12 space cm space space space space space space space space space space space space space space minus negative negative negative box enclose 3 end cell row cell rightwards double arrow from space space box enclose 1 comma straight space box enclose 2 , box enclose 3 end cell row cell rightwards double arrow 4 = left parenthesis 12 over QR right parenthesis squared end cell row cell rightwards double arrow 2 = 12 over QR end cell row cell rightwards double arrow space QR space equals space 6 space cm end cell end table end style$

So, the correct option is (c).

Question 9

The areas of two similar triangles are 121cm² and 64cm² respectively. If the median of the first triangle is 12.1cm, then the corresponding median of the other triangle is

(a) 11cm (b)8.8cm (c)11.1cm (d)8.1cm

Solution 9

$begin mathsize 11px style table attributes columnalign left end attributes row cell space space space fraction numerator area space of space straight capital delta subscript straight 1 over denominator area space of space straight capital delta subscript 2 end fraction equals left parenthesis fraction numerator median space of space straight capital delta subscript straight 1 straight space over denominator median space of space straight capital delta subscript 2 end fraction right parenthesis squared end cell row cell rightwards double arrow 121 over 64 equals left parenthesis fraction numerator 12.1 over denominator straight M end fraction right parenthesis squared end cell row cell rightwards double arrow space 12.1 divided by straight M space equals 11 over 8 end cell row cell rightwards double arrow straight M equals fraction numerator 12.1 cross times straight space 8 over denominator 11 end fraction end cell row cell space space space space space space equals space 8.8 space cm end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 10

In an equilateral triangle ABC if AD⊥ BC, then AD²

(a) CD² (b) 2CD²(c) 3CD²(d) 4CD²

Solution 10

$begin mathsize 11px style table attributes columnalign left end attributes row cell BD straight space equals straight space DC equals AB over 2 end cell row cell rightwards double arrow space space AD to the power of straight 2 plus BD squared equals AB squared end cell row cell rightwards double arrow space space AD to the power of straight 2 plus CD squared equals left parenthesis 2 CD right parenthesis squared straight space end cell row cell rightwards double arrow space space AD to the power of straight 2 equals 4 CD squared minus CD squared straight space end cell row cell rightwards double arrow space space space AD to the power of straight 2 equals 3 CD squared end cell end table end style$

So, the correct option is (c).

Question 11

In an equilateral triangle ABC if AD⊥ BC, then

(a) 5AB²= 4AD² (b) 3AB²= 4AD²

Solution 11

$begin mathsize 11px style table attributes columnalign left end attributes row cell BD equals DC equals AB over 2 end cell row cell and text AD end text to the power of text 2 end text end exponent plus BD squared equals AB squared end cell row cell rightwards double arrow text AD end text to the power of text 2 end text end exponent plus left parenthesis AB squared over 4 right parenthesis equals text AB end text to the power of text 2 end text end exponent end cell row cell rightwards double arrow text AD end text to the power of text 2 end text end exponent equals fraction numerator 3 AB squared over denominator 4 end fraction end cell row cell rightwards double arrow text end text 3 AB squared equals 4 AD squared end cell end table end style$

So, the correct option is (b).

Question 12

In an isosceles triangle ABC if AC=BC and AB²=2AC² then ∠C

(a) 30⁰ (b)45⁰ (c)90⁰ (d)60⁰

Solution 12

$begin mathsize 11px style table attributes columnalign left end attributes row cell AC equals BC space space space and space AB to the power of straight 2 equals 2 AC squared end cell row cell space space space space space space space space space space space space space space space space space space space space space AB to the power of straight 2 equals AC squared plus AC squared end cell row cell space space space space space space space space space space space space space space space space space space space space space AB to the power of straight 2 equals AC squared plus BC squared end cell row cell It space means space angle straight C space is space 90 to the power of straight 0 end cell row cell So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end cell end table end style$

Question 13

$begin mathsize 11px style increment ABC space is space an space isosceles space triangle space in space which space angle straight C equals 90 degree. space If space AC equals space 6 space cm space then space AB equals space space left parenthesis straight a right parenthesis 6 square root of 2 space cm space space space left parenthesis straight b right parenthesis 6 cm space space space left parenthesis straight c right parenthesis space 2 square root of 6 straight space cm space space left parenthesis straight d right parenthesis straight 4 square root of 2 space cm space end style$

Solution 13

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB squared equals AC squared plus BC squared space space space and space AC space equals space BC equals 6 end cell row cell space space space space space space equals space 6 to the power of straight 2 plus space 6 to the power of straight 2 end cell row cell space space space space space space equals space 72 end cell row cell AB space equals space square root of 72 equals space 6 square root of 2 space cm end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Chapter 7 - Triangles Excercise 7.134

Question 1

$begin mathsize 11px style If space in space two space triangles space ABC space and space DEF comma space angle straight A equals angle straight E comma space angle straight B equals angle straight F comma space then space which space of space the space following space is space not space true ? table attributes columnalign left end attributes row cell left parenthesis straight a right parenthesis BC over DF equals AC over DE text end text left parenthesis straight b right parenthesis AB over DE equals BC over DF text end text end cell row cell left parenthesis straight c right parenthesis AB over EF equals AC over DF text end text left parenthesis straight d right parenthesis BC over DF equals AB over EF text end text end cell end table end style$

Solution 1

$begin mathsize 11px style angle straight A equals angle straight E space and space angle straight B equals angle straight F comma space so space 3 rd space angle space of space t h e space t riangle space must space be space equal. space Hence space by space AAA space similarity space criteria space increment ABC space ∿ space increment EFD Hence AB over EF equals BC over FD equals AC over ED On space seeing space options comma space only space left parenthesis straight b right parenthesis space is space incorrect. space So space false space option space is space left parenthesis straight b right parenthesis. end style$

So, the correct option is (b).

Question 2

In fig, measures of ∠D and ∠F are respectively

(a) 50°, 40°(b) 20°, 30° (c) 40°, 50° (d) 30°, 30°

Solution 2

$begin mathsize 11px style table attributes columnalign left end attributes row cell In straight space ΔABC comma space space angle straight A space plus space angle straight B plus angle straight c equals 180 degree end cell row cell thin space thin space thin space thin space thin space thin space thin space rightwards double arrow space space angle straight A space plus space 30 plus 20 equals 180 end cell row cell space space space space space space space rightwards double arrow straight space angle straight A space equals space 130 degree end cell row cell also straight space AB over EF equals AC over DE space space and space angle straight A space equals space angle straight E end cell row cell Hence space by space SAS space similarity space criterion comma space ΔBAC tilde ΔFED end cell row cell and space space angle straight B space equals space angle straight F space space space and space space angle straight C space equals space angle straight D end cell row cell rightwards double arrow space space angle straight F space equals 30 degree space space and space angle straight D space equals space 20 degree end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 3

In fig, the value of x for which DE||BC is

(a) 4 (b)1 (c)3 (d)2

Solution 3

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB straight space equals straight space AD plus DB end cell row cell space space space space equals straight x plus 3 straight space plus straight space 3 straight x plus 19 end cell row cell AB straight space equals straight space 4 straight x straight space plus straight space 22 end cell row cell AC straight space equals straight space AE straight space plus straight space EC equals straight space straight x straight space plus straight space 3 straight x straight space plus 4 end cell row cell AC straight space equals straight space 4 straight x straight space plus 4 end cell row cell table attributes columnalign left end attributes row cell space DE parallel to AE end cell row cell So comma straight space angle ADE space equals space angle straight B space space and space angle AED space equals space angle straight C space end cell row cell Hence space by space AA comma space space ΔABC tilde ΔADE end cell row cell so comma space AD over AB equals AE over AC end cell row cell rightwards double arrow space space fraction numerator straight x plus 3 over denominator 4 straight x plus 22 end fraction equals fraction numerator straight x over denominator 4 straight x plus 4 end fraction end cell row cell rightwards double arrow space 4 straight x to the power of straight 2 plus 16 straight x plus 12 equals 4 straight x squared plus 22 straight x end cell row cell rightwards double arrow space 6 straight x space equals space 12 end cell row cell rightwards double arrow space straight x equals space 2 end cell end table end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Question 4

In fig, if ∠ADE = ∠ABC, then CE=

(a) 2 (b)5 (c) $begin mathsize 12px style 9 over 2 end style$ (d)3

Solution 4

$begin mathsize 11px style In space increment ADE space and space increment ABC space angle ADE space equals space angle ABC space left parenthesis given right parenthesis angle DAE space equals space angle BAC space left parenthesis common space angle right parenthesis Hence space increment ADE space and space increment ABC space left parenthesis By space AA space criterion right parenthesis table attributes columnalign left end attributes row cell So comma text end text AD over AB equals AE over AC end cell row cell rightwards double arrow text end text fraction numerator AD over denominator AD plus DB end fraction equals fraction numerator AE over denominator AE plus EC end fraction end cell row cell rightwards double arrow text end text 2 over 5 equals fraction numerator 3 over denominator 3 plus EC end fraction end cell row cell rightwards double arrow text 6+2EC = 15 end text end cell row cell rightwards double arrow text EC= end text 9 over 2 end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 5

$begin mathsize 11px style table attributes columnalign left end attributes row cell In text end text fig comma text end text RS parallel to DB parallel to PQ. text If CP=PD=11 cm and DR = RA=3 cm end text end cell row cell text Then the value of x and y respectively end text end cell row cell left parenthesis straight a right parenthesis 12 comma 10 text end text left parenthesis straight b right parenthesis 14 comma 6 text end text left parenthesis straight c right parenthesis 10 comma 7 text end text left parenthesis straight d right parenthesis 16 comma 8 end cell end table end style$

Solution 5

$begin mathsize 11px style table attributes columnalign left end attributes row cell PQ parallel to BD parallel to SR end cell row cell So comma text end text straight capital delta text ASR end text tilde straight capital delta text ABD end text end cell row cell text Hence end text AR over AD equals RS over DB end cell row cell text end text rightwards double arrow text end text 3 over 6 equals straight y over straight x end cell row cell text end text rightwards double arrow text x=2y ---- end text box enclose 1 end cell row cell Only text end text option text end text left parenthesis straight d right parenthesis satisfies text end text box enclose 1 end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Chapter 7 - Triangles Excercise 7.135

Question 1

$begin mathsize 11px style table attributes columnalign left end attributes row cell In text end text fig comma text PB end text parallel to text CF and DP end text parallel to text EF then end text AD over DE equals end cell row cell left parenthesis straight a right parenthesis 3 over 4 text end text left parenthesis straight b right parenthesis 1 third text end text left parenthesis straight c right parenthesis 1 fourth text end text left parenthesis straight d right parenthesis 2 over 3 end cell end table end style$

Solution 1

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB equals 2 end cell row cell AC equals 8 end cell row cell as text end text BP parallel to CF end cell row cell Hence text end text angle text B= end text angle text C and end text angle text APB = end text angle AFC end cell row cell So text by AAA end text end cell row cell straight capital delta text ABP is similar to end text straight capital delta text ACF end text end cell row cell text Hence end text AB over AC equals AP over AF text ----- end text box enclose 1 end cell row cell Aslo text given DF end text parallel to text EF end text end cell row cell text Hence end text angle text D= end text angle straight E text amd end text angle text APD = end text angle text AFE end text end cell row cell text So by AAA end text end cell row cell straight capital delta text APD is similar to end text straight capital delta text AEF end text end cell row cell text so end text AP over AF equals AD over AE text ---- end text box enclose 2 end cell row cell from text end text box enclose 1 text & end text box enclose 2 end cell row cell text end text AD over AE equals AB over AC end cell row cell rightwards double arrow text end text AD over AE equals 2 over 8 end cell row cell rightwards double arrow text end text AE over AD minus 1 equals 8 over 2 minus 1 end cell row cell rightwards double arrow text end text fraction numerator AE minus AD over denominator AD end fraction equals 3 text end text rightwards double arrow DE over AD equals 3 text or end text AD over DE equals 1 third end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 2

$begin mathsize 11px style straight A space chord space of space circle space of space radius space 10 space cm space subtends space straight a space right space angle space at space the space centre. space The space length space of space the space chord space left parenthesis in space cm right parenthesis space is left parenthesis straight a right parenthesis 5 square root of 2 text end text left parenthesis straight b right parenthesis 10 square root of 2 text end text left parenthesis straight c right parenthesis fraction numerator 5 over denominator square root of 2 end fraction text end text left parenthesis straight d right parenthesis 10 square root of 3 end style$

Solution 2

$begin mathsize 11px style OA space and space OB space are space radius space of space circle space space space So space OA space equals space OB equals space 10 space cm space given space angle AOB space equals space 90 degree space space Hence space increment AOB space is space straight a space right space angled space triangle. end style$

$Error converting from MathML to accessible text.$

Question 3

A vertical stick 20 m long cast a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of tower is

(a) 100 m (b) 120 m (c) 25 m (d) 200 m

Solution 3

$begin mathsize 11px style table attributes columnalign left end attributes row cell from space figure space 1 space space space tanθ space equals space 20 over 10 equals 2 space space space space space space space space minus negative negative box enclose 1 end cell row cell from space figure space 2 space space space tanθ space equals space straight h over 50 space space space space space space space space minus negative negative box enclose 2 space space space space space end cell row cell from space space box enclose 1 space & space box enclose 2 end cell row cell space space space space space space space space space straight h over 50 equals 2 space space equals space straight h equals 100 space straight m end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 4

Two isosceles triangles have equal angles and their areas are in ratio 16:25. The ratio of their corresponding heights is

(a) 4:5 (b) 5:4 (c) 3:2 (d) 5:7

Solution 4

$begin mathsize 11px style As space given space angles space are space equal. space So space by space AAA space criteria space the space given space triangles space are space similar. space Given space ratio space of space area space is space 16 colon 25. space So comma space ratio space of space this space corresponding space heights space table attributes columnalign left end attributes row cell equals text end text square root of 16 colon square root of 25 end cell row cell equals text end text 4 colon 5 end cell end table end style$

So, the correct option is (a).

Question 5

∆ABC is such that AB=3 cm, BC=2 cm and CA=2.5 cm.

If ∆DEF ∿ ∆ABC and EF=4cm, then perimeter of ∆DEF is

(a) 7.5cm (b) 15cm (c) 22.5cm (d) 30 cm

Solution 5

$Error converting from MathML to accessible text.$

So, the correct option is (b).

Chapter 7 - Triangles Excercise 7.136

Question 1

In ∆ABC, a line XY parallel to BC cuts AB at X and AC at Y.

If BY bisects ∠XYC, then

(a) BC=CY (b) BC=BY (c)BC≠CY (d) BC≠BY

Solution 1

$begin mathsize 11px style table attributes columnalign left end attributes row cell Given space XY parallel to BC end cell row cell Hence space angle XYB space equals space angle YBC end cell row cell By space alterate space Interior space angle space property end cell row cell In straight space ΔYBC end cell row cell angle space BYC space equals space angle YBC end cell row cell And space sides space opposite space to space equal space angles space are space also space equal. end cell row cell so space box enclose CY equals BC end enclose end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 2

$begin mathsize 11px style table attributes columnalign left end attributes row cell In straight space ΔABC comma space angle straight A equals 90 to the power of straight 0 comma space AB equals space 5 cm space and space AC equals 12 cm. end cell row cell If space AD perpendicular BC space then space AD equals end cell row cell left parenthesis straight a right parenthesis straight space 13 over 2 cm space space space left parenthesis straight b right parenthesis straight space 60 over 13 cm space space space left parenthesis straight c right parenthesis straight space 13 over 60 cm space space space space space left parenthesis straight d right parenthesis straight space fraction numerator 2 root index blank of 15 over denominator 13 end fraction cm end cell end table end style$

Solution 2

$begin mathsize 11px style table attributes columnalign left end attributes row cell In straight space ΔABC end cell row cell space space space AB to the power of straight 2 plus AC squared equals BC squared end cell row cell rightwards double arrow space CB to the power of straight 2 equals space 5 to the power of straight 2 plus 12 squared end cell row cell rightwards double arrow space CB to the power of straight 2 equals space 169 end cell row cell rightwards double arrow space BC equals 13 end cell row cell Let space BD equals straight x space space space space space space then space DC space equals space BC minus BD end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 13 minus straight x end cell row cell In space ΔBDA end cell row cell BD to the power of straight 2 plus AD squared equals AB squared end cell row cell rightwards double arrow space straight x to the power of straight 2 plus AD squared equals 5 squared end cell row cell rightwards double arrow straight space AD squared equals 25 minus straight x squared space space space space space space space space minus negative negative box enclose 1 end cell row cell In space ΔCDA straight space end cell row cell space space space space CD to the power of straight 2 plus AD squared equals AC squared end cell row cell rightwards double arrow straight space left parenthesis 13 minus straight x right parenthesis squared plus AD squared equals 12 squared end cell row cell rightwards double arrow space AD to the power of straight 2 equals space 12 to the power of straight 2 minus straight space left parenthesis 13 minus straight x right parenthesis squared space space space space minus negative negative negative box enclose 2 end cell row cell From space space space box enclose 1 space space & space space box enclose 2 end cell row cell rightwards double arrow space 25 minus straight x to the power of straight 2 space equals space 12 to the power of straight 2 minus left parenthesis 13 minus straight x right parenthesis squared end cell row cell rightwards double arrow space 25 minus straight x to the power of straight 2 equals 144 minus left parenthesis 169 plus straight x squared minus 26 straight x right parenthesis end cell row cell rightwards double arrow space 25 minus straight x to the power of straight 2 equals 144 minus 169 minus straight x squared plus 26 straight x end cell row cell rightwards double arrow space 26 straight x space equals space 25 straight x 2 end cell row cell rightwards double arrow space straight x equals fraction numerator 25 straight space straight x straight space 2 over denominator 26 end fraction equals 25 over 13 end cell row cell Hence space AD to the power of straight 2 equals 25 minus straight x squared end cell row cell space space space space space space space space space space space space space space space space equals space 25 left parenthesis 1 minus 25 over 169 right parenthesis end cell row cell space space space space space space space space space space space space space space space space equals space 25 left parenthesis 144 over 169 right parenthesis end cell row cell space space space space space space space space AD space space space space equals space fraction numerator 5 cross times 12 over denominator 13 end fraction equals 60 over 13 cm end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 3

In a ∆ABC, perpendicular AD from A on BC meets BC at D. If BD=8 cm, DC=2cm and AD=4cm then

(a) ∆ABC is isosceles (b) ∆ABC is equilateral

Solution 3

$begin mathsize 11px style table attributes columnalign left end attributes row cell space space space space AC squared equals AD squared plus DC squared end cell row cell space space space space AC squared equals 4 squared plus 2 squared end cell row cell space space space space AC equals square root of 20 equals 2 square root of 5 end cell row cell space space space space BD squared plus AD squared equals AB squared end cell row cell rightwards double arrow straight space AB squared equals 8 squared plus 4 squared end cell row cell space space space space space space space space space space space equals space 80 end cell row cell rightwards double arrow thin space space AB space equals space square root of 80 equals 4 square root of 5 end cell row cell space space space space space BC space equals space 10 end cell row cell From space sides space of space triangle end cell row cell space space space space space AC to the power of straight 2 plus AB squared equals BC squared end cell row cell Hence straight space ΔABC straight space is space right space angled space at space straight A. end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Question 4

$begin mathsize 11px style In space straight a space increment ABC comma space point space straight D space is space on space side space AB space and space point space straight E space is space on space side table attributes columnalign left end attributes row cell AC comma straight space such straight space BCED straight space is straight space straight a straight space trapezium. straight space If straight space DE colon BC equals 3 colon 5 comma straight space end cell row cell then straight space Area straight space left parenthesis ΔADE right parenthesis colon straight space Area straight space left parenthesis square BCED right parenthesis equals end cell row cell left parenthesis straight a right parenthesis space 3 colon 4 space space space space space space left parenthesis straight b right parenthesis 9 colon 16 space space space left parenthesis straight c right parenthesis 3 colon 5 space space space space space left parenthesis straight d right parenthesis 9 colon 25 end cell end table end style$

Solution 4

$begin mathsize 11px style table attributes columnalign left end attributes row cell Given space DECB space is space trapezium comma space end cell row cell so space DE parallel to BC end cell row cell Hence space ΔADE space is space similar space to space ΔABC end cell row cell Given space DE over BC equals 3 over 5 end cell row cell Hence straight space fraction numerator area straight space left parenthesis ΔADE right parenthesis over denominator area straight space left parenthesis ΔABC right parenthesis end fraction equals left parenthesis 3 over 5 right parenthesis squared equals 9 over 25 end cell row cell rightwards double arrow straight space fraction numerator area straight space left parenthesis ΔABC right parenthesis over denominator area straight space left parenthesis ΔADE right parenthesis end fraction equals 25 over 9 end cell row cell rightwards double arrow fraction numerator area straight space left parenthesis ΔABC right parenthesis over denominator area straight space left parenthesis ΔADE right parenthesis end fraction minus 1 equals 25 over 9 minus 1 end cell row cell rightwards double arrow fraction numerator area straight space left parenthesis ΔABC right parenthesis minus area straight space left parenthesis ΔADE right parenthesis over denominator area straight space left parenthesis ΔADE right parenthesis end fraction equals 16 over 9 end cell row cell rightwards double arrow fraction numerator area straight space left parenthesis square BCED right parenthesis over denominator area straight space left parenthesis ΔADE right parenthesis end fraction equals 16 over 9 end cell row cell rightwards double arrow fraction numerator area straight space left parenthesis ΔADE right parenthesis over denominator area straight space left parenthesis square BCED right parenthesis end fraction equals 9 over 16 end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Question 5

If ABC is an isosceles triangle and D is a point on BC such that AD⊥ BC, then

(a)AB²-AD²=BD.DC (b) AB²-AD²=BD²-DC²

Solution 5

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB squared equals AD squared plus BD squared space space space space space space space space minus negative negative negative box enclose 1 end cell row cell AC squared equals AD squared plus DC squared space space space space space space space space minus negative negative negative box enclose 2 end cell row cell Given straight space ΔABC straight space is straight space an straight space isosceles space triangle space end cell row cell Hence space AB equals AC end cell row cell So space from space box enclose 1 space space & space space box enclose 2 end cell row cell BD equals DC end cell row cell rightwards double arrow straight space AB squared minus AD squared equals BD squared end cell row cell rightwards double arrow straight space AB squared minus AD squared equals BD straight space. straight space BD end cell row cell rightwards double arrow straight space AB squared minus AD squared equals BD straight space. space DC end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 6

$begin mathsize 11px style increment ABC space is space straight a space right space triangle comma space right space – angled space at space straight A space and space AD perpendicular space BC. space Then space space left parenthesis straight a right parenthesis left parenthesis AB over AC right parenthesis squared text end text left parenthesis straight b right parenthesis AB over AC text end text left parenthesis straight c right parenthesis left parenthesis AB over AD right parenthesis squared text end text left parenthesis straight D right parenthesis AB over AD end style$

Solution 6

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB squared plus AC squared equals BC squared end cell row cell AD squared plus DB squared equals AB squared end cell row cell AD squared plus DC squared equals AC squared end cell row cell straight capital delta space CAB space is space similar space to space straight capital delta space ADB comma space by space AAA space criterion space end cell row cell rightwards double arrow straight space AC over AD equals straight space AB over DB equals straight space BC over AB end cell row cell rightwards double arrow straight space AB over DB equals straight space BC over AB straight space rightwards double arrow straight space AB squared over BC equals space BD space space space space minus negative negative negative box enclose 1 end cell row cell Also straight space straight capital delta space ABC space is space similar space to space straight capital delta space DAC. end cell row cell Hence space AB over AD equals straight space BC over AC equals straight space AC over DC end cell row cell rightwards double arrow straight space BC over AC equals straight space AC over DC straight space rightwards double arrow straight space AC squared over BC equals space DC space space space space minus negative negative negative box enclose 2 end cell row cell From straight space box enclose 1 space space & space space box enclose 2 end cell row cell BD over DC equals straight space left parenthesis AB over AC right parenthesis squared end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Question 7

If E is a point on side CD of an equilateral triangle ABC such that BE ⊥CA, then AB²+BC²+CA²=

(a) 2BE² (b) 3BE² (c) 4BE²(d) 2BE²

Solution 7

$begin mathsize 11px style table attributes columnalign left end attributes row cell ΔABC space is space an space equilateral space triangle space ABC space such space that space BE perpendicular space CA comma end cell row cell Hence space AE equals EC space end cell row cell and space space space space BE to the power of straight 2 plus AE to the power of straight 2 equals BA to the power of straight 2 straight space end cell row cell rightwards double arrow space BE to the power of straight 2 = BA to the power of straight 2 minus left parenthesis AC over 2 right parenthesis squared end cell row cell = BA to the power of straight 2 minus left parenthesis BA over 2 right parenthesis squared end cell row cell BE to the power of straight 2 equals fraction numerator 3 space AB to the power of straight 2 over denominator 4 end fraction end cell row cell rightwards double arrow AB to the power of straight 2 equals straight 4 over 3 space BE to the power of straight 2 straight space end cell row cell and space AB space equals space BC space equals space CA end cell row cell So comma space AB to the power of straight 2 plus BC to the power of straight 2 plus CA to the power of straight 2 equals 3 AB squared end cell row cell = 3 cross times 4 over 3 straight space BE squared end cell row cell = 4 BE squared end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 8

$begin mathsize 11px style In space straight a space right space triangle space ABC space right minus angle space at space straight B comma space if space straight P space and space straight Q space are space points space on space the space sides space AB space and space AC space respectively comma space space Then table attributes columnalign left end attributes row cell left parenthesis straight a right parenthesis straight space AQ to the power of blank squared end exponent plus space CP squared equals space 2 left parenthesis AC squared plus PQ squared right parenthesis space space space space space space space space space space space left parenthesis straight b right parenthesis 2 left parenthesis AQ squared plus CP squared right parenthesis equals AC squared plus PQ squared end cell row cell left parenthesis straight c right parenthesis straight space AQ squared plus straight space CP squared equals straight space AC squared plus straight space PQ squared space space space space space space space space space space space space space space space left parenthesis straight d right parenthesis space space AQ plus CP equals 1 half left parenthesis AC plus PQ right parenthesis end cell end table end style$

Solution 8

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB squared plus BC squared equals AC squared space space space space space minus negative negative negative box enclose 1 end cell row cell BQ squared plus BA squared equals AQ squared space space space space space minus negative negative negative box enclose 2 end cell row cell BC squared plus BP squared equals PC squared space space space space space minus negative negative negative box enclose 3 end cell row cell BP squared plus BQ squared equals PQ squared space space space space space minus negative negative negative box enclose 4 end cell row cell on space adding space box enclose 2 space & space box enclose 3 end cell row cell AQ squared plus PC squared equals BQ squared plus BA squared plus BC squared plus BP squared end cell row cell thin space space space space space space space space space space space space space equals space BQ squared plus BP squared plus BA squared plus BC squared end cell row cell space space space space space space space space space space space space space space equals space PQ squared plus AC squared end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Question 9

If ∆ABC ∿∆DEF such that DE=3cm, EF=2cm, DF=2.5cm, BC=4cm then perimeter of ∆ABC is

(a)18cm (b)20cm (c)12cm (d)15cm

Solution 9

$begin mathsize 11px style table attributes columnalign left end attributes row cell AB over DE equals BC over EF equals CA over FD end cell row cell rightwards double arrow straight space AB over 3 equals 4 over 2 equals fraction numerator CA over denominator 2.5 end fraction end cell row cell rightwards double arrow straight space AB over 3 equals 4 over 2 space space space and space space space rightwards double arrow straight space 4 over 2 equals fraction numerator CA over denominator 2.5 end fraction end cell row cell rightwards double arrow space AB space equals space 6 space space space space and space space AC space equals space 5 end cell row cell perimeter space equals space AB space plus space BC space plus space CA end cell row = 6 + 4 + 5 row cell space space space space space space space space space space space space space space equals space 15 space cm end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Question 10

If ∆ABC ∿ ∆DEF such that AB=9.1 cm and DE=6.5. If the perimeter of ∆DEF is 25 cm, then the perimeter of ∆ABC is

(a) 36 cm (b) 30 cm (c) 34 cm (d)35cm

Solution 10

$begin mathsize 11px style table attributes columnalign left end attributes row cell space space space space space AB over DE equals fraction numerator perimeter space of space ΔABC over denominator perimeter space of space ΔDEF end fraction end cell row cell rightwards double arrow straight space fraction numerator 9.1 over denominator 6.5 end fraction equals fraction numerator perimeter space of space ΔABC over denominator 25 end fraction end cell row cell rightwards double arrow straight space perimeter space of space ΔABC equals fraction numerator 9.1 over denominator 6.5 end fraction space straight x space 25 end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 35 space cm end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Question 11

In an isosceles triangle ABC, if AB=AC=25 cm and BC=14 cm, then the measure of altitude from A on BC is

(a) 20cm (b) 22cm (c) 18 cm (d)24 cm

Solution 11

$begin mathsize 11px style table attributes columnalign left end attributes row cell Let space BD space equals space straight x space space then space DC space equals space 14 minus straight x end cell row cell Also space space space AD to the power of straight 2 space plus space BD to the power of straight 2 equals AB squared end cell row cell rightwards double arrow space space space space AD to the power of straight 2 plus straight x squared equals 25 squared minus negative negative negative box enclose 1 end cell row cell And space space AD to the power of straight 2 plus DC squared equals AC squared end cell row cell rightwards double arrow space space space space AD to the power of straight 2 plus left parenthesis 14 minus straight x right parenthesis squared equals 25 squared minus negative negative negative box enclose 2 end cell row cell from space space box enclose 1 space space & space box enclose 2 end cell row cell rightwards double arrow space space straight x to the power of straight 2 space space equals space left parenthesis 14 minus straight x right parenthesis squared end cell row cell rightwards double arrow space space space straight x space equals space 14 space minus space straight x end cell row cell rightwards double arrow space space 2 straight x space equals 14 end cell row cell rightwards double arrow space straight x equals 7 end cell row cell Hence space AD to the power of straight 2 plus 7 squared equals 25 squared end cell row cell AD squared equals 625 minus 49 equals 576 end cell row cell AD equals 24 cm end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

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Chapter 7 : Triangles - Rd Sharma Solutions for Class 10 Maths CBSE

Chapter 7 - Triangles Excercise Ex. 7.1

Chapter 7 - Triangles Excercise Ex. 7.2

Chapter 7 - Triangles Excercise Ex. 7.3

Chapter 7 - Triangles Excercise Ex. 7.4

Chapter 7 - Triangles Excercise Ex. 7.5

Chapter 7 - Triangles Excercise Ex. 7.6

Chapter 7 - Triangles Excercise Ex. 7.7

Chapter 7 - Triangles Excercise Rev. 7

Chapter 7 - Triangles Excercise 7.131

Chapter 7 - Triangles Excercise 7.132

Chapter 7 - Triangles Excercise 7.133

Chapter 7 - Triangles Excercise 7.134

Chapter 7 - Triangles Excercise 7.135

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