Chapter 15 : Similarity (With Applications to Maps and Models) - Selina Solutions for Class 10 Maths ICSE

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Chapter 15 - Similarity (With Applications to Maps and Models) Excercise Ex. 15(A)

Question 1

In the figure, given below, straight lines AB and CD intersect at P; and AC BD. Prove that :

If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.

  

Solution 1

Question 2

In the figure, given below, straight lines AB and CD intersect at P; and AC BD. Prove that:

ΔAPC and ΔBPD are similar.

 

  

Solution 2

  

Question 3

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that :

Δ APB is similar to Δ CPD.

Solution 3

  

 

  

Question 4

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that :

PA x PD = PB x PC. 

Solution 4

 

  

Question 5

P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that :

DP : PL = DC : BL.

Solution 5

  

 

 

  

Question 6

P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that :

DL : DP = AL : DC.

Solution 6

  

 

 

  

Question 7

In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that :

Δ AOB is similar to Δ COD.

Solution 7

 

 

  

Question 8

In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that :

OA x OD = OB x OC.

Solution 8

 

 

 

 

  

Question 9

In Δ ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that :

CB : BA = CP : PA

Solution 9

 

  

 

 

 

  

Question 10

In Δ ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that :

AB x BC = BP x CA 

Solution 10

  

 

 

  

Question 11

  

Solution 11

 

  

Question 12

In the given figure, DE BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.

Write all possible pairs of similar triangles.

 

 

  

Solution 12

  

Question 13

In the given figure, DE BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.

Find lengths of ME and DM.

 

  

Solution 13

  

Question 14

In the given figure, AD = AE and AD2 = BD x EC.

Prove that: triangles ABD and CAE are similar. 

 

  

Solution 14

  

Question 15

In the given figure, AB DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.

 

  

Solution 15

  

Question 16

Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.

Solution 16

  

 

 

  

Question 17

State, true or false:

(i) Two similar polygons are necessarily congruent.

(ii) Two congruent polygons are necessarily similar.

(iii) All equiangular triangles are similar.

(iv) All isosceles triangles are similar.

(v) Two isosceles-right triangles are similar.

(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.

(vii) The diagonals of a trapezium, divide each other into proportional segments.

Solution 17

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) True

Question 18

Given: GHE = DFE = 90o, DH = 8, DF = 12, DG = 3x - 1 and DE = 4x + 2.

Find: the lengths of segments DG and DE.

Solution 18

Question 19

D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA2 = CB CD.

Solution 19

Question 20

In the given figure, ∆ABC and ∆AMP are right angled at B and M respectively.

Given AC = 10 cm, AP = 15 cm and PM = 12 cm.

  1. ∆ ABC ~ ∆ AMP.
  2. Find AB and BC.

Solution 20

(i)  In ∆ ABC and ∆ AMP,

 BAC=  PAM [Common]

 ABC=  PMA [Each = 90°]

∆ ABC ~ ∆ AMP [AA Similarity]

(ii)

 

 

 

 

 

 

 

 

Question 21

Given: RS and PT are altitudes of PQR. Prove that:

(i)

(ii) PQ QS = RQ QT.

Solution 21

(i)

(ii)

Since, triangles PQT and RQS are similar.

Question 22

Given: ABCD is a rhombus, DPR and CBR are straight lines. Prove that: DP CR = DC PR.

 

Solution 22

Hence, DP CR = DC PR

Question 23

Given: FB = FD, AE FD and FC AD. Prove:

Solution 23

Question 24

In PQR, Q = 90o and QM is perpendicular to PR. Prove that:

(i) PQ2 = PM PR

(ii) QR2 = PR MR

(iii) PQ2 + QR2 = PR2

Solution 24

(i) In PQM and PQR,

PMQ = PQR = 90o

QPM = RPQ (Common)

(ii) In QMR and PQR,

QMR = PQR = 90o

QRM = QRP (Common)

(iii) Adding the relations obtained in (i) and (ii), we get,

Question 25

In ABC, B = 90o and BD AC.

(i) If CD = 10 cm and BD = 8 cm; find AD.

(ii) If AC = 18 cm and AD = 6 cm; find BD.

(iii) If AC = 9 cm and AB = 7 cm; find AD.

Solution 25

(i) In CDB,

1 + 2 +3 = 180o

1 + 3 = 90o ..... (1)(Since, 2 = 90o)

3 + 4 = 90o .....(2) (Since, ABC = 90o)

From (1) and (2),

1 + 3 = 3 + 4

1 = 4

Also, 2 = 5 = 90o

Hence, AD = 6.4 cm

 

 

(iii)

Question 26

In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL: LP = 2: 3. QL produced meets RS at M and PS produced at N.

Find the lengths of PN and RM.

Solution 26

Question 27

In quadrilateral ABCD, diagonals AC and BD intersect at point E such that AE: EC = BE: ED. Show that ABCD is a trapezium.

Solution 27

Given, AE: EC = BE: ED

Draw EF || AB

In ABD, EF || AB

Using Basic Proportionality theorem,

Thus, in DCA, E and F are points on CA and DA respectively such that

Thus, by converse of Basic proportionality theorem, FE || DC.

But, FE || AB.

Hence, AB || DC.

Thus, ABCD is a trapezium.

Question 28

In triangle ABC, AD is perpendicular to side BC and AD2 = BD DC. Show that angle BAC = 90o.

Solution 28

Given, AD2 = BD DC

So, these two triangles will be equiangular.

Question 29

In the given figure, AB || EF || DC; AB = 67.5 cm, DC = 40.5 cm and AE = 52.5 cm.

(i) Name the three pairs of similar triangles.

(ii) Find the lengths of EC and EF.

Solution 29

(i) The three pair of similar triangles are:

BEF and BDC

CEF and CAB

ABE and CDE

(ii) Since, ABE and CDE are similar,

Since, CEF and CAB are similar,

Question 30

In the given figure, QR is parallel to AB and DR is parallel to QB. Prove that: PQ2 = PD PA.

Solution 30

Given, QR is parallel to AB. Using Basic proportionality theorem,

Also, DR is parallel to QB. Using Basic proportionality theorem,

From (1) and (2), we get,

Question 31

Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E. Prove that: EL = 2BL.

Solution 31

1 = 6 (Alternate interior angles)

2 = 3 (Vertically opposite angles)

DM = MC (M is the mid-point of CD)

So, DE = BC (Corresponding parts of congruent triangles)

Also, AD = BC (Opposite sides of a parallelogram)

AE = AD + DE = 2BC

Now, 1 = 6 and 4 = 5

Question 32

In the figure, given below, P is a point on AB such that AP: PB = 4: 3. PQ is parallel to AC.

(i) Calculate the ratio PQ: AC, giving reason for your answer.

(ii) In triangle ARC, ARC = 90o and in triangle PQS, PSQ = 90o. Given QS = 6 cm, calculate the length of AR.

Solution 32

(i) Given, AP: PB = 4: 3.

Since, PQ || AC. Using Basic Proportionality theorem,

Now, PQB = ACB (Corresponding angles)

QPB = CAB (Corresponding angles)

(ii) ARC = QSP = 90o

ACR = SPQ (Alternate angles)

Question 33

In the right-angled triangle QPR, PM is an altitude. Given that QR = 8 cm and MQ = 3.5 cm, calculate the value of PR.

Solution 33

We have:

Question 34

In the figure, given below, the medians BD and CE of a triangle ABC meet at G. Prove that:

(i) and

(ii) BG = 2 GD from (i) above.

Solution 34

(i) Since, BD and CE are medians.

AD = DC

AE = BE

Hence, by converse of Basic Proportionality theorem,

ED || BC

In EGD and CGB,

(ii) Since,

In AED and ABC,

From (1),

Chapter 15 - Similarity (With Applications to Maps and Models) Excercise Ex. 15(B)

Question 1

In the following figure, point D divides AB in the ratio 3 : 5. Find :

  

  

Solution 1

Given space that space AD over DB equals 3 over 5

Now, DE is parallel to BC.

Then, by Basic proportionality theorem, we have

AD over DB equals AE over EC
rightwards double arrow AE over EC equals 3 over 5AD over DB equals AE over EC
rightwards double arrow AE over EC equals 3 over 5 

Question 2

In the following figure, point D divides AB in the ratio 3 : 5. Find :

  

  

Solution 2

  

Question 3

In the following figure, point D divides AB in the ratio 3 : 5. Find :

  

  

Solution 3

  

Question 4

In the following figure, point D divides AB in the ratio 3 : 5. Find :

DE = 2.4 cm, find the length of BC.

  

Solution 4

  

Question 5

In the following figure, point D divides AB in the ratio 3 : 5. Find :

BC = 4.8 cm, find the length of DE.

  

Solution 5

  

Question 6

In the given figure, PQ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :

  

  

Solution 6

  

Question 7

In the given figure, PQ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :

PQ

  

Solution 7

  

Question 8

In the given figure, PQ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :

If AP = x, then the value of AC in terms of x.   

Solution 8

  

Question 9

A line PQ is drawn parallel to the side BC of Δ ABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.

Solution 9

  

 

 

  

Question 10

In Δ ABC, D and E are the points on sides AB and AC respectively.

Find whether DE BC, if

AB = 9cm, AD = 4cm, AE = 6cm and EC = 7.5cm. 

Solution 10

  

 

 

 

  

Question 11

In Δ ABC, D and E are the points on sides AB and AC respectively.

Find whether DE BC, if AB = 6.3 cm, EC = 11.0 cm, AD =0.8 cm and EA = 1.6 cm.

Solution 11

 

 

 

 

  

 

Question 12

In the given figure, Δ ABC ~ Δ ADE. If AE: EC = 4 : 7 and DE = 6.6 cm, find BC. If 'x' be the length of the perpendicular from

  

 

A to DE, find the length of perpendicular from A to BC in terms of 'x'.

Solution 12

  

 

 begin mathsize 14px style increment ABC tilde increment ADE comma
rightwards double arrow AE over AC equals DE over BC
rightwards double arrow 4 over 11 equals fraction numerator 6.6 over denominator BC end fraction
rightwards double arrow BC equals fraction numerator 11 cross times 6.6 over denominator 4 end fraction equals 18.15 space cm end style

  

Question 13

A line segment DE is drawn parallel to base BC of Δ ABC which cuts AB at point D and AC at point E. If AB = 5BD and EC = 3.2 cm, find the length of AE.

Solution 13

  

 

 

 

 

  

Question 14

In the figure, given below, AB, CD and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3. cm, calculate the values of x and y.

 

  

 

Solution 14

 

Question 15

In the figure, given below, PQR is a right-angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.

 

  

Solution 15

  

Question 16

In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E.

Prove that : PE = 2 PD 

 

  

Solution 16

  

Question 17

The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

  

Solution 17

  

Chapter 15 - Similarity (With Applications to Maps and Models) Excercise Ex. 15(C)

Question 1

(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.

(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.

Solution 1

We know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

 

(i) Required ratio =

(ii) Required ratio =

Question 2

A line PQ is drawn parallel to the base BC of  ABC which meets sides AB and AC at points P and Q respectively. If AP = PB; find the value of:

(i)

(ii)

Solution 2

 

(i) AP = PB

 

(ii)

Question 3

The perimeters of two similar triangles are 30 cm and 24 cm. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.

Solution 3

Let

Question 4

In the given figure, AX: XB = 3: 5.

Find:

(i) the length of BC, if the length of XY is 18 cm.

(ii) the ratio between the areas of trapezium XBCY and triangle ABC.

 

Solution 4

Given,

(i)

(ii)

Question 5

ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP: AB.

Solution 5

From the given information, we have:

Question 6

In the given triangle PQR, LM is parallel to QR and PM: MR = 3: 4.

Calculate the value of ratio:

(i)

(ii)

(iii)

Solution 6

(i)

(ii) Since  LMN and  MNR have common vertex at M and their bases LN and NR are along the same straight line

 

(iii) Since  LQM and  LQN have common vertex at L and their bases QM and QN are along the same straight line

Question 7

The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel; PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ. Calculate:

(i) the length of AP,

(ii) the ratio of the areas of triangle APQ and triangle ABC.

 

Solution 7

(i)

(ii)

Question 8

In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP: PC = 1: 2. DP produced meets AB produces at Q. Given the area of triangle CPQ = 20 cm2. Calculate:

(i) area of triangle CDP,

(ii) area of parallelogram ABCD.

 

Solution 8

 

Question 9

In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm2, Area of trapezium BCED = 24 cm2 and DE = 14 cm. Calculate the length of BC. Also, find the area of triangle BCD.

 

Solution 9

Question 10

The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersects at point P. If AP: CP = 3: 5, find:

(i)  APB :  CPB(ii)  DPC :  APB

(iii)  ADP :  APB(iv)  APB :  ADB

 

Solution 10

(i) Since  APB and  CPB have common vertex at B and their bases AP and PC are along the same straight line

(ii) Since  DPC and  BPA are similar

(iii) Since  ADP and  APB have common vertex at A and their bases DP and PB are along the same straight line

(iv) Since  APB and  ADB have common vertex at A and their bases BP and BD are along the same straight line

Question 11

In the given figure, ABC is a triangle. DE is parallel to BC and  .

(i) Determine the ratios

(ii) Prove that  DEF is similar to  CBF. Hence, find  .

(iii) What is the ratio of the areas of  DEF and  BFC?

Solution 11

 

(i) Given, DE || BC and

In  ADE and  ABC,

 A =  A(Corresponding Angles)

 ADE =  ABC(Corresponding Angles)

  (By AA- similarity)

  ..........(1)

Now

Using (1), we get .........(2)

(ii) In  DEF and  CBF,

 FDE =  FCB(Alternate Angle)

 DFE =  BFC(Vertically Opposite Angle)

  DEF   CBF(By AA- similarity)

 using (2)

 .

(iii) Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, therefore

Question 12

In the given figure,  B =  E,  ACD =  BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the  ABC and  DEC.

Solution 12

Chapter 15 - Similarity (With Applications to Maps and Models) Excercise Ex. 15(D)

Question 1

A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A' B' C' Calculate :

the length of AB, if A' B' = 6 cm. 

Solution 1

  

Question 2

A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A' B' C' Calculate :

the length of C' A' if CA = 4 cm. 

Solution 2

  

Question 3

A triangle LMN has been reduced by scale factor 0.8 to the triangle L' M' N'. Calculate:

the length of M' N', if MN = 8 cm.

Solution 3

  

Question 4

A triangle LMN has been reduced by scale factor 0.8 to the triangle L' M' N'. Calculate:

the length of LM, if L' M' = 5.4 cm.

Solution 4

  

Question 5

A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find : OC', if OC = 21 cm

Also, state the value of : 

(a)   

(b)   

Solution 5

Given that triangle ABC is enlarged and the scale factor is m = 3 to the triangle A'B'C'.

OC = 21 cm

So, (OC)3 = OC'

i.e. 21 x 3 = OC'

i.e. OC' = 63 cm


  

Question 6

A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find :

A' B', if AB = 4 cm.

Solution 6

  

Question 7

A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find :

BC, if B' C' = 15 cm. 

Solution 7

  

Question 8

A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find :

OA, if OA' = 6 cm. 

Solution 8

  

Question 9

A model of an aeroplane is made to a scale of 1 : 400. Calculate :

the length, in cm, of the model; if the length of the aeroplane is 40 m. 

Solution 9

  

Question 10

A model of an aeroplane is made to a scale of 1 : 400. Calculate :

the length, in m, of the aeroplane, if length of its model is 16 cm.

Solution 10

  

Question 11

The dimensions of the model of a multistorey building are 1.2 m x 75 cm x 2 m. If the scale factor is 1 : 30; find the actual dimensions of the building.

Solution 11

  

Question 12

On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and  ABC = 90°.

Calculate :

the actual lengths of AB and BC in km. 

Solution 12

  

Question 13

On a map drawn to a scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.

Calculate :

the area of the plot in sq. km.

Solution 13

  

Question 14

A model of a ship is made to a scale 1 : 300.

 i. The length of the model of the ship is 2 m. Calculate the length of the ship.

 ii. The area of the deck of the ship is 180,000 m2. Calculate the area of the deck of the model.

 iii. The volume of the model is 6.5 m3. Calculate the volume of the ship. 

Solution 14

 

  

Chapter 15 - Similarity (With Applications to Maps and Models) Excercise Ex. 15(E)

Question 1

In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.

 

  

 

  

Solution 1

  

Question 2

In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.

 

  

 

  

Solution 2

  

Question 3

In the following figure, ABCD to a trapezium with AB DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: PE

  

Solution 3

Question 4

In the following figure, ABCD to a trapezium with AB DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: CE

 

 

  

 

Solution 4

  

Question 5

In the following figure, ABCD to a trapezium with AB DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: AF

  

Solution 5

  

Question 6

In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

 

  

  

Solution 6

  

Question 7

Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: .

Solution 7

  

  

Question 8

  

Solution 8

  

  

Question 9

 

Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that:.

Solution 9

 

  

Question 10

In the following figure, AXY = AYX. If show that triangle ABC is isosceles.

 

 

  

Solution 10

  

Question 11

In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown. 

Prove that:

Solution 11

Join AR.

In  ACR, BX || CR. By Basic Proportionality theorem,

In  APR, XQ || AP. By Basic Proportionality theorem,

From (1) and (2), we get,

Question 12

  

  

Solution 12

  

Question 13

In the figure given below, AB EF CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : EF

 

 

  

Solution 13

  

Question 14

In the figure given below, AB EF CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : AC

 

  

Solution 14

  

Question 15

In ΔABC, ABC = DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

Prove that ΔACD is similar to ΔBCA.

 

  

Solution 15

  

Question 16

In ΔABC, ABC = DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

Find BC and CD.

  

Solution 16

  

Question 17

In ΔABC, ABC = DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

Find area of ΔACD : area of ΔABC.

  

Solution 17

  

Question 18

In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle QRP is similar to triangle ABC.

Solution 18

In ABC, PR || BC. By Basic proportionality theorem,

Also, in PAR and ABC,

Similarly,

Question 19

In the following figure, AD and CE are medians of ABC. DF is drawn parallel to CE. Prove that:

(i) EF = FB,

(ii) AG: GD = 2: 1

Solution 19

(i)

(ii) In AFD, EG || FD. Using Basic Proportionality theorem,

… (1)

Now, AE = EB (as E is the mid-point of AB)

AE = 2EF (Since, EF = FB, by (i))

From (1),

Hence, AG: GD = 2: 1.

Question 20

Two similar triangles are equal in area. Prove that the triangles are congruent.

Solution 20

Let us assume two similar triangles as ABC PQR

Question 21

The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their:

(i) medians. (ii) perimeters. (iii) areas.

Solution 21

The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Required ratio = 3: 5

(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Required ratio = 3: 5

(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

Required ratio = (3)2 : (5)2 = 9: 25

Question 22

The ratio between the areas of two similar triangles is 16: 25. Find the ratio between their:

(i) perimeters. (ii) altitudes. (iii) medians.

Solution 22

The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

So, the ratio between the sides of the two triangles = 4: 5

(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

Question 23

The following figure shows a triangle PQR in which XY is parallel to QR. If PX: XQ = 1: 3 and QR = 9 cm, find the length of XY.

Further, if the area of PXY = x cm2; find, in terms of x, the area of:

(i) triangle PQR. (ii) trapezium XQRY.

Solution 23

In PXY and PQR, XY is parallel to QR, so corresponding angles are equal.

Hence, (By AA similarity criterion)

 (i) We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

 (ii) Ar (trapezium XQRY) = Ar (PQR) - Ar (PXY)

= (16x - x) cm2

= 15x cm2

Question 24

On a map drawn to a scale of 1: 20000, a rectangular plot of land ABCD is measured as AB = 24 cm and BC = 32 cm. Calculate:

(i) the diagonal distance of the plot in kilometre.

(ii) the area of the plot in sq. km.

Solution 24

Scale :- 1 : 20000

1 cm represents 20000 cm= = 0.2 km

(i)

=

= 576 + 1024 = 1600

AC = 40 cm

 

Actual length of diagonal = 40 0.2 km = 8 km

(ii)

1 cm represents 0.2 km

1 cm2 represents 0.2 0.2

The area of the rectangle ABCD = AB BC

= 24 32 = 768

Actual area of the plot = 0.2 0.2 768 = 30.72 km2

Question 25

The dimensions of the model of a multistoreyed building are 1 m by 60 cm by 1.20 m. If the scale factor is 1: 50, find the actual dimensions of the building. Also, find:

(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq. cm.

(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90 m3.

Solution 25

The dimensions of the building are calculated as below.

Length = 1 50 m = 50 m

Breadth = 0.60 50 m = 30 m

Height = 1.20 50 m = 60 m

Thus, the actual dimensions of the building are 50 m 30 m 60 m.

(i)

Floor area of the room of the building

(ii)

Volume of the model of the building

Question 26

In a triangle PQR, L and M are two points on the base QR, such that LPQ = QRP and RPM = RQP. Prove that:

(i)

(ii)

(iii)

Solution 26

(i)

(ii)

(iii)

Question 27

A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to DEF such that the longest side of DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of DEF.

Solution 27

Triangle ABC is enlarged to DEF. So, the two triangles will be similar.

Longest side in ABC = BC = 6 cm

Corresponding longest side in DEF = EF = 9 cm

Scale factor = = 1.5

Question 28

Two isosceles triangles have equal vertical angles. Show that the triangles are similar.

If the ratio between the areas of these two triangles is 16: 25, find the ratio between their corresponding altitudes.

Solution 28

Let ABC and PQR be two isosceles triangles.

Then,

Also, A = P (Given)

Let AD and PS be the altitude in the respective triangles.

We know that the ratio of areas of two similar triangles is equal to the square of their corresponding altitudes.

Question 29

In ABC, AP: PB = 2: 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:

(i) area APO : area ABC.

(ii) area APO : area CQO.

Solution 29

In triangle ABC, PO || BC. Using Basic proportionality theorem,

(i)

(ii)

Question 30

The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively. Show that:

Solution 30

Question 31

In the given figure, ABC is a triangle with EDB = ACB. Prove that ABC EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of BED = 9 cm2. Calculate the :

(i) length of AB.

(ii) area of ABC.

Solution 31

In ABC and EBD,

ACB = EDB (given)

ABC = EBD (common)

(by AA- similarity)

(i) We have,

(ii)

Question 32

In the given figure, ABC is a right angled triangle with mBAC = 90°

  1. Prove that ∆ADB ~ ∆CDA
  2. If BD = 18 cm and CD = 8 cm, find AD.
  3. Find the ratio of the area of ∆ADB to the area of ∆CDA.
Solution 32

left parenthesis i right parenthesis space L e t space angle C A D equals x
rightwards double arrow m angle D A B equals 90 degree minus x
rightwards double arrow m angle D B A equals 180 degree minus open parentheses 90 degree plus 90 degree minus x close parentheses equals x
rightwards double arrow angle C A D equals angle D B A space space space space space space space.... left parenthesis 1 right parenthesis
I n space triangle A D B space a n d space triangle C D A comma
angle A D B equals angle C D A space space space space space.... left square bracket E a c h space 90 degree right square bracket
angle A B D equals angle C A D space space space space space.... left square bracket F R o m space left parenthesis 1 right parenthesis right square bracket
therefore space triangle A D B tilde triangle C D A space space space space.... left square bracket B y space A. A. right square bracket

left parenthesis i i right parenthesis space S i n c e space t h e space c o r r e s p o n d i n g space s i d e s space o f space s i m i l a r space t r i a n g l e s space a r e space p r o p o r t i o n a l comma space w e space h a v e
fraction numerator B D over denominator A D end fraction equals fraction numerator A D over denominator C D end fraction
rightwards double arrow fraction numerator 18 over denominator A D end fraction equals fraction numerator A D over denominator 8 end fraction
rightwards double arrow space A D squared equals 18 cross times 8 equals 144
rightwards double arrow A D equals 12 space c m

left parenthesis i i i right parenthesis space T h e space r a t i o space o f space t h e space a r e a s space o f space t w o space s i m i l a r space t r i a n g l e s space i s space e q u a l space t o space t h e space r a t i o space o f space t h e space s q u a r e s
o f space t h e i r space c o r r e s p o n d i n g space s i d e s.
rightwards double arrow fraction numerator A r left parenthesis triangle A D B right parenthesis over denominator A r left parenthesis triangle C D A right parenthesis end fraction equals fraction numerator A D squared over denominator C D squared end fraction equals 12 squared over 8 squared equals 144 over 64 equals 9 over 4 equals 9 space colon space 4 space

Question 33

In the given figure, AB and DE are perpendiculars to BC.

Prove that : ΔABC ~  ΔDEC 

 

  

Solution 33

  

Question 34

In the given figure, AB and DE are perpendiculars to BC.

If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.

 

  

Solution 34

  

Question 35

In the given figure, AB and DE are perpendiculars to BC.

Find the ratio of the area of a ΔABC : area of ΔDEC.

 

  

Solution 35

 

 

  

 

 

 

Question 36

ABC is a right angled triangle with ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :

 

  

 

 

ΔADE ~ ΔACB.

Solution 36

  

Question 37

ABC is a right angled triangle with ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :

 

  

 

 

If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

Solution 37

  

Question 38

ABC is a right angled triangle with ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :

 

  

 

 

Find, area of ΔADE : area of quadrilateral BCED.

Solution 38

Question 39

Given: AB || DE and BC || EF. Prove that:

(i)

(ii)

 

Solution 39

(i) In AGB, DE || AB , by Basic proportionality theorem,

.... (1)


 

In GBC, EF || BC, by Basic proportionality theorem,

.... (2)


 

From (1) and (2), we get,


 

(ii)

From (i), we have:

Question 40

PQR is a triangle. S is a point on the side QR of ΔPQR such that PSR = QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.

i. Prove ΔPQR ∼ ΔSPR.

ii. Find the lengths of QR and PS.

iii.

  

Solution 40

i.

In PQR and SPR,

PSR = QPR … given

PRQ = PRS … common angle

 PQR SPR  (AA Test)

ii. Find the lengths of QR and PS.

Since ∆PQR SPR … from (i)

iii.

  

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