# R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 7 - Triangles

## Chapter 4 - Triangles MCQ

Correct option: (b)

Recall that the diagonals of a trapezium divide each other proportionally.

Note that this happens even in a parallelogram, square and rectangle, but without additional information it is not possible to be sure.

Correct option: (a)

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram as shown below.

* *

Correct option: (b)

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.

Correct option: (b)

Clearly, option (b) is false since the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

## Chapter 4 - Triangles Ex. 7E

Two triangles are said to be similar to each other if:

(i) their corresponding angles are equal, and

(ii) their corresponding sides are proportional.

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct point, then the other sides are divided in the same ratio.

If a line divides any two sides of a triangle in the same ratio then the line must be parallel to the third side.

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side.

If in any two triangles, the corresponding angles are equal, then their corresponding sides are proportional and hence the triangles are similar.

If two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar.

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.

Similar figures have the same shape but need not have the same size.

Since all circles irrespective of the radii will have the same shape, all will be similar.

So, the statement is true.

Two rectangles are similar if their corresponding sides are proportional.

So, the statement is false.

Two triangles are said to be similar to each other if:

(i) their corresponding angles are equal, and

(ii) their corresponding sides are proportional.

So, the statement is false.

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram as shown.

It may or may not be a rhombus.

So, the statement is false.

## Chapter 4 - Triangles Ex. 7A

(i)In _{}ABC, DE || BC, AD = 3.6 cm, AB = 10 cm, AE = 4.5 cm

Hence, AC = 12.5 cm and EC = 8cm

(ii)In _{}ABC, DE || BC, AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm

Hence, AD = 7.7 cm

(iii)In _{}ABC, DE || BC, AC = 6.6 cm, _{}

?

Hence, AE = 2.4 cm

(iv)In _{}ABC, DE || BC, Given _{}

Hence AE = 4 cm

(i)D and E are points on the sides AB and AC respectively of a _{}ABC such that DE || BC, AD = x cm, DB = (x - 2) cm,

AE = (x + 2) cm, EC = (x - 1) cm

_{ }

Hence, x = 4

(ii)In _{}ABC, DE || BC, AD = 4 cm, DB = (x - 4) cm, AE = 8 cm, EC = (3x - 19) cm

Hence, x = 11

(iii)In _{}ABC, DE || BC, AD = (7x - 4) cm, AE = (5x - 2) cm, DB = (3x + 4)cm, EC = 3x cm

_{ }_{ }

_{ }

Given: A _{}ABC in which D and E are points on the sides AB and AC respectively.

To prove: DE ||BC

Proof:

(i)AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm

Since D and E are the points on AB and AC respectively.

Hence, by the converse of Thales theorem DE || BC

(ii)AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm, AE = 4.2 cm

Since D and E are points on AB and AC respectively.

_{}

_{}

Hence, by the converse of Thales theorem DE is not parallel to BC.

(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm, EC = 4 cm

Since D and E are the points on AB and AC respectively.

_{ }

Therefore, _{}(each is equal to 1.4)

Hence by the converse of Thales theorem DE || BC

(iv)AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm, AC = 10 cm

Since D and E are points on the side AB and AC respectively.

_{}

Hence, by the converse of Thales theorem DB is not parallel to BC

(i) AB = 6.4 cm, AC = 8 cm, BD = 5.6 cm

Let BC = x

Now, DC = (BC - BD)

= (x - 5.6) cm

In ABC, AD is the base for of A

So, by the angle bisector theorem, We have

Hence, BC = 12.6 cm and DC = (12.6 - 5.6) cm = 7 cm

(ii) AB = 10 cm, AC = 14 cm, BC = 6cm

Let BD = x,

DC = (BC - BD) = (6 - x) cm

In ABC, AD is the bisector of ??A

So, By angle bisector theorem,

Hence, BD = 2.5 cm and DC = (6 - 2.5) cm = 3.5 cm

_{}

(iii)AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm

DC = BC - BD = (6 - 3.2) cm = 2.8 cm

Let AC = x,

In ABC, AD is the bisector of A

So, by the angle bisector theorem we have

_{}

Hence, AC = 4.9 cm

(iv)AB = 5.6 cm, AC = 4 cm, DC = 3 cm

Let BD = x,

In ABC, AD is the bisector of A

So, by the angle bisector theorem, we have

_{}

Hence, BD = 4.2 cm

So BC = BD + AC = (4.2 + 3) cm

_{}BC = 7.2 cm

Let ABCD be the trapezium and let E and F be the midpoints of AD and BC respectively.

Const: Produce AD and BC to meet at P

In PAB, DC || AB

We know that CD || AB in trap ABCD and its diagonals intersect at O.

Since the diagonals of a trapezium divides each other proportionally therefore, we have

?

Given: ABC and DBC lie on the same side of BC. P is a point on BC, PQ || AB and PR || BD are drawn meeting AC at Q and CD at R respectively.

To Prove: QR || AD

Proof: In ABC

_{}

Hence, in ACD, Q and R the points in AC and CD such that

_{}

_{}QR || AD(by the converse of Thales theorem)

Hence proved.

Given BD = CD and OD = DX

Join BX and CX

Thus, the diagonals of quad OBXC bisect each other

_{}OBXC is a parallelogram

_{}BX || CF and so, OF || BX

Similarly, CX || OE

In ABX, OF || BX

_{}

Given: ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that_{}. PQ produced meets BC at R.

To prove: R is the midpoint of BC

Construction: Join BD

Proof: Since the diagonals of a || gm bisect each other at S such that

_{}

_{}Q is the midpoint of CS

So, PQ || DS.

Therefore, QR || SB.

In CSB, Q is the midpoint of CS and QR || SB.

So R is the midpoint of BC.

Given: ABC is a triangle in which AB = AC. D and E are points on AB and AC respectively such that AD = AE

To prove: The points B, C, E and D are concyclic.

Proof: AB = AC (given)

AD = AE (given)

_{}Quad BCEA is cyclic

Hence, the point B, C, E, D are concyclic

## Chapter 4 - Triangles Ex. 7B

(i)In ABC and PQR

_{}A = _{}Q = 50°

_{}B = _{}P = 60°

_{}C = _{}R = 70°

_{}ABC ~QPR(by AAA similarity)

(ii)In ABC and EFD

_{}A = _{}D = 70°

SAS: Similarity condition is not satisfied as _{}A and _{}D are not included angles.

(iii)CAB _{}QRP (SAS Similarity)

_{}

(iv) In EFD and PQR

FE = 2cm, FD = 3 cm, ED = 2.5 cm

PQ = 4 cm, PR = 6 cm, QR = 5 cm

_{} FED ~PQR (SSS similarity)

ODC ~ OBC

_{}BOC = 115^{o}

_{}CDO = 70^{o}

_{(i) }DOC = (180^{o} - _{}BOC)

= (180^{o} - 115^{o})

= 65^{o}

_{(ii) }OCD = 180^{o} - CDO - DOC

_{ }OCD = 180^{o} - (70^{o} + 65^{o})

= 45^{o}

(iii) Now, ABO ~ ODC

_{ }AOB = _{}COD (vert. Opp _{}s) = 65^{o}

_{ }OAB = _{}OCD = 45^{o}

_{(iv) }OBA = _{}ODC(alternate angles) = 70^{o}

So, _{}OAB = 45^{o} and _{}OBA = 70^{o}

Given: OAB _{}OCD

AB = 8 cm, BO = 6.4 cm, CD = 5 cm, OC = 3.5 cm

_{}

Given: _{}ADE = _{}B, AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm

Proof:

In ADE and ABC,

_{}A = _{}A (common)

ADE = B (given)

Therefore, ADE ABC (AA Criterion)

_{}

Hence, DE = 2.8 cm

ABC and PQR are similar triangles, therefore corresponding sides of both the triangles are proportional.

_{}

Hence, AB = 16 cm

ABC and DEF are two similar triangles, therefore corresponding sides of both the triangles are proportional.

Hence, _{}

Let perimeter of ABC = x cm

_{}

Hence, perimeter of ABC = 35 cm

Given: AB = 100 cm, BC = 125 cm, AC = 75 cm

Proof:

In BAC and BDA

_{}BAC = _{}BDA = 90^{o}

_{}B = _{}B (common)

BAC _{}BDA(by AA similarities)

_{}

Therefore, AD = 60 cm

Given that AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm

In CBA and CDB

_{}CBA = _{}CDB = 90^{o}

_{}C = _{}C (Common)

Therefore, CBA _{} CDB (by AA similarities)

_{}

Hence, BC = 8.1 cm

Given that BD = 8 cm, AD = 4 cm

In DBA and DCB, we have

_{}BDA = _{}CDB = 90^{o}

_{ }_{}DBA = _{}DCB [each = 90^{o} - _{}A]

_{}DBA _{}DCB (by AAA similarity)

_{}

Hence, CD = 16 cm

Given: P is a point on AB.

Then, AB = AP + PB = (2 + 4) cm = 6 cm

Also Q is a point on AC.

Then, AC = AQ + QC = (3 + 6) cm = 9 cm

_{}

Thus, in APQ and ABC

_{}A = _{}A (common)

And_{}

_{}APQ ~ ABC(by SAS similarity)

_{}

Hence proved.

Given: ABCD is a parallelogram and E is point on BC. Diagonal DB intersects AE at F.

To Prove: AF × FB = EF × FD

Proof: In AFD and EFB

_{A}FD = _{}EFB (vertically opposite _{}s)

DAF = BEF (Alternate _{}s)

_{}

Hence proved.

In the given figure: DB AB, AC BC and DB || AC

_{}

AB is the transversal

_{}_{}DBE = _{}BAC [Alternate _{}s]

In BDE and ABC

_{}DEB = _{}ACB = 90^{o}

_{}DBE = _{}BAC

_{}~ _{}[By AA similarity]

_{}

Hence proved.

Let AB be the vertical stick and let AC be its shadow.

Then, AB = 7.5 m and AC = 5 m

Let DE be the vertical tower and let DF be its shadow

Then,DF = 24 m, Let DE = x meters

Now, in BAC and EDF,

BAC ~ EDF by SAS criterion

_{}

Therefore, height of the vertical tower is 36 m.

In ACP and BCQ

CA = CB

_{}_{}CAB = _{}CBA

_{}

_{}ACP _{}BCQ

_{}1 = _{}2 (given)

_{ } (given)_{}

Also, _{}2 = _{}1 _{}

Therefore, by SAS similarity criterion ACB ~ DCE

Given: ABCD is a quadrilateral in which AD = BC. P, Q , R, S are the midpoints of AB, AC, CD and BD.

To prove: PQRS is a rhombus

Proof: In ABC,

Since P and Q are mid points of AB and AC

Therefore, PQ || BC and _{ }(Mid-point theorem)

Similarly,

_{}SP || RQ and PQ || SR and PQ = RQ = SP = SR

Hence,PQRS is a rhombus.

## Chapter 4 - Triangles Ex. 7C

Given: ABC _{}DEF,

area of ABC= _{}and area of DEF = 121_{}

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

_{}

Hence, BC = 11.2 cm

Given: ?ABC _{}PQR,

areaof ABC = 9 and area of PQR = 16.

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

_{}

Hence, QR = 6 cm

Given: ABC ~PQR,

area of ABC = 4 area of PQR.

Let area of PQR = x. Then area of ABC = 4x.

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

_{}

_{}

Hence. QR = 6 cm

Given: ABC _{} DEF such that ar(ABC) = 169_{} and ar(DEF) = 121 _{}

We know that the ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.

_{}

Hence, the longest side of smallest triangle side is 22 cm.

Given: ABC DEF

ar(ABC) = 100 and ar(DEF) = 49

Let AL and DM be the corresponding altitude of ABC and DEF respectively such that AL = 5 cm and let DM = x cm

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

_{}

Therefore, the required altitude is 3.5 cm

Given: ABC _{}DEF

Let AL and DM be the corresponding altitudes of ABC and DEF respectively such that AL = 6 cm and DM = 9 cm.

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

_{}

Hence, ratio of their areas = 4 : 9

Given: ABC _{}DEF such that

ar(ABC) = 81_{} and ar(DEF) = 49_{}

Let AL and DM be the corresponding altitudes of ABC and DEF respectively, such that AL = 6.3 cm and Let DM= x cm

We know that the ratio of the area of two similar triangles is equal to the ratio of the square of corresponding altitudes:

_{}

Hence, the required altitude 4.9 cm

Given: ABC _{} DEF such that ar(ABC) = 100 cm and ar(DEF) = 64_{}

Let AP and DQ be the corresponding medians of ABC and DEF respectively such that DQ = 5.6cm.

Let AP = x cm.

We know that the ratio of the areas of two similar triangle is equal be the ratio of the squares of their corresponding medians.

Hence, AP = 7 cm

Given: AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm

AB = AP + PB = (1 + 3) cm = 4 cm

AC = AQ + QC = (1.5 + 4.5) cm = 6 cm

In APQ and ABC, we have

_{}APQ = _{}ABC (corresponding _{}s)

_{}AQP = _{}ACB (corresponding _{}s)

_{}APQ _{}ABC [by AA similarity]

Hence proved.

Given DE || BC

DE = 3 cm and BC = 6 cm

ar(ADE) = 15

In ADE and ABC, we have

_{}

In BAC and ADC, we have

_{}BAC = _{}ADC = 90^{o} (AD _{}BC)

_{}ACB = _{}DCA (common)

BAC _{}ADC

_{}

Therefore, the ratio of the areas of ABC and ADC = 169:25

Let DE = 3x and BC = 5x

In ADE and ABC, we have

_{}ADE = _{}ABC (corres. _{}s)

_{}AED = _{}ACB (corres. _{}s)

_{}ADE _{}ABC (by AA similarity)

Let, ar(ADE) = 9x^{2} units

Then, ar(ABC) = 25x^{2} units

_{}

Therefore, ratio of ar(ADE) to the ar(trap BCED) = 9:16

In ABC, D and E are midpoint of AB and AC respectively.

So, DE|| BC and

Now, in ADE and ABC, we have

_{}ADE = _{}ABC (corres. _{}s)

_{}AED = _{}ACB (corres. _{}s)

_{}ADE _{}ABC (by AA similarity)

Let AD = x and AB = 2x

Therefore, the ratio of the areas of ADE and ABC = 1:4

## Chapter 4 - Triangles Ex. 7D

For a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.

(i)Let a = 9cm, b = 16 cm and c = 18 cm. Then

_{}

Hence the given triangle is not right angled.

(ii)Let a = 7cm, b = 24 cm and c = 25 cm, Then

_{}

Hence, the given triangle is a right triangle.

(iii)Let a = 1.4 cm, b = 4.8 cm, and c = 5 cm

_{}

Hence, the given triangle is a right triangle

(iv)Let a = 1.6 cm, b = 3.8 cm and c = 4 cm

_{}

Hence, the given triangle is not a right triangle

(v)Let p = (a - 1) cm, q = _{}cm and r = (a + 1) _{}

_{}

Hence, the given triangle is a right triangle

Starting from A, let the man goes from A to B and from B to C, as shown in the figure.

Then,

AB = 80 m, BC = 150 m and_{}ABC = 90^{o}

From right ABC, we have

_{}

Hence, the man is 170m north-east from the starting point.

Starting from O, let the man goes from O to A and then A to B as shown in the figure.

Then,

OA = 10 m, AB = 24 m and OAB = 90^{o}

Using Pythagoras theorem:

_{}

Hence, the man is 26 m south-west from the starting position.

Let AB be the building and CB be the ladder.

Then,

AB = 12 m, CB = 13 m and CAB = 90^{o}

By Pythagoras theorem, we have

Hence, the distance of the foot of the ladder from the building is 5 m.

Let AB be the wall where window is at B, CB be the ladder and AC be the distance between the foot of the ladder and wall.

Then,

AB = 20 m, AC = 15 m, and CAB = 90^{o}

By Pythagoras theorem, we have

Hence, the length of ladder is 25 m.

Let AB and CD be the given vertical poles.

Then,

AB = 9 m, CD = 14 m and AC = 12 m

Const: Draw, BE || AC.

Then,

CE = AB = 9m and BE = AC = 12 m

_{}DE = (CD - CE)

= (14 - 9)

= 5 m

In right BED, we have

_{}

Hence, the distance between their tops is 13 m.

In PQR, _{}QPR = 90^{o}, PQ = 24 cm, and QR = _{}

In POR, PO = 6 cm, QR = 8cm and _{}POR = 90^{o}

In POR,

In PQR,

By Pythagoras theorem, we have

Hence, _{}

(sum of square of two sides equal to square of greatest side)

Hence, PQR is a right triangle which is right angled at P.

Given: ABC is an isosceles triangle with AB = AC = 13 _{}

Const: Draw altitude from A to BC (AL _{}BC).

Now, AL = 5 cm

In ALB,

_{}ALB = 90^{o}

In ALC,

_{ }ALC = 90^{o}

Given: ABC in which AB = AC = 2a units and BC = a units

Const: Draw AD _{}BC then D is the midpoint of BC.

In ABC

_{}

In an equilateral triangle all sides are equal.

Then, AB = BC = AC = 2a units

Const: Draw an altitude AD _{}BC

Given BC = 2a. Then, BD = a

In ABD,

_{}ADB = 90^{o}

Hence, length of each altitude is _{}

ABC is an equilateral triangle in which all side are equa.

Therefore, AB = BC = AC = 12 cm

If BC = 12 cm

Then, BD = DC = 6 cm

In ADB,

Hence the height of the triangle is _{}

Let ABCD is the given rectangle, let BD is a diagonal making a ADB.

_{}BAD = 90^{o}

Using Pythagoras theorem:

_{}

Hence, length of diagonal DB is 34 cm.

Let ABCD be the given rhombus whose diagonals intersect at O.

Then AC = 24 cm and BD = 10 cm

We know that the diagonals of a rhombus bisect each other at right angles.

From right AOB, we have

Hence, each side of a rhombus 13 cm

Given: ABC in which D is the midpoint of BC. AE BC and AC > AB.

Then, BD = CD and AED = 90^{o},

Then, _{}ADE < 90^{o} and ADC > 90^{o}

In AED,

Putting value of _{}from (1) in (2), we get

_{}

Const: Draw a perpendicular AE from A

Thus, AE BC

Proof:

In ABC,AB = AC

And AE is a bisector of BC

Then,BE = EC

In right angle triangles AED and ACE

_{}

Hence proved.

Given: D is the midpoint of side BC, AE _{}BC, BC = a, AC = b, AB = c, ED = x, AD = p and AE = h

In AEC, AEC = 90^{o}

(i) In AEC, AEC = 90^{o}

_{}

_{}

_{}

(ii)In ABE, ABE = 90^{o}

_{}

(iii)Adding (1) and (2), we get

_{}

(iv)Subtracting (2) from (1), we get

_{}

ABC is an isosceles triangle right angled at B,

Let AB = BC = x cm

By Pythagoras theorem,

_{}

## Chapter 4 - Triangles FA

### Other Chapters for CBSE Class 10 Maths

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- Probability### R S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects

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