# RS Aggarwal Solutions Class 10 Chapter 4 Triangles

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## RS Aggarwal Solutions Class 10 Chapter 4 Triangles

### Exercise 4A

Question 1

Solution:

From given triangle, points D and E are on the sides AB and AC respectively such that DE || BC.

(i)AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm.

By Thale’s Theorem:

Here DB = AB – AD = 10 – 3.6 = 6.4

=> EC = 4.6/3.6 x 6.4

or EC = 8

And, AC = AE + EC

AC = 4.5 + 8 = 12.5

(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.

By Thale’s Theorem:

AD/DB + 1 = AE/EC + 1

(AD + DB)/DB = (AE + EC)/EC

AB/DB = AC/EC

or DB = (ABxEC)/ AC

= (13.3 x 5.1)/11.90

= 5.7

=> BD = 5.7

And, AD = AB – DB

(iii) AD/DB = 4/7 or AD = 4 cm, DB = 7 cm, and AC = 6.6

By Thale’s Theorem:

AD/DB + 1 = AE/EC + 1

(AD + DB)/DB = (AE + EC)/EC

(4+7)/7 = AC/EC = 6.6/EC

EC = (6.6 x 7)/11

= 4.2

And, AE = AC – EC

AE = 6.6 – 4.2

AE = 2.4 cm

(iv)

AD/AB = 8/15 or AD = 8 cm, AB = 15 cm, and EC = 3.5 cm

By Thale’s Theorem:

8/15 = AE/(AE+EC) = AE/(AE+3.5)

8(AE + 3.5) = 15AE

7 AE = 28

or AE = 4 cm

Question 2

Solution:

From figure, D and E are the points on the sides AB and AC respectively and DE || BC

(i) AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.

x/(x-2) = (x+2)/(x-1)

x(x-1) = (x+2)(x-2)

Solving above equation, we get

x = 4 cm

(ii) AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.

4/(x-4) = 8/(3x-19)

4(3x-19) = 8(x-4)

Solving, we get x = 11 cm

(iii)

AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) and EC = 3x cm.

(7x-4)/(3x + 4) = (5x – 2)/3x

(7x – 4) (3x) = (5x – 2) (3x + 4)

21x² – 12x – 15x² – 20x + 6x = -8

6x² – 26x + 8 = 0

(x – 4) (3x – 1) = 0

Either x – 4 = 0 or (3x – 1) = 0

=> x = 4 or 1/3 (not possible)

So, x = 4

Question 3:

Solution:

From figure, D and E are the points on the sides AB and AC of ∆ABC

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

and AE/EC = 4.8/8 = 3/5

=> DE || BC

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

AD = AB – BD = 11.7 – 6.5 = 5.2 cm and

EC = AC – AE = 11.2 – 4.2 = 7 cm

AE/EC = 4.2/7 = 3/5

=> DE is not parallel to BC

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

DB = AB – AD = 10.8 – 6.3 = 4.5 cm

AE =AC – EC = 9.6 – 4 = 5.6 cm

AD /DB = 6.3/4.5 = 7/5

AE/EC = 5.6/4 = 7/5

=> DE || BC

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

DB = AB – AD = 12 – 7.2 = 4.8 cm and

EC = AC – AE = 10 – 6,4 =3.6 cm

AD /DB = 7.2/4.8 = 3/2 and

AE/EC = 6.4/3.6 = 16/9

=> DE is not parallel to BC

Question 4:

Solution:

(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

AD bisects ∠A, we can apply angle-bisector theorem in ∆ABC,

BD/DC = AB/AC

Substituting given values, we get

5.6/DC = 6.4/8

DC = 7 cm

(ii) If AB = 10 cm, AC = 14 cm and BC – 6 cm, find BD and DC.

By angle-bisector theorem

BD/DC = AB/AC = 10/14

Let BD = x cm and DC = (6-x) (As BC = 6 cm given)

x/(6-x) = 10/14

14x = 10(6 – x)

14x = 60 – 10x

14x + 10x = 60

or x = 2.5

Or BD = 2.5

Then DC = 6 – 2.5 = 3.5 cm

(iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC.

BD/DC = AB/AC

Here DC = BC – BD = 6 – 3.2 = 2.8

=> DC = 2.8

3.2/2.8 = 5.6/AC

=> AC = 4.9 cm

(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

BD/DC = AB/AC

BD/3 = 5.6/4

=> BD = 4.2

Now, BC = BD + DC = 4.2 + 3 = 7.2

BC is 7.2 cm.

Question 5:

Solution:

M is a point on the side BC of a parallelogram ABCD

(i)Consdier ∆DMC and ∆NMB,

By By AAA-similarity:

∆DMC ∼ ∆NMB

From similarity of the triangle:

DM/MN = DC/BN

(ii)

From (i), DM/MN = DC/BN

DM/MN + 1 = DC/BN + 1

(DM+MN)/MN = (DC+BN)/BN

Since AB = CD

(DM+MN)/MN = (AB+BN)/BN

DN/DM = AN/DC

Hence proved.

Question 6:

Solution:

Here, AB || DC

M and N are the mid points of sides AD and BC respectively.

MN is joined.

To prove : MN || AB or DC.

Produce AD and BC to meet at P.

Now, In ∆PAB

DC ||AB

PD/DA = PC/CB

PD/2PM = PC/2CN

M and N are midpoints of AD and BC respectively.

PD/PM = PC/CN

MN||DC But DC ||AB

Therefore, MN || DC ||AB

Question 7:

Solution:

From given statement:

EO || AB || DC

By thales theorem: AE/ED = AO/OC …(1)

In Δ DAB,

EO || AB

So, By thales theorem: DE/EA = DO/OB …(2)

From (1) and (2)

AO/OC = DO/OB

(5x – 7) / (2x + 1) = (7x-5) / (7x+1)

(5x – 7)(7x + 1) = (7x – 5)(2x + 1)

35x^2 + 5x – 49x – 7 = 14x^2 – 10x + 7x – 5

35x^2 – 14x^2 – 44x + 3x – 7 + 5 = 0

21x^2 – 42x + x – 2 = 0

21(x – 2) + (x – 2) = 0

(21x + 1)(x – 2) = 0

Either (21x + 1) = 0 or (x – 2) = 0

x = -1/21 (does not satisfy) or x = 2

=> x = 2.

Question 8:

Solution:

In ∆ABC, M and N are points on the sides AB and AC respectively such that BM = CN and if ∠B = ∠C.

We know that, sides opposite to equal angles are equal.

AB = AC

BM = CN ( given)

AB – BM = AC – CN

=> AM = An

From ∆ABC

AM/MB = AN/NC

Therefore, MN ||BN

### Exercise 4B

Question 1:

Solution:

Two triangles are similarity of their corresponding angles are equal and corresponding sides are proportional.

(i) In ∆ABC and ∆PQR

∠A = ∠Q = 50°

∠B = ∠P = 60° and

∠C = ∠R = 70°

∆ABC ~ ∆QPR (By AAA)

(ii) In ∆ABC and ∆DEF

In ∆ABC and ∆DEF

AB = 3 cm, BC = 4.5

DF = 6 cm, DE = 9 cm

∆ABC is not similar to ∆DEF

(iii) In ∆ABC and ∆PQR

In ∆ABC and ∆PQR

AC = 8 cm BC = 6 cm

Included ∠C = 80°

PQ = 4.5 cm, QR = 6 cm

and included ∠Q = 80°

AC/QR = 8/6 = 4/3

and BC/PQ = 6/4.5 = 4/3

=> AC/QR = BC/PQ

and ∠C = ∠Q = 80°

∆ABC ~ ∆PQR (By SAS)

(iv)In ∆DEF and ∆PQR

DE = 2.5, DF = 3 and EF = 2

PQ = 4, PR = 6 and QR = 5

DE/QR = 2.5/5 = 1/2

DF/PR = 3/6 = 1/2

and EF/PQ = 2/4 = 1/2

∆DEF ~ ∆PQR (By SSS)

(v) In ∆ABC and ∆MNR

∠A = 80°, ∠C = 70°

So, ∠B = 180° – (80° + 70°) = 30°

∠M = 80°, ∠N = 30°, and ∠R = 180° – (80° + 30°) = 70°

Now, in ∆ABC

∠A = ∠M – 80°, ∠B = ∠N = 30°

and ∠C = ∠R = 70°

∆ABC ~ ∆MNR (By AAA or AA)

Question 2:

Solution:

Here ∆ODC ~ ∆OBA, so

∠D = ∠B = 70°

∠C = ∠A

∠COD = ∠AOB

(i) But ∠DOC + ∠BOC = 180° (Linear pair)

∠DOC + 115°= 180°

∠DOC = 180° – 115° = 65°

(ii) ∠DOC + ∠CDO + ∠DCO = 180° (Angles of a triangle)

65° + 70° + ∠DCO = 180°

135° + ∠DCO = 180°

∠DCO = 180° – 135° = 45°

(iii) ∠AOB = ∠DOC = 65° (vertically opposite angles)

∠OAB = ∠DCO = 45° (Since ∆ODC ~ ∆OBA)

(iv) ∠OBA = ∠CDO = 70° (Since ∆ODC ~ ∆OBA)

Question 3:

Solution:

Since ∆OAB ~ ∆OCD

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

Let OD = y and OA = x

Question 4:
Solution:

From given figure,

To prove:

find DE

Given: AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm

∠A = ∠A (common)

Again,

Question 5:
Solution:

Form given statement: ∆ABC ~ ∆PQR,

PQ = 12 cm

Perimeter of ∆ABC = AB + BC + CA = 32 cm

Perimeter of ∆PQR = PQ + QR + RP = 24 cm

Now,

Question 6:
Solution:

Form given statement: ∆ABC ~ ∆DEF

BC = 9.1 cm, EF = 6.5 cm and Perimeter of ∆DEF = 25 cm

Perimeter of ∆ABC is 35 cm

Question 7:
Solution:

∠CAB = 90° and AD ⊥ BC

If AC = 75 cm, AB = 1 m or 100 cm, BC = 1.25 m or 125 cm

∠BDA = ∠BAC = 90°

∠DBA = ∠CBA [common angles]

By AA

∆BDA ~ ∆BAC

And,

Question 8:
Solution:

From given:

∠ABC = 90°, BD ⊥ AC.

AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm

In ∆ABC and ∆BDC,

∠ABC = ∠BDC (each 90°)

∠BCA = ∠BCD (common angles)

∆ABC ~ ∆BDC (AA axiom)

So, corresponding sides are proportional

AB/BD = BC/CD

=> 5.7/3.8 = BC/5.4

=> BC = 8.1

### Exercise 4C

Question 1:

Solution:

Area of ∆ABC = 64 cm² and

area of ∆DEF = 121 cm²

EF = 15.4 cm

Question 2:
Solution:

The areas of two similar triangles ABC and PQR are in the ratio 9 : 16.

BC = 4.5 cm

Question 3:
Solution:

∆ABC ~ ∆PQR

ar (∆ABC) = 4ar (∆PQR)

Question 4:
Solution:

Areas of two similar triangles are 169 cm² and 121 cm² (given)

Longest side of largest triangle = 26 cm

Let longest side of smallest triangle is x cm

Longest side of smallest triangle is 22 cm

Question 5:
Solution:

Area of ∆ABC = 100 cm²

area of ∆DEF = 49 cm²

Altitude of ∆ABC is 5 cm

AL ⊥ BC and DM ⊥ EF

Let DM = x cm

Or x = 3.5

Altitude of smaller triangle is 3.5 cm

Question 6:
Solution:

Corresponding altitudes of two similar triangles are 6 cm and 9 cm (given)

We know that the areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.

Ratio in the areas of two similar triangles = (6)² : (9)² = 36 : 81 = 4 : 9

Question 7:
Solution:

Areas of two similar triangles are 81 cm² and 49 cm²

Altitude of the first triangle = 6.3 cm

Let altitude of second triangle = x cm

Area of ∆ABC = 81 cm² and area of ∆DEF = 49cm²

Altitude AL = 6 – 3 cm

Let altitude DM = x cm

Altitude of second triangle is 4.9 cm

Question 8:
Solution:

Areas of two similar triangles are 100 cm² and 64 cm²

Median DM of ∆DEF = 5.6 cm

Let median AL of ∆ABC = x

Corresponding median of the other triangle is 7 cm.

Question 9:
Solution:

In ∆ABC, PQ is a line which meets AB in P and AC in Q.

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

Now, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3

=> AP/PB = AQ/QC

From figure: AB = AP + PB = 1+3 = 4 cm

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

In ∆APQ and ∆ABC,

AP/AB = AQ/AC

angle A (common)

∆APQ and ∆ABC are similar triangles.

Now,

Which implies,

area of ∆APQ = 1/16 of the area of ∆ABC

Hence Proved.

Question 10:
Solution:

DE || BC

DE = 3 cm, BC = 6 cm

Now,

In ∆ABC

DE ||BC. Therefore triangles, ∆ADE and ∆ABC are similar.

Area of ∆ABC is 60 cm2.

### Exercise 4D

Question 1:
Solution:

A given triangle to be right-angled, if it satisfies Pythagorean Theorem. That is, the sum of the squares of the two smaller sides must be equal to the square of the largest side.

(i) 9 cm, 16 cm, 18 cm

Longest side = 18

Now (18)² = 324

and (9)² + (16)² = 81 + 256 = 337

324 ≠ 337

It is not a right triangle.

(ii) 1 cm, 24 cm, 25 cm

Longest side = 25 cm

(25)² = 625

and (7)² x (24)² = 49 + 576 = 625

625 = 625

It is a right triangle.

(iii) 1.4 cm, 4.8 cm, 5 cm

Longest side = 5 cm

(5)² = 25

and (1.4)² + (4.8)² = 1.96 + 23.04 = 25.00 = 25

25 = 25

It is a right triangle.

(iv) 1.6 cm, 3.8 cm, 4 cm

Longest side = 4 cm

(4 )² = 16

and (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00 = 17

16 ≠ 17

It is not a right triangle.

(v) (a- 1) cm, 2√a cm, (a + 1) cm

Longest side = (a + 1) cm

(a + 1)² = a² + 2a + 1

and (a – 1)² + (2 √a )² = a² – 2a + 1 + 4a = a² + 2a + 1

a² + 2a + 1 = a² + 2a + 1

It is a right triangle.

Question 2:
Solution:

A man goes 80 m from O to east side and reaches A, then he goes 150 m due north from A and reaches B.

Draw a figure based on given instructions:

From right ∆OAB,

By Pythagoras Theorem:

OB2 = OA2+ AB2

= (80)² + (150)²

= 6400 + 22500

= 28900

or OB = √28900 = 170

Man is 170 m away from the starting point.

Question 3:

Solution:

A man goes 10 m due south from O and reaches A and then 24 m due west from A and reaches B.

Draw a figure based on given instructions:

From right ∆OAB,

By Pythagoras Theorem:

OB2 = OA2+ AB2

= (10)² + (24)²

= 676

or OB = 26

Man is 26 m away from the starting point.

Question 4:
Solution:

Height of the window = 12 m

Length of a ladder = 13 m

In the figures,

Let AB is ladder, A is window of building AC

By Pythagoras Theorem:

AB2 = AC2 + BC2

(13)² = (12)² + x²

169 = 144 + x²

x² = 169 – 144 = 25

or x = 5

Distance between foot of ladder and building = 5 m.

Question 5:
Solution:

Height of window AC = 20 m

Let length of ladder AB = x m

Distance between the foot of the ladder and the building (BC) = 15 m

In the figure:

By Pythagoras Theorem:

AB2 = AC2 + BC2

x² = 20² + 15²

= 400 + 225

= 625

or x = 25

Length of ladder is 25 m

Question 6:
Solution:

Height of first pole AB = 9 m and

Height of second pole CD = 14 m

Let distance between their tops CA = x m

From A, draw AE || BD meeting CD at E.

Then EA = DB = 12 m CE = CD – ED = CD – AB = 14-9 = 5 m

In right ∆AEC,

AC² = AE² + CE²

= 122 + 52

= 144 + 25

= 169

or AC = 13

Distance between pole’s tops is 13 m

Question 7:

Solution:

Length of wire = AC = 24 m

Height of the pole = AB = 18 m

Let Distance between the base of the pole and other end of the wire = BC = x m

In right ∆ABC,

By Pythagoras Theorem:

AC2 = AB2 + BC2

(24)² = (18)² + x²

576 = 324 + x²

x² = 576 – 324 = 252

or x = 6v7

BC is 6v7m

Question 8:
Solution:

In ∆PQR, O is a point in it such that

OP = 6 cm, OR = 8 cm and ∠POR = 90°

PQ = 24 cm, QR = 26 cm

Now,

In ∆POR, ∠O = 90°

PR² = PO² + OR²

= (6)² + (8)²

= 36 + 64

= 100

PR = 10

Greatest side QR is 26 cm

QR² = (26)² = 676

and PQ² + PR² = (24)² + (10)²

= 576 + 100

= 676

Which implies, 676 = 676

QR² = PQ² + PR²

∆PQR is a right angled triangle and right angle at P.

Question 9:
Solution:

In isosceles ∆ABC, AB = AC = 13 cm

Consider AL is altitude from A to BC and AL = 5 cm

Now, in right ∆ALB

AB2 = AL2 + BL2

(13)² = (5)² + BL²

169 = 25 + BL²

BL² = 169 – 25 = 144

or BL = 12

Since L is midpoint of BC, then

BC = 2 x BC = 2 x 12 = 24

BC is 24 cm

Question 10:

Solution:

In an isosceles ∆ABC in which AB = AC = 2a units, BC = a units

AD is the altitude. Therefore, D is the midpoint of BC

=> BD = a/2

By Pythagoras theorem,

### Exercise 4E

Question 1:

Solution:

Two properties for similarity of two triangles are:

(i) Angle-Angle-Angle (AAA) property.

(ii) Angle-Side-Angle (ASA) property.

Question 2:
Solution:

In a triangle, if a line parallel to one side is drawn, it will divide the other two sides proportionally.

Question 3:
Solution:

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

Question 4:
Solution:

The line joining the midpoints of two sides of a triangle, is parallel to the third side.

Question 5:
Solution:

In two triangles, if three angles of the one triangle are equal to the three angles of the other, the triangles are similar.

Question 6:
Solution:

In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar.

Question 7:
Solution:

In two triangles, if three sides of the one are proportional to the corresponding sides of the other, the triangles are similar.

Question 8:
Solution:

In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar.

Question 9:
Solution:

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

Question 10:
Solution:

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

Question 11:
Solution:

The ratio of their areas will be 1 : 4.

Question 12:
Solution:

In two triangles ∆ABC and ∆PQR,

AB = 3 cm, AC = 6 cm, ∠A = 70°

PR = 9 cm, ∠P = 70° and PQ= 4.5 cm

Now,

∠A = ∠P = 70° (Same)

AC/PR = 6/9 = 2/3 and

AB/PQ = 3/4.5 = 2/3

=> AC/PR = AB/PQ

Both ∆ABC and ∆PQR are similar.

Question 13
Solution:

∆ABC ~ ∆DEF (given)

2AB = DE, BC = 6 cm (given)

∠E = ∠B and ∠D = ∠A and ∠F = ∠C

2AB = DE

=> AB/DE = 1/2

Therefore,

AB/DE = BC/EF

1/2 = 6/EF

or EF = 12 cm

Question 14:
Solution:

From figure:

DE || BC

AD = x cm, DB = (3x + 4) cm

AE = (x + 3) cm and EC = (3x + 19) cm

In ∆ABC

x/(3x+4) = (x+3)/(3x+19)

3x2 + 19x – 3x2– 9x – 4x = 12

x = 2

Question 15:
Solution:

Let AB is the ladder and A is window.

Then, AB = 10 m and AC = 8 m

Let BC = x

In right ∆ABC,

By Pythagoras Theorem:

AB2 = AC2 + BC2

(10)² = 8² + x²

100 = 64 + x²

x² = 100 – 64 = 36

or x = 6

Therefore, Distance between foot of ladder and base of the wall is 6 m.

Complete RS Aggarwal Solutions Class 10

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