# R S Aggarwal Solutions for Class 10 Maths Chapter 7 Triangles

R S Aggarwal Solutions for Class 10 Chapter 7 Triangles benefit the students in understanding concepts thoroughly. All questions are created by subject experts using step-by-step problem solving approach. Class 10 R S Aggarwal Solutions help the students to analyze their shortcomings so that they overcome and clear their doubts on concepts. Chapter 7 triangle is based on triangles and its important theorem results. Students can download free pdf and practice various problems on triangles.

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#### Exercise 7A

Question 1: D and E are points on the sides AB and AC respectively of a âˆ†ABC such that DE || BC.

(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.

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

(iii) If AD/DB = 4/7 and AC = 6.6 cm, find AE.

(iv) if AD/AB = 8/15 and EC = 3.5 cm, find AE.

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 Thaleâ€™s Theorem:

Here DB = AB – AD = 10 â€“ 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 Thaleâ€™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 â€“ DB

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

By Thaleâ€™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 â€“ EC

AE = 6.6 â€“ 4.2

AE = 2.4 cm

(iv)

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

By Thaleâ€™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: D and E are points on the sides AB and AC respectively of a âˆ†ABC such that DE || BC. Find the value of x, when

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

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

(iii) AD = (7x â€“ 4) cm, AE = (5x â€“ 2) cm, DB = (3x + 4) and EC = 3x cm.

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 â€“ 2) cm, AE = (x + 2) cm and EC = (x â€“ 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 â€“ 4) cm, AE = 8 cm and EC = (3x â€“ 19) cm.

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

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

Solving, we get x = 11 cm

(iii)

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

(7x-4)/(3x + 4) = (5x â€“ 2)/3x

(7x â€“ 4) (3x) = (5x â€“ 2) (3x + 4)

21xÂ² â€“ 12x â€“ 15xÂ² â€“ 20x + 6x = -8

6xÂ² â€“ 26x + 8 = 0

(x â€“ 4) (3x â€“ 1) = 0

Either x â€“ 4 = 0 or (3x â€“ 1) = 0

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

So, x = 4

Question 3: D and E are points on the sides AB and AC respectively of a âˆ†ABC. In each of the following cases, determine whether DE || BC or not.

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

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

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

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

Solution:

From figure, D and E are the points on the sides AB and AC of âˆ†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: In a âˆ†ABC, AD is the bisector of âˆ A.

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

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

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

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

Solution:

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

AD bisects âˆ A, we can apply angle-bisector theorem in âˆ†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 â€“ 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 â€“ x)

14x = 60 â€“ 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 â€“ BD = 6 â€“ 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: M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that

(i) DM/MN = DC/BN (ii) DN/DM = AN/DC

Solution:

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

(i)Consdier âˆ†DMC and âˆ†NMB,

 âˆ DCM = âˆ NBM alternate angles âˆ DMC = âˆ NMB vertically opposite angles âˆ CDM = âˆ MNB alternate angles

By By AAA-similarity:

âˆ†DMC âˆ¼ âˆ†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: Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.

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 âˆ†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: In the adjoining figure, ABCD is a trapezium in which CD || AB and its diagonals intersect at O. If AO = (5x â€“ 7) cm, OC = (2x + 1) cm, BO = (7x â€“ 5) cm and OD = (7x + 1) cm, find the value of x.

Solution:

From given statement:

EO || AB || DC

By thales theorem: AE/ED = AO/OC …(1)

In Î” 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 â€“ 7)(7x + 1) = (7x â€“ 5)(2x + 1)

35x^2 + 5x â€“ 49x â€“ 7 = 14x^2 â€“ 10x + 7x â€“ 5

35x^2 â€“ 14x^2 â€“ 44x + 3x â€“ 7 + 5 = 0

21x^2 â€“ 42x + x â€“ 2 = 0

21(x â€“ 2) + (x â€“ 2) = 0

(21x + 1)(x â€“ 2) = 0

Either (21x + 1) = 0 or (x â€“ 2) = 0

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

=> x = 2.

Question 8: In a âˆ†ABC, M and N are points on the sides AB and AC respectively such that BM = CN. If âˆ B = âˆ C then show that MN || BC.

Solution:

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

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

AB = AC

BM = CN ( given)

AB – BM = AC – CN

=> AM = An

From âˆ†ABC

AM/MB = AN/NC

Therefore, MN ||BN

#### Exercise 7B

Question 1: In each of the given pairs of triangles, find which pair of triangles are similar. State the similarity criterion and write the similarity relation in symbolic form:

Solution:

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

(i) In âˆ†ABC and âˆ†PQR

âˆ A = âˆ Q = 50Â°

âˆ B = âˆ P = 60Â° and

âˆ C = âˆ R = 70Â°

âˆ†ABC ~ âˆ†QPR (By AAA)

(ii) In âˆ†ABC and âˆ†DEF

In âˆ†ABC and âˆ†DEF

AB = 3 cm, BC = 4.5

DF = 6 cm, DE = 9 cm

âˆ†ABC is not similar to âˆ†DEF

(iii) In âˆ†ABC and âˆ†PQR

In âˆ†ABC and âˆ†PQR

AC = 8 cm BC = 6 cm

Included âˆ C = 80Â°

PQ = 4.5 cm, QR = 6 cm

and included âˆ Q = 80Â°

AC/QR = 8/6 = 4/3

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

=> AC/QR = BC/PQ

and âˆ C = âˆ Q = 80Â°

âˆ†ABC ~ âˆ†PQR (By SAS)

(iv)In âˆ†DEF and âˆ†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

âˆ†DEF ~ âˆ†PQR (By SSS)

(v) In âˆ†ABC and âˆ†MNR

âˆ A = 80Â°, âˆ C = 70Â°

So, âˆ B = 180Â° â€“ (80Â° + 70Â°) = 30Â°

âˆ M = 80Â°, âˆ N = 30Â°, and âˆ R = 180Â° â€“ (80Â° + 30Â°) = 70Â°

Now, in âˆ†ABC

âˆ A = âˆ M â€“ 80Â°, âˆ B = âˆ N = 30Â°

and âˆ C = âˆ R = 70Â°

âˆ†ABC ~ âˆ†MNR (By AAA or AA)

Question 2: In the given figure, âˆ†ODC ~ âˆ†OBA, âˆ BOC = 115Â° and âˆ CDO = 70Â°. Find:

(i) âˆ DOC

(ii) âˆ DCO

(iii) âˆ OAB

(iv) âˆ OBA

Solution:

Here âˆ†ODC ~ âˆ†OBA, so

âˆ D = âˆ B = 70Â°

âˆ C = âˆ A

âˆ COD = âˆ AOB

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

âˆ DOC + 115Â°= 180Â°

âˆ DOC = 180Â° â€“ 115Â° = 65Â°

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

65Â° + 70Â° + âˆ DCO = 180Â°

135Â° + âˆ DCO = 180Â°

âˆ DCO = 180Â° â€“ 135Â° = 45Â°

(iii) âˆ AOB = âˆ DOC = 65Â° (vertically opposite angles)

âˆ OAB = âˆ DCO = 45Â° (Since âˆ†ODC ~ âˆ†OBA)

(iv) âˆ OBA = âˆ CDO = 70Â° (Since âˆ†ODC ~ âˆ†OBA)

Question 3: In the given figure, âˆ†OAB ~ âˆ†OCD. If AB = 8 cm, BO = 6.4 cm, OC = 3.5 cm and CD = 5 cm, find (i) OA (ii) DO.

Solution:

Since âˆ†OAB ~ âˆ†OCD

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

Let OD = y and OA = x

Question 4: In the given figure, if âˆ ADE = âˆ B, show that âˆ†ADE ~ âˆ†ABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE.

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

âˆ ADE = âˆ B (given)

âˆ A = âˆ A (common)

Again,

Question 5: The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ = 12 cm, find AB.

Solution:

Form given statement: âˆ†ABC ~ âˆ†PQR,

PQ = 12 cm

Perimeter of âˆ†ABC = AB + BC + CA = 32 cm

Perimeter of âˆ†PQR = PQ + QR + RP = 24 cm

Now,

Question 6: The corresponding sides of two similar triangles ABC and DEF are BC = 9.1 cm and EF = 6.5 cm. If the perimeter of âˆ†DEF is 25 cm, find the perimeter of âˆ†ABC.

Solution:

Form given statement: âˆ†ABC ~ âˆ†DEF

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

Perimeter of âˆ†ABC is 35 cm

Question 7: In the given figure, âˆ CAB = 90Â° and AD âŠ¥ BC. Show that âˆ†BDA ~ âˆ†BAC. If AC = 75 cm, AB = 1 m and BC = 1.25 m, find AD.

Solution:

âˆ CAB = 90Â° and AD âŠ¥ BC

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

âˆ BDA = âˆ BAC = 90Â°

âˆ DBA = âˆ CBA [common angles]

By AA

âˆ†BDA ~ âˆ†BAC

And,

Question 8: In the given figure, âˆ ABC = 90Â° and BD âŠ¥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

Solution:

From given:

âˆ ABC = 90Â°, BD âŠ¥ AC.

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

In âˆ†ABC and âˆ†BDC,

âˆ ABC = âˆ BDC (each 90Â°)

âˆ BCA = âˆ BCD (common angles)

âˆ†ABC ~ âˆ†BDC (AA axiom)

So, corresponding sides are proportional

AB/BD = BC/CD

=> 5.7/3.8 = BC/5.4

=> BC = 8.1

#### Exercise 7C

Question 1: âˆ†ABC ~ âˆ†DEF and their areas are respectively 64 cmÂ² and 121 cmÂ². If EF = 15.4 cm, find BC.

Solution:

Area of âˆ†ABC = 64 cmÂ² and

area of âˆ†DEF = 121 cmÂ²

EF = 15.4 cm

Question 2: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

Solution:

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

BC = 4.5 cm

Question 3: âˆ†ABC ~ âˆ†PQR and ar (âˆ†ABC) = 4ar (âˆ†PQR). If BC = 12 cm, find QR.

Solution:

âˆ†ABC ~ âˆ†PQR

ar (âˆ†ABC) = 4ar (âˆ†PQR)

Question 4: The areas of two similar triangles are 169 cmÂ² and 121 cmÂ² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

Solution:

Areas of two similar triangles are 169 cmÂ² and 121 cmÂ² (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: âˆ†ABC ~ âˆ†DEF and their areas are respectively 100 cmÂ² and 49 cmÂ². If the altitude of âˆ†ABC is 5 cm, find the corresponding altitude of âˆ†DEF.

Solution:

Area of âˆ†ABC = 100 cmÂ²

area of âˆ†DEF = 49 cmÂ²

Altitude of âˆ†ABC is 5 cm

AL âŠ¥ BC and DM âŠ¥ EF

Let DM = x cm

Or x = 3.5

Altitude of smaller triangle is 3.5 cm

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

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)Â² : (9)Â² = 36 : 81 = 4 : 9

Question 7: The areas of two similar triangles are 81 cmÂ² and 49 cmÂ² respectively. If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the other.

Solution:

Areas of two similar triangles are 81 cmÂ² and 49 cmÂ²

Altitude of the first triangle = 6.3 cm

Let altitude of second triangle = x cm

Area of âˆ†ABC = 81 cmÂ² and area of âˆ†DEF = 49cmÂ²

Altitude AL = 6 â€“ 3 cm

Let altitude DM = x cm

Altitude of second triangle is 4.9 cm

Question 8: The areas of two similar triangles are 100 cmÂ² and 64 cmÂ² respectively. If a median of the smaller triangle is 5.6 cm, find the corresponding median of the other.

Solution:

Areas of two similar triangles are 100 cmÂ² and 64 cmÂ²

Median DM of âˆ†DEF = 5.6 cm

Let median AL of âˆ†ABC = x

Corresponding median of the other triangle is 7 cm.

Question 9: In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of âˆ†APQ is 1/16 of the area of âˆ†ABC.

Solution:

In âˆ†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 âˆ†APQ and âˆ†ABC,

AP/AB = AQ/AC

angle A (common)

âˆ†APQ and âˆ†ABC are similar triangles.

Now,

Which implies,

area of âˆ†APQ = 1/16 of the area of âˆ†ABC

Hence Proved.

Question 10: In the given figure, DE || BC. If DE = 3 cm, BC = 6 cm and ar (âˆ†ADE) = 15 cmÂ², find the area of âˆ†ABC.

Solution:

DE || BC

DE = 3 cm, BC = 6 cm

Now,

In âˆ†ABC

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

Area of âˆ†ABC is 60 cm2.

#### Exercise 7D

Question 1: The sides of certain triangles are given below. Determine which of them are right triangles.

(i) 9 cm, 16 cm, 18 cm

(ii) 1 cm, 24 cm, 25 cm

(iii) 1.4 cm, 4.8 cm, 5 cm

(iv) 1.6 cm, 3.8 cm, 4 cm

(v) (a â€“ 1) cm, 2âˆša cm, (a + 1) cm

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)Â² = 324

and (9)Â² + (16)Â² = 81 + 256 = 337

324 â‰  337

It is not a right triangle.

(ii) 1 cm, 24 cm, 25 cm

Longest side = 25 cm

(25)Â² = 625

and (7)Â² x (24)Â² = 49 + 576 = 625

625 = 625

It is a right triangle.

(iii) 1.4 cm, 4.8 cm, 5 cm

Longest side = 5 cm

(5)Â² = 25

and (1.4)Â² + (4.8)Â² = 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 )Â² = 16

and (1.6)Â² + (3.8)Â² = 2.56 + 14.44 = 17.00 = 17

16 â‰  17

It is not a right triangle.

(v) (a- 1) cm, 2âˆša cm, (a + 1) cm

Longest side = (a + 1) cm

(a + 1)Â² = aÂ² + 2a + 1

and (a â€“ 1)Â² + (2 âˆša )Â² = aÂ² â€“ 2a + 1 + 4a = aÂ² + 2a + 1

aÂ² + 2a + 1 = aÂ² + 2a + 1

It is a right triangle.

Question 2: A man goes 80 m due east and then 150 m due north. How far is he from the starting point?

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 âˆ†OAB,

By Pythagoras Theorem:

OB2 = OA2+ AB2

= (80)Â² + (150)Â²

= 6400 + 22500

= 28900

or OB = âˆš28900 = 170

Man is 170 m away from the starting point.

Question 3: A man goes 10 m due south and then 24 m due west. How far is he from the starting point?

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 âˆ†OAB,

By Pythagoras Theorem:

OB2 = OA2+ AB2

= (10)Â² + (24)Â²

= 676

or OB = 26

Man is 26 m away from the starting point.

Question 4: A 13-m-long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.

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)Â² = (12)Â² + xÂ²

169 = 144 + xÂ²

xÂ² = 169 â€“ 144 = 25

or x = 5

Distance between foot of ladder and building = 5 m.

Question 5: A ladder is placed in such a way that its foot is at a distance of 15 m from a wall and its top reaches a window 20 m above the ground. Find the length of the ladder.

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

xÂ² = 20Â² + 15Â²

= 400 + 225

= 625

or x = 25

Length of ladder is 25 m

Question 6: Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

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 â€“ ED = CD â€“ AB = 14-9 = 5 m

In right âˆ†AEC,

ACÂ² = AEÂ² + CEÂ²

= 122 + 52

= 144 + 25

= 169

or AC = 13

Distance between pole’s tops is 13 m

Question 7: 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:

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 âˆ†ABC,

By Pythagoras Theorem:

AC2 = AB2 + BC2

(24)Â² = (18)Â² + xÂ²

576 = 324 + xÂ²

xÂ² = 576 â€“ 324 = 252

or x = 6v7

BC is 6v7m

Question 8: In the given figure, O is a point inside a âˆ†PQR such that âˆ POR = 90Â°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that âˆ†PQR is right-angled.

Solution:

In âˆ†PQR, O is a point in it such that

OP = 6 cm, OR = 8 cm and âˆ POR = 90Â°

PQ = 24 cm, QR = 26 cm

Now,

In âˆ†POR, âˆ O = 90Â°

PRÂ² = POÂ² + ORÂ²

= (6)Â² + (8)Â²

= 36 + 64

= 100

PR = 10

Greatest side QR is 26 cm

QRÂ² = (26)Â² = 676

and PQÂ² + PRÂ² = (24)Â² + (10)Â²

= 576 + 100

= 676

Which implies, 676 = 676

QRÂ² = PQÂ² + PRÂ²

âˆ†PQR is a right angled triangle and right angle at P.

Question 9: âˆ†ABC is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.

Solution:

In isosceles âˆ†ABC, AB = AC = 13 cm

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

Now, in right âˆ†ALB

AB2 = AL2 + BL2

(13)Â² = (5)Â² + BLÂ²

169 = 25 + BLÂ²

BLÂ² = 169 â€“ 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: Find the length of altitude AD of an isosceles âˆ†ABC in which AB = AC = 2a units and BC = a units.

Solution:

In an isosceles âˆ†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 7E

Question 1: State the two properties which are necessary for given two triangles to be similar.

Solution:

Two properties for similarity of two triangles are:

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

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

Question 2: State the basic proportionality theorem.

Solution:

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

Question 3: State the converse of Thalesâ€™ theorem.

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: State the midpoint theorem.

Solution:

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

Question 5: State the AAA-similarity criterion.

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: State the AA-similarity criterion.

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: State the SSS-criterion for similarity of triangles.

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: State the SAS-similarity criterion.

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: State Pythagorasâ€™ theorem.

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: State the converse of Pythagoras theorem.

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: If D, E and F are respectively the midpoints of sides AB, BC and CA of âˆ†ABC then what is the ratio of the areas of âˆ†DEF and âˆ†ABC?

Solution:

The ratio of their areas will be 1 : 4.

Question 12: Two triangles ABC and PQR are such that AB = 3 cm, AC = 6 cm, âˆ A = 70Â°, PR = 9 cm, âˆ P = 70Â° and PQ = 4.5 cm. Show that âˆ†ABC ~ âˆ†PQR and state the similarity criterion.

Solution:

In two triangles âˆ†ABC and âˆ†PQR,

AB = 3 cm, AC = 6 cm, âˆ A = 70Â°

PR = 9 cm, âˆ P = 70Â° and PQ= 4.5 cm

Now,

âˆ A = âˆ P = 70Â° (Same)

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

AB/PQ = 3/4.5 = 2/3

=> AC/PR = AB/PQ

Both âˆ†ABC and âˆ†PQR are similar.

Question 13: If âˆ†ABC ~ âˆ†DEF such that 2AB = DE and BC = 6 cm, find EF.

Solution:

âˆ†ABC ~ âˆ†DEF (given)

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

âˆ E = âˆ B and âˆ D = âˆ A and âˆ F = âˆ C

2AB = DE

=> AB/DE = 1/2

Therefore,

AB/DE = BC/EF

1/2 = 6/EF

or EF = 12 cm

Question 14: In the given figure, DE || BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Solution:

From figure:

DE || BC

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

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

In âˆ†ABC

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

3x2 + 19x – 3x2– 9x – 4x = 12

x = 2

Question 15: A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution:

Let AB is the ladder and A is window.

Then, AB = 10 m and AC = 8 m

Let BC = x

In right âˆ†ABC,

By Pythagoras Theorem:

AB2 = AC2 + BC2

(10)Â² = 8Â² + xÂ²

100 = 64 + xÂ²

xÂ² = 100 â€“ 64 = 36

or x = 6

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

## R S Aggarwal Solutions For Class 10 Maths Chapter 7 Exercises:

Get detailed solutions for all the questions listed under the below exercises:

Exercise 7 A Solutions

Exercise 7 B Solutions

Exercise 7 C Solutions

Exercise 7 D Solutions

Exercise 7 E Solutions

## R S Aggarwal Solutions for Class 10 Maths Chapter 7 Triangles

In this chapter students will study important concepts of triangles as listed below:

• Triangles introduction
• Congruent figures
• Similar figures
• Similar Polygons
• Equiangular triangles
• Similar Triangles
• Important Theorems on Triangles:
• Thales’ Theorem and it’s converse
• Midpoint theorem
• Angle-Bisector Theorem
• Criteria for similarity of two triangles
• Ratio of the Areas of two similar triangles
• Pythagoras Theorem
• Similar Triangles
• Thales’ Theorem and it’s converse
• Midpoint theorem

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 7 Triangles

1. Students will learn how to solve problems on triangles with the help of various solved problems.

2. Easy and simple language is used.

3. All questions are solved using step-by-step problem solving approach.