# RS Aggarwal Solutions Chapter 4 Triangles Exercise - 4B Class 10 Maths

 Chapter Name RS Aggarwal Chapter 4 Triangles Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 4AExercise 4CExercise 4DExercise 4E Related Study NCERT Solutions for Class 10 Maths

### Exercise 4B Solutions

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

(i)

(ii)
(iii)
(iv)
(v)

(i) We have:°

∠BAC = ∠PQR = 50°

∠ABC = ∠QPR = 60°

∠ACB = ∠PRQ = 70°

Therefore, by AAA similarity theorem, ∆ABC – QPR

(ii) We have:

AB/DF = 3/6 = 1/2 and BC/DE = 4.5/9 = 1/2

But, ∠ABC ≠ ∠EDF (Included angles are not equal)

Thus, this triangles are not similar.

(iii) We have:

CA/QR = 8/6 = 4/3 and CB/PQ = 6/4.5 = 4/3

⇒ CA/QR = CB/PQ

Also, ∠ACB = ∠PQR = 80°

Therefore, by SAS similarity theorem, ∆ACB - ∆RQP.

(iv) We have

DE/QR = 2.5/5 = 1/2

EF/PQ = 2/4 = 1/2

DF/PR = 3/6 = 1/2

⇒ DE/QR = EF/PQ = DF/PR

Therefore, by SSS similarity theorem, ∆FED - ∆PQR

(v) In ∆ ABC,

∠A + ∠B + ∠C = 180° (Angle Sum Property)

⇒ 80° + ∠B + 70° = 180°

⇒ ∠B = 30°

∠A = ∠M and ∠B = ∠N

Therefore, by AA similarity, ∆ABC - ∆MNR

2. In the given figure, ∆ODC~∆OBA, ∠BOC = 115° and ∠CDO = 70°

Find (i) ∠DCO (ii) ∠DCO (iii) ∠OAB (iv) ∠OBA.

Solution

(i) It is given that DB is a straight line.

Therefore,

∠DOC + ∠COB = 180°

∠DOC = 180° − 115° = 65°

(ii) In ∆ DOC, we have:

∠ODC + ∠DCO + ∠DOC = 180°

Therefore,

70° + ∠DCO + 65° = 180°

∠DCO = 180° − 70° − 65° = 45°

(iii) It is given that ∆ODC - ∆OBA

Therefore,

∠OAB = ∠OCD = 45°

(iv) Again, ∆ODC - ∆OBA

Therefore,

∠OBA = ∠ODC = 70°

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

Solution

(i) Let OA be X cm.

∵ ∆OAB - ∆OCD

∴ OA/OC = AB/CD

⇒ x/3.5 = 8/5

⇒ x = (8 × 3.5)/5 = 5.6

Hence, OA = 5.6 cm

(ii) Let OD be Y cm

∵ ∆ OAB - ∆ OCD

∴ AB/CD = OB/OD

⇒ 8/5 = 6.4/y

⇒ y = (6.4 × 5)/8 = 4

Hence, DO = 4 cm

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

Solution

Given :

∠ADE = ∠ABC and ∠A = ∠A

Let DE be X cm

Therefore, by AA similarity theorem, ∆ADE - ∆ABC

⇒ 3.8/(3.6 + 2.1) = x/4.2

⇒ x = (3.8 × 4.2)/5.7 = 2.8

Hence, DE = 2.8 cm

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

Solution

It is given that triangles ABC and PQR are similar.

Therefore,

Perimeter(∆ABC)/Perimeter of (∆PQR) = AB/PQ

⇒ 32/24 = AB/12

⇒ AB = (32 × 12)/24 = 16 cm

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

Solution

It is given that ∆ABC - ∆DEF.

Therefore, their corresponding sides will be proportional.

Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.

⇒ Perimeter of ∆ABC/Perimeter of ∆DEF = BC/EF

Let the perimeter of ∆ABC be X cm

Therefore,

x/25 = 9.1/6.5

⇒ x = (9.1 × 25)/6.5

Thus, the perimeter of ∆ABC is 35 cm.

7. In the given figure, ∠CAB = 90° and AD⊥BC. Show that ∆BDA ~ ∆BAC. If AC = 75cm, AB = 1m and BC = 1.25m, find AD.

Solution

In ∆ BDA and ∆ BAC, we have:

∠BDA = ∠BAC = 90°

∠DBA = ∠CBA (Common)

Therefore, by AA similarity theorem, ∆BDA - ∆BAC

= 0.6 m or 60 cm

8. In the given figure, ∠ABC = 90° and BD⊥AC.

If AB = 5.7cm, BD = 3.8cm and CD = 5.4cm, find BC.

Solution

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

In ∆ BDC and ∆ ABC, we have :

∠ABC = ∠BBC = 90° (Given)

∠C = ∠C (Common)

By AA similarity theorem, we get:

∆ BDC- ∆ ABC

AB/BD = BC/DC

⇒ 5.7/3.8 = BC/5.4

⇒ BC = (5.7/3.8 × 5.4)

= 8.1

Hence, BC = 8.1 cm

9. In the given figure, ABC = 90° and BDAC.

If BD = 8cm, AD = 4cm, find CD.

Solution

It is given that ABC is a right angled triangle

and BD is the altitude drawn from the right angle to the hypotenuse.

In ∆ DBA and ∆ DCB, we have:

∠BDA = ∠CDB

∠DBA = ∠DCB = 90°

Therefore, by AA similarity theorem, we get:

∆DBA - ∆DCB

CD = (8 × 8)/4 = 16 cm

10. P and Q are points on the sides AB and AC respectively of a ∆ABC. If AP = 2cm, PB = 4cm, AQ = 3cm and QC = 6cm, show that BC = 3PQ.

Solution

We have:

AP/AB = 2/6 = 1/3 and AQ/AC = 3/9 = 1/3

⇒ AP/AB = AQ/AC

In ∆APQ and ∆ABC, we have:

AP/AB = AQ/AC

∠A = ∠A

Therefore, by AA similarity theorem, we get: ∆APQ - ∆ABC

Hence, PQ/BC = AQ/AC = 1/3

⇒ PQ/BC = 1/3

BC = 3PQ

This completes the proof.

11. ABCD is parallelogram and E is a point on BC.

If the diagonal BD intersects AE at F, prove that

AF × FB = EF × FD.

Solution

We have:

∠AFD = ∠EFB (Vertically Opposite angles)

∵ DA || BC

∴ ∠DAF = ∠BEF (Alternate angles)

∆ DAF ~ ∆ BEF (AA similarity theorem)

⇒ AF/EF = FD/FB

Or, AF × FB = FD × EF

This completes the proof.

12. In the given figure, DBBC, DEAB and ACBC.

Prove that BE/DE = AC/BC

Solution

In ∆BED and ∆ACB, we have:

∠BED = ∠ACB = 90°

∵ ∠B + ∠C = 180°

∴ BD || AC

∠EBD = ∠CAB (Alternate angles)

Therefore, by AA similarity theorem, we get:

∆BED ~ ∆ACB

⇒ BE/AC = DE/BC

⇒ BE/DE = AC/BC

This completes the proof.

13. A vertical pole of length 7.5 cm casts a shadow 5m long on the ground and at the same time a tower casts a shadow 24m long. Find the height of the tower.

Solution

Let AB be the vertical stick and BC be its shadow.

Given:

AB = 7.5 m, BC = 5 m

Let PQ be the tower and QR be its shadow.

Given:

QR = 24 m

Let the length of PQ be x m.

In ∆ ABC and ∆ PQR, we have:

∠ABC = ∠PQR = 90°

∠ACB = ∠PRQ (Angular elevation of the Sun at the same time)

Therefore, by AA similarity theorem, we get:

∆ABC ~ ∆PQR

⇒ AB/BC = PQ/QR

⇒ 7.5/5 = x/24

x = 7.5/5 × 24 = 36 cm

Therefore, PQ = 36 m

Hence, the height of the tower is 36 m.

14. In an isosceles ∆ABC, the base AB is produced both ways in P and Q such that AP × BQ = AC2.

Prove that ∆ACP ~ ∆BCQ.

Solution

Disclaimer: It should be ∆APC ~ ∆BCQ instead of ∆ACP ~ ∆BCQ

It is given that ∆ABC is an isosceles triangle.

Therefore,

CA = CB

⇒ ∠CAB = ∠CBA

⇒ 180° - ∠CAB = 180° − ∠CBA

⇒ ∠CAP = ∠CBQ

Also,

AP × BQ = AC2

⇒ AP/AC = AC/BQ

⇒ AP/AC = BC/BQ (∵ AC = BC)

Thus, by SAS similarity theorem, we get

∆APC ~ ∆BCQ

This completes the proof.

15. In the given figure, ∠1 = ∠2 and AC/BD = CB/CE.

Prove that ∆ACB ~ ∆DCE.

Solution

We have:

∠1 = ∠2

Also, ∠1 = ∠2

i.e., ∠DBC = ∠ACB

Therefore, by SAS similarity theorem, we get :

∆ACB - ∆DCE

If P, Q, R, S be the midpoints of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Solution

In ∆ ABC, P and Q are mid points of AB and AC respectively.

So, PQ || BC and PQ = 1/2BC …(1)

QR = 1/2 AD = 1/2 BC …(2)

Now in ∆BCD,

SR = 1/2BC ...(3)

Similarly in ∆ABD,

PS = 1/2AD = 1/2BC ...(4)

Using (1), (2), (3), and (4).

PQ = QR = SR = PS

Since, all sides are equal

Hence, PQRS is a rhombus.

17. In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that

(a) ΔPAC ~ ΔPDB

(b) PA.PB = PC.PD

Solution

Given: AB and CD are two chords

To Prove:

(a) ∆ PAC ~ ∆PDB

(b) PA.PB = PC.PD

Proof: In ∆ PAC and ∆ PDB

∠APC = ∠DPB (Vertically Opposite angles)

∠CAP = ∠BDP (Angles in the same segment are equal)

By AA similarity criterion ∆PAC ~ ∆PDB

When two triangles are similar, then the ratios of lengths of their corresponding sides are proportional.

∴ PA/PD = PC/PB

⇒ PA. PB = PC. PD

18. Two chords AB and CD of a circle intersect at a point P outside the circle. Prove that:

(i) ∆ PAC ~ ∆ PDB

(ii) PA. PB = PC.PD

Solution

Given : AB and CD are two chords

To Prove:

(a) ∆PAC - ∆PDB

(b) PA. PB = PC.PD

Proof: ∠ABD + ∠ACD = 180°...(1) (Opposite angles of a cyclic quadrilateral are supplementary)

∠PCA + ∠ACD = 180°...(2) (Linear Pair Angles)

Using (1) and (2), we get

∠ABD + ∠PCA

∠A = ∠A  (Common)

By AA similarity-criterion ∆PAC - ∆PDB

When two triangles are similar, then the rations of the lengths of their corresponding sides are proportional.

∴ PA/PD = PC/PB

⇒ PA.PB = PC.PD

19. In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD AC, if DP AB and DQ BC then prove that

(a) DQ2 = DP.QC

(b) DP2 = DQ AP2

Solution

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other.

(a) Now using the same property in In ∆BDC, we get

∆CQD ~ ∆DQB

CQ/DQ = DQ/QB

⇒ DQ2 = QB.CQ

Now, Since all the angles in quadrilateral BQDP are right angles.

Hence, BQDP is a rectangle.

So, QB = DP and DQ = PB

∴ DQ2 = DP.CQ

(b) Similarly, ∆APD ~ ∆DPB

AP/DP = PD/PB

⇒ DP2 = AP. PB

⇒ DP2 = AP. DQ [∵ DQ = PB]