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Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.5 | Set 1
  • Last Updated : 05 Apr, 2021

Question 1. Sides of triangles are given below. Determine which of them are right triangles? In the case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

Solution:

(i) Given: 7cm, 24cm, 25cm

(25)2= 25 * 25 = 625

∴The are sides of a right ∆ and hypotenuse = 625  

(ii) Given: 3cm, 8cm, 6cm



(8)2 = 8 * 8 = 64  

They are not side of right triangle.

(iii) Given: 50cm, 80cm, 100cm

(100)2 = 100 * 100 = 10000

(50)2 + (80)2 = 2500 + 6400

Therefore, this is not right triangle.

(iv) Given: 13cm, 12cm, 5cm

(13)2 = 13 * 13 = 169

(12)2 + (5)2 = 169

Therefore, this is the sides of right triangle and hypotenuse = 169

Question 2. PQR is a triangle right-angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.

Solution:

Given: A right ∆PQR right-angled at P and PM⊥QR

To show: PM2 = QM * MR

PM * PM = QM * MR

In ∆PMQ and ∆PMR

∠1 =∠2           -(each 90°)

∠1 +∠2 +∠3 = 180°

90° + ∠4 + ∠5 = 180°



∠4 +∠5 = 180° – 90°

∠3 + ∠4 = 180° – 90°

∠4 +∠5 = ∠3 +∠4

∴∠5 =∠3  

∴∆PMQ ~ ∆RMP  

\frac{PM}{RM} = \frac{MQ}{MP} = \frac{QP}{PR}     

\frac{PM}{RM} = \frac{MQ}{MP}

MP2 = MQ.PM

Question 3. In Fig., ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB2 = BC × BD

(ii) AC2 = BC × DC

(iii) AD2 = BD × CD

Solution:

Given: A right ∆ABD, right angled at A and AC⊥ BD

To show:  

(i)AB2 = BC × BD

(ii) AC2 = BC × DC

(iii) AD2 = BD × CD

(i) AB2 = BC × BD

AB * AB = BC.BD

In ∆ ABC and ∆ABD

∠B =∠B          -(common)

∠BCA = ∠A          -(each 90°)

∴ ∆ABC~∆DBA  

\frac{AB}{DB} = \frac{BC}{BA} = \frac{CA}{AD}

AB2 = BC * DC

(ii) AC2 = BC * DC

AC * AC = BC * DC

In ∆ABC and ∆ACD

∠ACB = ∠ACD          -(each 90°)

In ∆ ACB

∠1 + ∠2 + ∠3 = 180°

∠1 + ∠2 + 90° = 180°

∠1 + ∠2 = 90°           -(1)

∠1 + ∠3 = 90°          -(2)

∠1 + ∠2 =∠1 + ∠3

∴ ∠2 = ∠3  

∴ ∆ABC~∆DAC  

\frac{AB}{DA} = \frac{BC}{AC} = \frac{CA}{CD}

AC2 = BC * CD

(iii) AD2 = BD × CD

In ∆ABD and ∆ACD

∠A =∠ACD          -(each 90)

∠D =∠D          -(common)

∴ ∆ABD~∆CAD          -(AA similarity)  

\frac{AB}{CA} = \frac{BD}{AD} = \frac{DA}{DC}

AD2 = BD * CD

Question 4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.

Solution:

Given: ∆ABC right angled at B

To prove: AB2 = 2AC2

According to Pythagoras theorem,

(AB)2 = (AC)2 + (BC)2

(AB)2 = (AC)2 + (AC)2           -(BC = AC)

AB2 = 2AC2

Question 5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

Solution:

Given: ∆ABC is an isosceles triangle in which  

AC = BC

AB2 = 2AC2

To prove: ABCD is right ∆  

AB2 = 2AC2

AB2 = AC2 + AC2

∴ By the, converse of Pythagoras theorem ∆ACB is a right ∆ at C.  

Question 6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Solution:

Let ABC is an equilateral triangle of each 2a units

Construction: Draw AD ⊥ BC

In right ∆ ADB

(AB)2 = (AD)2 + (BD)2

(2a)2 = (AD)2 + (a)2

(4a)2 = (AD)2 + a2

4a2 – a2 = (AD)2

3a2 = (AB)2

√3 a2 = AD               ∴ Each of its altitude= √3 a unit

√3 a = AD  

Question 7. Prove that the sum of the squares of the sides of the rhombus is equal to the sum of the squares of its diagonals.

Solution:

Given: ABCD is a rhombus

To prove: AB2 + BC2 + CD2 + DA2

Poof: ABCD is rhombus and let diagonals AC and BD bisect each other at O.

∴ ∠AOB =∠BOC =∠COD =∠DOA = 90°  

In ∆AOB  

AB2 = AC2 + BO2

AB2 = (1/2AC2) + (1/2BD2)

AB2 = 1/4(AC2 + BD2)

4AB2 = AC2 + BD2          -(1)

Similarly, in ∆BOC  

4BC2 = AC2 + BD2          -(2)

4CD2 = AC2 + BD2          -(3)

4AD2 = AC2 + BD2          -(4)

Adding 1, 2, 3, and 4

4AB2 + 4BC2 + 4CD2 + 4DA2 = 4(AC2 + BD2)

4(AB2 + BC2 + CD2 + DA2) = 4(AC2 + BD2)

∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD 

Question 8. In Fig. 6.54, O is a point in the interior of a triangle diagonals.

(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Solution: 

(i) In right ∆AFO, by Pythagoras theorem  

OA2 = AP2 + OF2          -(1) 

In right ∆ODB, by Pythagoras theorem

OB2 = BD2 + OD2          -(2)          

In right ∆OEC, by Pythagoras theorem

OC2 = CE2 + OE2          -(3)   

Adding 1, 2, and 3

OA2 + OB2 + OC2 = AF2 + BD2 + CF2 + OD2 + OE2

OA2 + OB2 + OC2 – OE2 – OD2 – OE2 = AF2 + BD2 + CE2

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2

OB2 = BD2 + OD2          -(1) 

OC2 = DC2 + OD2          -(2) 

Subtracting (1) from (2) we get,

OB2 – OC2 = BD2 + OD2 – (DC2 + OD2)

= BD2 + OD2 – DC2 – OD2

= BD2 – DC2          -(3)      

Similarly,  

OC2 – AD2 = CE2 – EA2          -(4) 

AO2 = BO2 = AF2 – FB2          -(5) 

Adding 3, 4, and 5, we get,

OB2 – OC2 + OC2 – AO2 + AO2 – BO2 = BD2 – DC2 + CE2 – EA2 + AF2 – FB2

0 + DC2 + EA2 + FB2 = BD2 + CD2 + AF2

Question 9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution:

AB is wall  = 8m

AC is ladder = 10m

BC = ?

In right ∆ABC, by Pythagoras theorem

(AC)2 = (AB)2 + (BC)2

(10)2 = (8)2 + (BC)2

100 = 64 + (BC)2

100 – 64 = (BC)2

36 = (BC)2

√36 = BC  

√(6 * 6) = BC

6 = BC  

Hence, the foot of the ladder is at a distance of 6 m from the base of the wall.

Question 10. 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:

In fig.  

AB is pole = 18m

AC is wire = 24m

BC = ?

In right ∆ABC, by Pythagoras theorem

(AC)2 = (AB)2 + (BC)2

(24)2 = (18)2 + (BC)2

576 = 324 + (BC)2

576 – 324 = (BC)2

252 = (BC)2

√252 = BC  

√(2 * 2 * 3 * 3 * 7) = BC 

2 * 3√7 = BC

6√7 = BC

∴ Hence, the stake may be placed at distance of 6√7 from the base of the pole.

Chapter 6 Triangles – Exercise 6.5 | Set 2

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