# Ncert Solutions For Class 10 Maths Ex 6.5

## Ncert Solutions For Class 10 Maths Chapter 6 Ex 6.5

Question 1:

Sides of the triangles are as follows:

1. i) 7 cm, 25 cm, 24 cm
2. ii) 3 cm, 6 cm, 8 cm

iii)           50 cm, 100 cm, 80 cm

1. iv) 5 cm, 12 cm, 13 cm

Determine which of them are right-angled triangles. Write the length of its hypotenuse.

Solution:

1. i) Sides of the triangle given are 7 cm, 25 cm and 24 cm.

Squaring the length of these sides we get 49 cm, 625 cm and 576 cm.

However, 49 + 576 = 625

(7)2 + (24)2 = (25)2

It satisfies the Pythagoras theorem. Hence, it is a right-angled triangle.

Length of Hypotenuse = 25 cm.

1. ii) Sides of the triangle given are 3 cm, 6 cm and 8 cm.

Squaring the length of these sides we get 9 cm, 36 cm and 64 cm.

However, 9 + 36 ≠ 64

(7)2 + (24)2 ≠ (25)2

It does not satisfy the Pythagoras theorem. Hence, it is not a right-angled triangle.

iii)           Sides of the triangle given are 50 cm, 100 cm and 80 cm.

Squaring the length of these sides we get 2500 cm, 10000 cm and 6400 cm.

However, 2500 + 6400 ≠ 10000

(50)2 + (80)2 ≠ (100)2

It does not the Pythagoras theorem. Hence, it is not a right-angled triangle.

1. iv) Sides of the triangle given are 5 cm, 12 cm and 13 cm.

Squaring the length of these sides we get 25 cm, 144 cm and 169 cm.

However, 25 + 144 = 169

(5)2 + (12)2 = (13)2

It satisfies the Pythagoras theorem. Hence, it is a right-angled triangle.

Length of Hypotenuse = 13 cm.

Question 2:

ABC is a right-angled triangle at point A. P is a point on BC such that AP ⊥ BC. Prove that AP2 = BP x PC.

Solution:

Given: ΔABC is a right-angled triangle at point A. P is appointed on BC such that AP ⊥ BC.

To prove: AP2 = BP x PC

Proof:

In ΔABC, we have

AB2 = AP2 + BP2 [By Pythagoras theorem]

Or, AP2 = AB2 – BP2 ……….. (i)

In ΔAPC, we have

AC2 = AP2 + PC2 [By Pythagoras theorem]

Or, AP2 = AC2 – PC2 ……….. (ii)

Adding (i) and (ii), we get

2AP2 = (AB2 – BP2) – (AC2 – PC2)

= BC2 – BP2 – PC2   [ BC2 = AB2 + AC2]

= (BP + PC)2 – BP2 – PC2

= 2BP x PC

∴ AP2 = BP x PC

Question 3:

In the given figure, PQM is a right-angled triangle at P. Also PR ⊥ QM.

Show that

i) PQ2 = QR x QM

ii) PR2 = QR x MR

Solution:

i) In ΔPMQ and ΔRPQ, we have

∠MPQ = ∠PRQ = 90o

∠PQM = ∠RQP (Common angle)

∴ΔPMQ ~ ΔRPQ [AA similarity criterion]

⇒ PQ/RQ = QM/PQ

⇒ PQ2 = QR x QM

ii) Let ∠RPQ = x

In ΔRQP,

∠RQP = 180o – 90o – x

∠RQP = 90o – x

Similarly, in ΔRPM

∠RPM = 90o – ∠RQP

= 90o – x

∠RMP = 180o – 90o – (90o – x)

∠RMP = x

In ΔRQP and ΔRPM, we have

∠RQP = ∠RPM

∠RPQ = ∠RMP

∠PRQ = ∠MRP = 90o

∴ ΔRQP ~ ΔRPM [By AAA similarity criterion]

⇒PR/MR = QR/PR

⇒PR2 = QR x MR

Question 4:

PQR is an isosceles right-angled triangle at point R. Prove that PQ2 = 2PR2.

Solution:

Given: ΔPQR is an isosceles triangle right angled at R.

In ΔPRQ, ∠R = 90o

PR = QR (Given)

PQ2 = PR2 +QR2 [By Pythagoras theorem]

= PR2 + PR2 [Since, PR = QR]

PQ2 = 2PR2

Question 5:

PQR is an isosceles triangle with PR = QR. Given that PQ2 = 2PR2. Prove that PQR is a right-angled triangle.

Solution:

Given that ΔPQR is an isosceles triangle having PR = QR and PQ2 = 2PR2

In ΔPRQ,

PR = QR (Given)

PQ2 = 2PR2 (Given)

= PR2 + PR2

= PR2 + QR2 [Since, PR = QR]

Hence, by Pythagoras theorem ΔPQR is a right-angle triangle.

Question 6:

PQR is an equilateral triangle of side 2a. Find each of its altitudes.

Solution:

Given: PQR is an equilateral triangle of side 2a.

Draw PS ⊥ QR

In ΔPSQ and ΔPSR, we have

PQ = PR [Given]

PS = PS [Given]

∠PSQ = ∠PSR = 90o

Therefore, ΔPSQ ≅ ΔPSR by RHS congruence.

In right-angled ΔPSQ,

(PQ)2 = (PS)2 + (QD)2

(2a)2 = (PS)2 + a2

⇒ (PS)2 = 4a2 – a2

⇒ (PS)2 = 3a2

⇒ PS = 3a$\sqrt{3}a$

Question 7:

Sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Prove it.

Solution:

Given: PQRS is a rhombus whose diagonals are PR and QS. They intersect at O.

To prove: PQ2 + QR2 + RS2+ PS2 = PR2 + QS2

Since the diagonals of as rhombus bisect each other ar tight angles.

Therefore, PO = RO and QO = SO

In ΔPOQ,

∠POQ = 90o

PQ2 = PO2 + QO2 ……. (i) [By Pythagoras]

Similarly,

PS2 = PO2 + SO2 ……. (ii)

RS2 = SO2 + RO2 ……. (iii)

QR2 = RO2 + QO2 ……. (iv)

Adding equation (i) + (ii) + (iii) + (iv) we get,

PQ2 + PS2 + RS2 + QR2 = 2(PO2 + QO2+ RO2 + SO2)

= 4PO2 + 4QO2 [Since, PO = RO and QO = SO]

= (2PO)2 + (2QO2) = PR2 + QS2

Question 8:

In the given figure, O is a point in the interior of a triangle PQR.

OS ⊥ QR, OT ⊥ PR and OU ⊥ PQ. Show that

i) OP2 + OQ2 + OR2 – OS2 – OT2 – OU2 = PU2 + QS2 + RT2

ii) PU2 + QS2 + RT2 = PT2 + RS2 + QU2

Solution:

Join OP, OQ and OR

i) Applying Pythagoras theorem in ΔPOU, we have

OP2 = OU2 + PU2

Similarly, in ΔQOS

OQ2 = OS2 + QS2

Similarly, in ΔROT

OR2 = OT2 + TR2

OP2 + OQ2 + OR2 = OU2 + PU2 + OS2 + QS2 + OT2 + RT2

OP2 + OQ2 + OR2 – OS2 – OT2 – OU2 = PU2 + QS2 + RT2.

ii) PU2 + QS2 + RT2 = (OP2 – OT2) + (OR2 – OS2) + (OQ2 – OU2)

$∴$ PU2 + QS2 + RT2 = PT2 + RS2 + QU2.

Question 9:

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

Solution:

Let QP be the wall and PR be the ladder,

Therefore, by Pythagoras theorem, we have

PR2 = PQ2 + QR2

102 = 82 + QR2

QR2 = 100 -64

QR2 = 36

QR = 6 m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Question 10:

An airplane leaves an airport and flies due north at a speed of 1,000 km/hr. At the same time, another airplane leaves the same airport and flies due west at a speed of 1,200 km/hr. How far apart will be the two planes after one and half hours?

Solution:

Speed of the first aeroplane = 1000 km/he

Distance covered by first aeroplane due north in one and half hours

(OA) = 1000 x 3/2 km = 1500 km

Sped of the second aeroplane = 1200 km/hr

Distance covered by second aeroplane due west in one and half hours

(OB) = 1200 x 3/2 km = 1800 km

In right angle ΔPOQ, we have

PQ2 = PO2 + OQ2

⇒ PQ2 = (1500)2 + (1800)2

⇒ PQ = 2250000+3240000$\sqrt{2250000+3240000}$

= 5490000$\sqrt{5490000}$

= 30061$300\sqrt{61}$ km

Hence, the distance between two aeroplanes will be 30061$300\sqrt{61}$ km.

Question 11:

Two planes of heights 6 m and 11m respectively stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution:

Let PQ and RS be the poles of height 6 m and 11 m respectively.

Therefore RO = 11 – 6 = 5 m

From the figure, it can be observed that PO = 12 m

Applying Pythagoras theorem for ΔPOR, we get

PO2 = OR2 + PR2

(12m)2 + (5m)2 = (AC)2

AC2 = (144+25) m = 169 m

AC = 13 m

Therefore, the distance between their tops is 13 m.

Question 12:

S and T are points on the sides RP and RQ respectively of a triangle PQR right angled at R. Prove that PT2 + QS2 = PQ2 + ST2

Solution:

Applying Pythagoras theorem in ΔPRT, we get

PR2 + RT2 = PT2 ………. (i)

Applying Pythagoras theorem in ΔQRS, we get

QR2 + RS2 = QS2 ………. (ii)

Adding (i) + (ii), we get

PR2 + RT2 + QR2 + RS2 = PT2 + QS2 ………. (iii)

Applying Pythagoras theorem in ΔRST, we get

ST2 = RS2 + RT2

Applying Pythagoras theorem in ΔPQR, we get

PQ2 = PR2 + RQ2

Putting these values in equation (iii), we get

ST2 + PQ2 = PT2 + QS2.

Question 13:

The perpendicular from P on side QR of a ΔPQR intersects QR at S such that SQ = 3RS. Prove that 2PQ2 = 2PR2 + QR2.

Solution:

Given that in ΔPQR, we have

PS ⊥ QR and SQ = 3RS

In right-angled triangles PSQ and PSR, we have

PQ2 = PS2 + QS2 ……. (i)

PR2 = PS2 + SR2 ……. (ii) [By Pythagoras theorem]

(ii) – (i), we get

PQ2 – PR2 = QS2 – SR2

= 9SR2 – SR2 [Since, SQ = 3SR]

= 8(QR/4)2 [Since, QR = SQ + RS = 3RS +RS = 4RS]

Therefore, PQ2 – PR2 = QR2/2

⇒ 2(PQ2 – PR2) = QR2

Therefore, 2PQ2 = 2PR2 + QR2.

Question 14:

In an equilateral triangle PQR, S is a point on side QR such that QS = 1/3 QR. Prove that 9PS2 = 7PQ2.

Solution:

Let the side of the equilateral triangle be a, and PT be the altitude of ΔPQR.

∴ QT = TR = QR/2 = a/2

And, PT = 3a2$\frac{\sqrt{3}a}{2}$

Given that, QS = 1/3 QR

∴ QS = a/3

ST = QT – QS = a/2 – a/3 = a/6

Applying Pythagoras theorem in ΔPST, we get

PS2 = PT2 + ST2

PS2 = (a32)2$(\frac{a\sqrt{3}}{2})^{2}$ + (a6)2$(\frac{a}{6})^{2}$

= 3a24$\frac{3a^{2}}{4}$ + (a236)$(\frac{a^{2}}{36})$

= 28a236$\frac{28a^{2}}{36}$

= 79$\frac{7}{9}$ AB2

Question 15:

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of its altitudes.

Solution:

Let the side of the equilateral triangle be a, and PO be the altitude of ΔPQR.

∴ QO = OR = QR/2 = a/2

Applying Pythagoras theorem in ΔPQO, we get

PQ2 = PO2 + QO2

a2=AE2+(a2)2$a^{2} = AE^{2} + (\frac{a}{2})^{2}$ AE2=a2a24$AE^{2} = a^{2} – \frac{a^{2}}{4}$ AE2=3a24$AE^{2} = \frac{3a^{2}}{4}$

4AE2 = 3a2

⇒ 4 x (Square of altitude) = 3 x (Square of one side)

Question 16:

Choose the correct solution and justify:

In ΔPQR, PQ = 6√3 cm, PR = 12 cm and QR = 6 cm.

The angle Q is:

i) 120o

ii) 60o

iii) 90o

iv) 45o

Solution:

Given that, PQ = 6√3 cm, PR = 12 cm and QR = 6 cm

We can observe that

PQ2 = 108

PR2 = 144

And, BC2 = 36

PQ2 + BC2 = PR2

The given triangle, ΔPQR is satisfying Pythagoras theorem.

Therefore, the triangle is a right-angled triangle at B.

∴ ∠B = 90o

Hence, the correct option is (iii)