# PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

## PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.5

Question 1.

Sides of triangles are given below. Determine which of them are right triangles. In 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) Let ∆ABC, with AB = 7 cm BC = 24 cm, AC = 25 cm

AB^{2} + BC^{2} = (7)^{2} + (24)^{2}

= 49 + 576 = 625

AC^{2} = (25)^{2} = 625

Now AB^{2} + BC^{2} = AC^{2}

∴ ∆ABC is right angled triangle. Hyp. AC = 25cm.

(ii) Let ∆PQR with PQ = 3 cm, QR = 8 cm PR = 6 cm

PQ^{2} + PR^{2} = (3)^{2} + (6)^{2}

= 9 + 36 = 45

QR^{2} = (8)^{2} = 64.

Here PQ^{2} + PR^{2} ≠ QR^{2}

∴ ∆PQR is not right angled triangle.

(iii) Let ∆MNP, with MN =50 cm, NP = 80 cm, MP = 100 cm

MN ^{2}+ NP^{2} = (50)^{2} + (80)^{2}

= 2500 + 6400 = 8900

MP^{2} = (100)^{2} = 10000

Here MP^{2} ≠ MN^{2} + NP^{2}.

∴ ∆MNP is not right angled triangle.

(iv) Let ∆ABC, AB = 13 cm, BC = 12 cm, AC = 5 cm

BC^{2} + AC^{2} = (12)^{2} + (5)^{2}

= 144 + 25 = 169

AB^{2} = (13)^{2} = 169

∴ AB^{2} = BC^{2} + AC^{2}

∆ABC is right angled triangle.

Hyp. AB = 13 cm.

Question 2.

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM^{2} = QM . MR.

Solution:

Given: ∆PQR is right angled at P and M is a point on QR such that PM ⊥ QR.

To prove : PM^{2} = QM × MR

Proof: ∠P = 90° (Given)

∴ ∠1 + ∠2 = 90°

∠M = 900 (Given)

In ∆PMQ,

∠1 + ∠3 + ∠5 = 180°

=> ∠1 + ∠3 = 90° [Angle Sum Property] ………….(2) [∠5 = 90°]

Question 3.

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

(i) AB^{2} = BC.BD

(ii) AC^{2} = BC.DC

(üi) AD^{2} = BD.CD.

Solution:

Given. A right angled ∆ABD in which right angled at A and AC ⊥ BD.

To Prove:

(i) AB^{2} = BC.BD

(ii) AC^{2} = BC.DC .

(iii) AD^{2} = BD.ÇD .

Proof. In ∆DAB and ∆DCA,

∠D = ∠D (common)

∠A = ∠C (each 90°)

∴ ∆DAB ~ ∆DCA [AA similarity]

In ∆DAB and ∆ACB,

∠B = ∠B (common)

∠A = ∠C . (each 90°)

∴ ∆DAB ~ ∆ACB, .

From (1) and (2),

∆DAB ~ ∆ACB ~ ∆DCA.

(i) ∆ACB ~ ∆DAB (proved)

Question 4.

ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}.

Solution:

Given: ABC is an isosceles triangle right angled at C.

To prove : AB^{2} = 2AC^{2}.

Proof: In ∆ACB, ∠C = 90° & AC = BC (given)

AB^{2} = AC^{2} + BC^{2}

[By using Pythagoras Theorem]

=AC^{2} + AC^{2} [BC = AC]

AB^{2} = 2AC^{2}

Hence proved.

Question 5.

ABC is an isosceles triangle with AC = BC. If AB^{2} = 2AC^{2}, prove that ABC is right triangle.

Solution:

Given: ∆ABC is an isosceles triangle AC = BC

To prove: ∆ABC is a right triangle.

Proof: AB^{2} = 2AC^{2} (given)

AB^{2} = AC^{2} + AC^{2}

AB^{2} = AC^{2} + BC^{2} [AC = BC]

∴ By Converse of Pythagoras Theorem,

∆ABC is right angled triangle.

Question 6.

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

Solution:

∆ABC is equilateral triangle with each side 2a

AD ⊥ BC

AB = AC = BC = 2a

∆ADB ≅ ∆ADC [By RHS Cong.]

∴ BD = DC = a [c.p.c.t]

In right angled ∆ADB

AB^{2} = AD^{2} + BD^{2}

(2a)^{2} = AD^{2} + (a)^{2}

4a^{2} – a^{2} = AD^{2}.

AD^{2} = 3a^{2}

AD = √3a.

Question 7.

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [Pb. 2019]

Solution:

Given: Rhombus, ABCD diagonal AC and BD intersect each other at O.

To prove:

AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}

Proof:The diagonals of a rhombus bisect each other at right angles.

∴ AO = CO, BO = DO

∴ ∠s at O are rt. ∠s

In ∆AOB, ∠AOB = 90°

∴ AB^{2} = AO^{2} + BO^{2} [By Pythagoras Theorem] …………..(1)

Similarly, BC^{2} = CO^{2} + BO^{2} ……………..(2)

CD^{2} = CO^{2} + DO^{2} ……………(3)

and DA^{2} = DO^{2} + AO^{2} ……………….(4)

Adding. (1), (2), (3) and (4), we get

AB^{2} + BC^{2} + CD^{2} + DA^{2} = 2AO^{2} + 2CO^{2} + 2BO^{2} + 2DO^{2}

= 4AO^{2} + 4BO^{2}

[∵ AO = CO and BO = DO]

= (2AO)^{2} + (2BO)^{2} = AC^{2} + BD^{2}.

Question 8.

In fig., O is a point In the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

(ii)AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

Solution:

Given: A ∆ABC in which OD ⊥ BC, 0E ⊥ AC and OF ⊥ AB.

To prove:

(i) AF^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

Construction: Join OB, OC and OA.

Proof: (i) In rt. ∠d ∆AFO, we have

OA^{2} = OF^{2} + AF^{2} [By Pythagoras Theorem]

or AF^{2} = OA^{2} – OF^{2} …………..(1)

In rt. ∠d ∆BDO, we have:

OB^{2} = BD^{2}+ OD^{2} [By Pythagoras Theorem]

⇒ BD^{2} = OB^{2} – OD^{2} …………..(2)

In rt. ∠d ∆CEO, we have:

OC^{2} = CE^{2} + OE^{2} [By Pythagoras Theorem]

⇒ CE^{2} = OC^{2} – OE^{2} ……………(3)

∴ AF^{2} + BD^{2} + CE^{2} = OA^{2} – OF^{2} + OB^{2} – OD^{2} + OC^{2} – 0E^{2}

[On adding (1), (2) and (3)]

= OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

which proves part (1).

Again, AF^{2} + BD^{2} + CE^{2} = (OA^{2} – OE^{2}) + (OC^{2} – OD^{2}) + (OB^{2} – OF^{2})

= AE^{2} + CD^{2} + BF^{2}

[∵AE^{2} = AO^{2} – OE^{2}

CD^{2} = OC^{2} – OD^{2}

BF^{2} = OB^{2} – OF^{2}].

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:

Height of window from ground (AB) = 8m.

Length of ladder (AC) = 10 m

Distance between foot of ladder and foot of wall (BC) = ?

In ∆ABC,

AB^{2} + BC^{2} = AC^{2} [By Pythagoras Theorem]

(8)^{2} + (BC)^{2} = (10)^{2}

64 + BC^{2} = 100

BC^{2} = 100 – 64

BC = √36

BC = 6 cm.

∴ Distance between fóot of ladder and foot of wall = 6 cm.

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:

Let AB is height of pole (AB) = 18 m

AC is length of wire = 24 m

Question 11.

An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two pLanes after 1½ hours?

Solution:

Speed of first aeroplane = 1000km/hr.

Question 12.

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

distance between their tops.

Solution:

Height of pole AB = 11 m

Height of pole (CD) = 6 m

Question 13.

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

Prove that AE^{2} + BD^{2} = AB^{2} + DE^{2}.

Solution:

Given: In right angled ∆ABC, ∠C = 90° ;

D and E are points on sides CA & CB respectively.

To prove: AE^{2} + BD^{2} = AB^{2} + DE^{2}

Proof: In rt. ∠d ∆BCA,

AB^{2} = BC^{2} + CA^{2} …………..(1) [By Pythagoras Theorem]

In rt. ∠d ∆ECD,

DE^{2} = EC^{2} + DC^{2} ……………….(2) [By Pythagoras Theorem]

In right angled triangle ∆ACE,

AE^{2} = AC^{2} + CE^{2} ……………….(3)

In right angled triangle ∆BCD

BD^{2} = BC^{2} + CD^{2} ……………….(4)

Adding (3) and (4),

AE^{2} + BD^{2} = AC^{2} + CE^{2} + BC^{2} + CD^{2}

= [AC^{2} + CB^{2}] + [CE^{2} + DC^{2}]

= AB^{2} + DE^{2}

[From (1) and (2)]

Hence 2 + BD^{2} = AB^{2} + DE^{2}.

Which is the required result.

Question 14.

The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD. Prove that 2AB^{2} = 2AC^{2} + BC^{2}.

Solution:

Given: ∆ABC, AD ⊥ BC

BD = 3CD.

To prove: 2AB^{2} = 2AC^{2} + BC^{2}.

Question 15.

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD^{2} = 7 AB^{2}.

Solution:

Given: Equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.

To prove: 9AD^{2} = 7 AB^{2}.

Construction: AB ⊥ BC.

Proof: ∆AMB ≅ ∆AMC [By R.HS. Rule since AM = AM and AB = AC]

Question 16.

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

altitudes.

Solution:

Given:

ABC is equilateral ∆ in which AB = BC = AC

To prove: 3 AB^{2} = 4 AD^{2}

Proof: In right angled ∆ABD,

AB^{2} = AD^{2} + BD^{2} (Py. theorem)

AB^{2} = A BD^{2} (Py. theorem)

Question 17.

Tick the correct answer and justify: In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6√3 cm. [The angles of B are respectively

(A) 120°

(B) 64°

(C) 90°

(D) 45°

Solution.

AC = 12 cm

AB = 6√3 cm

BC = 6 cm

AC^{2} = (12)^{2} = 144 cm

AB^{2} + BC^{2} = (6√3)^{2} + (6)^{2}

= 108 + 36

AB√3 + BC√3 = 144

∴ AB√3 + BC√3 = AC√3

Hence by converse of pythagoras theorem ∆ABC is right angred triangle right angle at B

∴ ∠B = 90°

∴ correct option is (C).

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