## EXERCISE 6.5

#### Page No 150:

#### 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

#### Answer:

(i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of these sides, we will obtain 49, 576, and 625.

49 + 576 = 625

Or,

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

We know that the longest side of a right triangle is the hypotenuse.

Therefore, the length of the hypotenuse of this triangle is 25 cm.

(ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will obtain 9, 64, and 36.

However, 9 + 36 ≠ 64

Or, 3^{2} + 6^{2} ≠ 8^{2}

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iii)Given that sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50^{2} + 80^{2} ≠ 100^{2}

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iv)Given that sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will obtain 169, 144, and 25.

Clearly, 144 +25 = 169

Or,

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

We know that the longest side of a right triangle is the hypotenuse.

Therefore, the length of the hypotenuse of this triangle is 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.

#### Answer:

#### Question 3:

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

(i) AB^{2} = BC × BD

(ii) AC^{2} = BC × DC

(iii) AD^{2} = BD × CD

#### Answer:

(i) In ,

∴ (AA similarity criterion)

(ii)

(iii)

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)

#### Question 4:

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

#### Answer:

Given that ΔABC is an isosceles triangle.

∴ AC = CB

Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

#### Question 5:

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

#### Answer:

Given that,

#### Question 6:

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

#### Answer:

Let AD be the altitude in the given equilateral triangle, ΔABC.

We know that altitude bisects the opposite side.

∴ BD = DC = *a*

In an equilateral triangle, all the altitudes are equal in length.

Therefore, the length of each altitude will be.

#### Question 7:

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

#### Answer:

In Î”AOB, Î”BOC, Î”COD, Î”AOD,

Applying Pythagoras theorem, we obtain

#### Page No 151:

#### Question 8:

In the following figure, 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}

#### Answer:

Join OA, OB, and OC.

(i) Applying Pythagoras theorem in ΔAOF, we obtain

Similarly, in ΔBOD,

Similarly, in ΔCOE,

(ii) From the above result,

#### 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.

#### Answer:

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,

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

#### 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?

#### Answer:

Let OB be the pole and AB be the wire.

By Pythagoras theorem,

Therefore, the distance from the base is m.

#### Question 11:

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

#### Answer:

Distance travelled by the plane flying towards north in

Similarly, distance travelled by the plane flying towards west in

Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes after, AB =

Therefore, the distance between these planes will be km after.

#### 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.

#### Answer:

Let CD and AB be the poles of height 11 m and 6 m.

Therefore, CP = 11 − 6 = 5 m

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

Applying Pythagoras theorem for ΔAPC, we obtain

Therefore, the distance between their tops is 13 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}

#### Answer:

Applying Pythagoras theorem in ΔACE, we obtain

#### Question 14:

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

#### Answer:

Applying Pythagoras theorem for ΔACD, we obtain

Applying Pythagoras theorem in ΔABD, we obtain

#### Question 15:

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

#### Answer:

Let the side of the equilateral triangle be *a*, and AE be the altitude of ΔABC.

∴ BE = EC = =

And, AE =

Given that, BD = BC

∴ BD =

DE = BE − BD =

Applying Pythagoras theorem in ΔADE, we obtain

AD^{2} = AE^{2} + DE^{2}

⇒ 9 AD^{2} = 7 AB^{2}

#### Question 16:

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

#### Answer:

Let the side of the equilateral triangle be *a*, and AE be the altitude of ΔABC.

∴ BE = EC = =

Applying Pythagoras theorem in ΔABE, we obtain

AB^{2} = AE^{2} + BE^{2}

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

#### Question 17:

Tick the correct answer and justify: In ΔABC, AB = cm, AC = 12 cm and BC = 6 cm.

The angle B is:

(A) 120° (B) 60°

(C) 90° (D) 45°

#### Answer:

Given that, AB =cm, AC = 12 cm, and BC = 6 cm

It can be observed that

AB^{2} = 108

AC^{2} = 144

And, BC^{2} = 36

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

The given triangle, ΔABC, is satisfying Pythagoras theorem.

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

∴ ∠B = 90°

Hence, the correct answer is (C).