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### Exercise 6.5 Page: 150

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) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

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

49 + 576 = 625

(7)2 + (24)2 = (25)2

Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.

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

Clearly, 9 + 36 ≠ 64

Or, 32 + 62 ≠ 82

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfies Pythagoras theorem.

(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.

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

However, 2500 + 6400 ≠ 10000

Or, 502 + 802 ≠ 1002

As you can see, 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 does not satisfies Pythagoras theorem.

Hence, it is not a right triangle.

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

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

Thus, 144 +25 = 169

Or, 122 + 52 = 132

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Hence, length of the hypotenuse of this triangle is 13 cm.

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, ΔPQR is right angled at P is a point on QR such that PM ⊥QR

We have to prove, PM2 = QM × MR

In ΔPQM, by Pythagoras theorem

PQ2 = PM2 + QM2

Or, PM2 = PQ2 – QM2 ……………………………..(i)

In ΔPMR, by Pythagoras theorem

PR2 = PM2 + MR2

Or, PM2 = PR2 – MR2 ………………………………………..(ii)

Adding equation, (i) and (ii), we get,

2PM2 = (PQ2 + PM2) – (QM2 + MR2)

= QR2 – QM2 – MR2        [∴ QR2 = PQ2 + PR2]

= (QM + MR)2 – QM2 – MR2

= 2QM × MR

∴ PM2 = QM × MR

3. In Figure, 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:

∠DAB = ∠ACB (Each 90°)

∠ABD = ∠CBA (Common angles)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB2 = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180° – 90° – x

∠CBA = 90° – x

= 90° –x

∠CDA = 180° – 90° – (90° – x)

∠CDA = x

In ΔCBA and ΔCAD, we have

∠CAB = ∠CDA

∠ACB = ∠DCA (Each 90°)

∴ ΔCBA ~ ΔCAD [AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC2 =  DC × BC

(iii) In ΔDCA and ΔDAB,

∠DCA = ∠DAB (Each 90°)

∴ ΔDCA ~ ΔDAB [AA similarity criterion]

⇒ DC/DA = DA/DA

⇒ AD2 = BD × CD

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

Solution:

Given, ΔABC is an isosceles triangle right angled at C.

In ΔACB, ∠C = 90°

AC = BC (By isosceles triangle property)

AB2 = AC2 + BC2 [By Pythagoras theorem]

= AC2 + AC2 [Since, AC = BC]

AB2 = 2AC2

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

Solution:

Given, ΔABC is an isosceles triangle having AC = BC and AB2 = 2AC2

In ΔACB,

AC = BC

AB2 = 2AC2

AB2 = AC+ AC2

= AC2 + BC[Since, AC = BC]

Hence, by Pythagoras theorem ΔABC is right angle triangle.

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

Solution:

Given, ABC is an equilateral triangle of side 2a.

AB = AC

Hence, BD = DC [by CPCT]

⇒ AD2 = 4a2 – a2

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

Solution:

Given,ABCD is a rhombus whose diagonals AC and BD intersect at O.

We have to prove, as per the question,

AB+ BC+ CD2 + AD= AC+ BD2

Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In ΔAOB,

∠AOB = 90°

AB2 = AO+ BO…………………….. (i) [By Pythagoras theorem]

Similarly,

DC2 = DO+ CO…………………….. (iii)

BC2 = CO+ BO…………………….. (iv)

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

AB+ AD+DC+BC2 = 2(AO+ BO+ DO+ CO2)

= 4AO+ 4BO[Since, AO = CO and BO =DO]

= (2AO)+ (2BO)2 = AC+ BD2

Hence, proved.

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

ABC, OD  BC, OE  AC and OF  AB. Show that:
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Solution:

Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA2 = OF2 + AF2

Similarly, in ΔBOD

OB2 = OD2 + BD2

Similarly, in ΔCOE

OC2 = OE2 + EC2

OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE+ EC2

OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.

(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)

∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

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:

Given, a ladder 10 m long reaches a window 8 m above the ground.

Let BA be the wall and AC be the ladder,

Therefore, by Pythagoras theorem,

AC2 =AB2 + BC2

102 = 82 + BC2

BC= 100 – 64

BC= 36

BC= 6m

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

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:

Given, 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.

Let AB be the pole and AC be the wire.

By Pythagoras theorem,

AC2 =AB2 + BC2

242 = 182 + BC2

BC= 576 – 324

BC= 252

BC= 6√7m

Therefore, the distance from the base is 6√7m.

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?

Solution:

Given,

Speed of first aeroplane = 1000 km/hr Distance covered by first aeroplane flying due north in

hours (OA) = 100 × 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr Distance covered by second aeroplane flying due west in

hours (OB) = 1200 × 3/2 km = 1800 km

In right angle ΔAOB, by Pythagoras Theorem,

AB2 =AO2 + OB2

⇒ AB2 =(1500)2 + (1800)2

⇒ AB = √(2250000 + 3240000)

= √5490000

⇒ AB = 300√61 km

Hence, the distance between two aeroplanes will be 300√61 km.

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:

Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And distance between the feet of the poles is 12 m.

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

Therefore, CP = 11 – 6 = 5m

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

By Pythagoras theorem for ΔAPC, we get,

AP2 =PC2 + AC2

(12m)2 + (5m)2 = (AC)2

AC2 = (144+25) m2 = 169 m2

AC = 13m

Therefore, the distance between their tops is 13 m.

13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.

Solution:

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

By Pythagoras theorem in ΔACE, we get

AC2 +CE2 = AE2 ………………………………………….(i)

In ΔBCD, by Pythagoras theorem, we get

BC2 +CD2 = BD2 ………………………………..(ii)

From equations (i) and (ii), we get,

AC2 +CE2 + BC2 +CD2 = AE2 + BD2 …………..(iii)

In ΔCDE, by Pythagoras theorem, we get

DE2 =CD2 + CE2

In ΔABC, by Pythagoras theorem, we get

AB2 =AC2 + CB2

Putting the above two values in equation (iii), we get

DE2 + AB2 = AE2 + BD2.

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2.

Solution:

Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

Subtracting equation (ii) from equation (i), we get

AB2 – AC2 = BD2 – DC2

= 9CD2 – CD2 [Since, BD = 3CD]

= 8CD2

= 8(BC/4)[Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB2 – AC2 = BC2/2

⇒ 2(AB2 – AC2) = BC2

⇒ 2AB2 – 2AC2 = BC2

∴ 2AB2 = 2AC2 + BC2.

15.  In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.

Solution:

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = 1/3BC

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

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given, BD = 1/3BC

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6 In ΔADE, by Pythagoras theorem,

⇒ 9 AD2 = 7 AB2

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.

Solution:

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

In ΔABE, by Pythagoras Theorem, we get

AB2 = AE2 + BE2

4AE2 = 3a2

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

Hence, proved.

17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°

(B) 60°
(C) 90°

(D) 45°

Solution:

Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB2 = 108

AC2 = 144

And, BC2 = 36

AB2 + BC2 = AC2

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).