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EXERCISE 6.5

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

Let the ∆ABC whose sides are AB, BC and CA.

Let AB = 7 cm BC = 24 cm and AC = 25 cm

For the right angle triangle one angle must be formed 90° and sum of square of

any two side must be equal to square the third side.

AB2 + BC2 = AC2

72 + 242 = 252

49 + 576 = 625

625 = 625

Therefore ∆ABC is a right angled triangle and AC is the hypotenuse.

(ii) 3 cm, 8 cm, 6 cm

Let the ∆ABC whose sides are AB, BC and CA.

Let AB = 3 cm BC = 8 cm and AC = 6 cm

For the right angle triangle one angle must be formed 90° and sum of square of any

two side must be equal to square the third side.

AB2 + AC2 = BC2

32 + 62 = 82

9 + 36 = 49

45 ≠ 49

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Therefore ∆ABC is not a right angled triangle.

(iii) 50 cm, 80 cm, 100 cm

Let the ∆ABC whose sides are AB, BC and CA.

Let AB = 50 cm BC = 80 cm and AC = 100 cm

For the right angle triangle one angle must be formed 90° and sum of square of any

two sides must be equal to square the third side.

AB2 + BC2 = AC2

502 + 802 = 1002

2500 + 6400 = 10000

8900 ≠ 10000

Therefore ∆ABC is not a right angled triangle.

(iv) 13 cm, 12 cm, 5 cm

Let the ∆ABC whose sides are AB, BC and CA.

Let AB = 5 cm BC = 12 cm and AC = 13 cm

For the right angle triangle one angle must be formed 90° and sum of square of any

two sides must be equal to square the third side.

AB2 + BC2 = AC2

52 + 122 = 132

25 + 144 = 169

169 = 169 : Therefore ∆ABC is a right angled triangle and AC is the hypotenuse.

2. PQR is a triangle right angled at P and M is apoint on QR such that PM ⊥

QR. Show that

PM2 = QM .MR.

Let ∠MPR = 𝑥

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In ∆MPR,

∠MRP + ∠MPR + ∠PMR = 180°

∠MRP = 180° − 90° − 𝑥

∠MRP = 90° − 𝑥

In ∆MPQ,

∠MPQ + ∠MPR = 90°

∠MPQ = 90° − 𝑥

∠MPQ + ∠MQP + ∠PMQ = 180°

∠MQP = 180° − 90° − (90° − 𝑥)

∠MQP = 𝑥

In ∆QMP and ∆PMR,

∠MQP = ∠MPR

∠MPQ = ∠MRP

∠PMQ = ∠RMP

∴ ∆QMP ~ ∆PMR (AA similarity rule)

QM

PM=

MP

MR

PM2 = QM × MR

3. In Figure, ABD is a triangle right angled at Aand AC ⊥ BD. Show that

(i) AB2 = BC . BD

In ∆ADB and ∆ACB,

∠DAB = ∠ACB = 90°

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∠ABD = ∠CBA (Common angle)

∴ ∆ADB ~ ∆ACB

BD

AB=

AB

CB

AB2 = BD × CB

(ii) AC2 = BC . DC

Let ∠CAB = 𝑥

In ∆CBA,

∠CBA + ∠CAB + ∠ACB = 180°

∠CBA = 180° − 90° − 𝑥

∠CBA = 90° − 𝑥

In ∆CAD,

∠CAB + ∠CAD = 90°

∠CAD = 90° − 𝑥

∠CDA + ∠CAD + ∠ACD = 180°

∠CDA = 180° − 90° − (90° − 𝑥)

∠CDA = 𝑥

In ∆CBA and ∆CAD,

∠ACB = ∠DCA = 90°

∠CAB = ∠CDA

∠CBA = ∠CAD

∴ ∆𝐶𝐴𝐵 ~ ∆CAD (AA A similarity rule)

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QM

PM=

MP

MR

PM2 = QM × MR

(iii) AD2 = BD . CD

In ∆CDA and ∆BAD,

∠DCB = ∠DAB = 90°

∠CDA = ∠ADB (Common angle)

∴ ∆DCA ~ ∆DAB (AA similarity rule)

DC

DA=

DA

BD

AD2 = BD × CD

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

According to question triangle ABC is right angled

isosceles triangle.

Therefore AC = BC

∠C = 90°

By using Pythagoras theorem in triangle ABC

AB2 = AC2 + BC2

AB2 = AC2 + AC2 (AC = AB)

2AC2 = AB2

5. ABC is an isosceles triangle with AC = BC. If AB2 = 2

AC2, prove that ABC is a right triangle.

According to question AB2 = 2AC2

AB2 = AC2 + AC2

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AB2 = BC2 + AC2

The above case satisfied for Pythagoras theorem so this is a right angle triangle.

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

According to question;

ABC is an equilateral whose sides are 2a and AD is the altitude.

AB = BC = AC = 2a

And BD = DC = a

By using Pythagoras theorem in ∆ABD where ∠D =

90°

AB2 = AD2 + BD2

(2a)2 = AD2 + (a)2

4a2 = AD2 + a2

AD2 = 4a2 − a2 = 3a2

AD = a√3

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.

Let PQRS are the vertices of the Rhombus.

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

In ∆POQ, ∆QOR, ∆ROS, ∆POS are the four triangles

in rhombus PQRS.

PQ2 = PO2 + OQ2 (i)

QR2 = OQ2 + OR2 (ii)

RS2 = OR2 + OS2 (iii)

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PS2 = OS2 + OP2 (iv)

Adding equation (i),(ii),(iii) and (iv)

PQ2 + QR2 + RS2 + PS2 = PO2 + OQ2 + OQ2 + OR2 + OR2 + OS2+OS2 + OP2

PQ2 + QR2 + RS2 + PS2 = 2(PO2 + OQ2 + OQ2 + OR2)

PQ2 + QR2 + RS2 + PS2 = 2[(PR

2)

2+ (

RS

2)

2+ (

PR

2)

2+ (

RS

2)

2]

PQ2 + QR2 + RS2 + PS2 = PR2 + QS2

8. In Figure, 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,

Construction: join OA, OB and OC

By using Pythagoras theorem in ∆AOF, we get

OA2 = OF2 + AF2 (i)

By using Pythagoras theorem in ∆BOD, we get

OB2 = OD2 + BD2 (ii)

By using Pythagoras theorem in ∆COE, we get

OC2 = OE2 + EC2 (iii)

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

OA2 + OB2 + 0C2 = OF2 + AF2 + 0D2 + BD2 + 0E2 + EC2

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

OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + EC2

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

AF2 + BD2 + EC2 = AE2 + CD2 + BF2

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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.

Let PQR is the right angle triangle.

Length of ladder is PR = 10m, the height of

window PQ = 8m and the distance between foot ladder

and the base of the wall QR = .

By using Pythagoras theorem

PR2 = PQ2 + QR2

102 = 82 + 𝑥2

𝑥2 = 100 − 64

𝑥2 = 36

𝑥 = √36 = 6m

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

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?

Let PQR is the right angle triangle.

Height of Pole is PQ = 18m, the length of wire PR =

24m and the distance between Base Pole and the

stake QR = .

By using Pythagoras theorem

PR2 = PQ2 + QR2

242 = 182 + 𝑥2

𝑥2 = 576 − 324

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𝑥2 = 252

𝑥 = 6√7 m

The distance of the Pole of the ladder from base of the stack is 6√7m.

11. An airplane leaves an airport and flies due north at a speed of 1000 km per

hour. At the same time, another airplane leaves the same airport and flies due

west at a speed of1200 km per hour. How far apart will be the two planes after

1𝟏

𝟐hours?

Distance travelled by the plane flying in North direction 1 and half hour = speed

×time = 1,000 × 3

2 = 1,500km

Distance travelled by the plane flying in West direction 1 and half hour = speed

×time = 1,200 × 3

2 = 1,800km

Let the distance travelled in north direction is OP = 1,500km and the distance

travelled in west direction is OQ =

1,800km

The distance between them PQ = 𝑥

By using Pythagoras theorem

PQ2 = OQ2 + OP2

PQ2 = (1,500)2 + (1,800)2

PQ = √5490000

PQ = 300√61 km

Therefore the distance between these planes after 1 and half hour 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.

Let the PQ= 6m and the RS= 11m are the height of the two Poles.

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Therefore RA = RS – AS = 11 – 6 = 5m

PA = QS = 12m and AS = PQ = 6 m

By using Pythagoras theorem

In triangle RPA

PR2 = PA2 + RA2

PR2 = (12)2 + (5)2

PR = √144 + 25

PR = 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.

In ∆BCD, C = 90°

By using Pythagorean Theorem;

BD2 = BC2 + CD2 (i)

In ∆ACE, C = 90°

By using Pythagorean Theorem;

AE2 = AC2 + CE2 (ii)

Adding equation (i) and equation (ii)

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

In ∆CDE, C = 90°

By using Pythagorean Theorem;

DE2 = CE2 + CD2 (iv)

In ∆ABC, C = 90°

By using Pythagorean Theorem;

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AB2 = AC2 + CB2 (v)

Adding equation (iv) and equation (v)

AB2 + DE2 = AC2 + CE2 + CB2 + CD2 (vi)

Equation (vi) use in equation (iii)

AE2 + BD2 = AB2 + DE2

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such

that DB = 3 CD(see Figure). Prove that 2 AB2 = 2 AC2 + BC2.

In ∆ABD, D = 90°

By using Pythagorean Theorem

AB2 = AD2 + DB2

AD2 = AB2 − DB2 (i)

In ∆ACD, D = 90°

By using Pythagorean Theorem

AC2 = AD2 + DC2

AD2 = AC2 − DC2 (ii)

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

AC2 − DC2 = AB2 − DB2 (iii)

According to question 3DC = DB

Therefore DC = DB

3 and DC =

BC

4

By using above value 3BC

4 = DC

Putting the above DC value in equation (iii) and we get

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AC2 − (BC

4)

2

= AB2 − (3BC

4)

2

16AC2 − BC2 = 16AB2 − BC2

16AB2 = 8BC2 + 16AC2

2 AB2 = 2 AC2 + BC2

15. In an equilateral triangle ABC, D is a point on side BC such that BD

=𝟏

𝟑BC. Prove that9 AD2 = 7 AB2.

Let ABC is the equilateral triangle whose sides is 𝑥, where AE is the altitude.

BE = EC =BC

2

And AE = 𝑎√3

2 and BD =

BC

3=

𝑎

3

DE=BE- BD = 𝑎

2−

𝑎

3=

𝑎

6

In triangle ADE

By using Pythagorean Theorem,

AD2 = AE2 + DE2

AD2 =(𝑎√3)

(2)2

2

+(𝑎)

(6)2

2

AD = 28𝑎2

36

AD = 7

9AB2

9AD2 = 7AB2

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.

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Let PQR is a equilateral triangle whose sides are PQ, RS and PR is 𝑥. Where PE is

the altitude of triangle PQR.

QE = ER = 𝑥

2

In triangle PQE

Using Pythagoras theorem

PQ2 = QE2 + PE2

𝑥2 = AE2 + (𝑥

2)

2

AE2 = 3𝑥2

4

4 × AE2 = 3𝑥2

4(square of altitude) = 3(Square of one side)

17. Tick the correct answer and justify: In Δ ABC, AB = 6√3 cm, AC = 12 cm

and BC = 6 cm.

The angle B is :

(A) 120° (B) 60° (C) 90° (D) 45°

According to question;

AB2 = 108 cm , AC2 = 144 cm and BC2 = 36 cm

AB2 + BC2 = AC2 (Satisfy Pythagoras theorem)

Therefore triangle ABC is right angle triangle and right

angle at B.

Answer (c)