Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

## Haryana Board 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, 26 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

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

Solution:

(i) The given sides of the triangle are 7 cm, 24 cm, 25 cm,

The longest side = 25cm.

The triangle is right angled if

(the longest side)^{2} = (sum of squares of other two sides) (by Pythagoras theorem)

(25)^{2} = (7)^{2} + (24)^{2}

625 = 49 + 576

625 = 625

The given sides make a right triangle and length of hypotenuse = 25cm.

(ii) The given sides of the triangle are 3cm, 8 cm, 6 cm.

The longest side = 8 cm

(the longest side)^{2} = (sum of the squares of other two sides) (by Pythagoras theorem)

(8)^{2} = (3)^{2} + (6)^{2}

64 = 9 + 36

But 64 â‰ 45

The given sides do not make a right triangle.

(iii) The given sides of the triangle Eire 50 cm, 80 cm, 100 cm.

The longest side = 100 cm,

(the longest side)^{2} = (sum of the squares of other two sides) (by Pythagoras theorem)

(100)^{2} = (50)^{2} + (80)^{2}

10000 = 2500 + 6400

But 10000 â‰ 8900

âˆ´ The given sides do not mEike a right triangle.

(iv) The given sides of the triangle are 13 cm, 12 cm, 5 cm.

The longest side = 13 cm.

(the longest side)^{2} = (sum of the squares of other two sides) (by Pythagoras theorem)

(13)^{2} = (12)^{2} + (5)^{2}

169 = 144 + 25

169 = 169

âˆ´ The given sides make a right triangle and length of hypotenuse = 13 cm.

Question 2.

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

Solution:

Given : A right âˆ†PQR in which âˆ P = 90Â° and PM âŠ¥ QR.

To Prove : PM^{2} = QM.MR

Proof : In right âˆ†QPR, PM âŠ¥ QR.

âˆ†RPM ~ âˆ†PQM [By theorem 6.7]

â‡’ \(\frac{\mathrm{MR}}{\mathrm{PM}}=\frac{\mathrm{PM}}{\mathrm{QM}}\)

[Corresponding sides of similar triangles sere proportional]

â‡’ PM^{2} = QM Ã— MR

Hence Proved.

Question 3.

In given figure, ABD is a triangle right angled at A and AC âŠ¥ BD. Show that

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

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

(iii) AD^{2} = BD.CD

Solution:

Given : A right triangle in which âˆ A = 90Â° sind AC âŠ¥ BD.

To Prove :

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

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

(iii) AD^{2} = BD.CD

Proof :

(i) In right âˆ†BAC, AC âŠ¥ BD.

âˆ†BAC ~ âˆ†BDA [By theorem 6.7]

â‡’ \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{AB}}\)

[Corresponding sides of similar triEingles are proportional]

â‡’ AB^{2} = BC.BD

Hence Proved.

(ii) Similsirly, âˆ†ABC~ âˆ†DAC [By theorem 6.7]

\(\frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)

[Corresponding sides of similar triEingles Eire proportional]

â‡’ AC^{2} = BC.CD

Hence Proved.

(iii) Similarly, âˆ†DAC ~ âˆ†DBA [By theorem 6.7]

\(\frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AD}}{\mathrm{BD}}\)

[Corresponding sides of similar triangles are proportional]

â‡’ AD^{2} = BD.CD

Hence Proved.

Question 4.

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

Solution:

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

i.e., AC = BC

To Prove : AB^{2} = 2AC^{2}

Proof : In right âˆ†ACB

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

â‡’ AB^{2} = AC^{2} + AC^{2} [âˆµ BC = AC]

â‡’ 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 a right triangle.

Solution:

Given: âˆ†ABC is an isosceles triangle in which AC = BC and AB^{2} = 2AC^{2}

To Prove : âˆ†ABC is a right triangle.

Proof : In âˆ†ABC, AC = BC and AB^{2} = 2AC^{2}

â‡’ AB^{2} = AC^{2} + AC^{2}

â‡’ AB^{2} = BC^{2} + AC^{2} [v AC = BC]

â‡’ ABC is right angle triangle.

[By converse of Pythagoras theorem]

Hence Proved.

Question 6.

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

Solution :

Side of equilateral âˆ† = 2a.

And AD is its altitude. We know that altitude bisects its corresponding side.

i.e., BD = CD = a

Now right triangle âˆ†ADC.

AC^{2} = AD^{2} + CD^{2} [By Pythagoras theorem]

â‡’ (2a)^{2} = AD^{2} + a^{2}

â‡’ 4a^{2} = AD^{2} + a^{2}

â‡’ AD^{2} = 4a^{2} – a^{2} = 3a^{2}

â‡’ AD = âˆš3a

Hence, Altitude = âˆš3a.

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.

Given : ABCD is a rhombus

Proof: We know that diagonals of a rhombus bisect each other at right angle

âˆ´ âˆ AOB = âˆ BOC = âˆ COD = âˆ AOD = 90Â° and AO = CO, OD = BO.

In right âˆ†AOB, AB^{2} = OA^{2} + OB^{2} [By Pythagoras theorem]

AB^{2} = (\(\frac{1}{2}\) AC)^{2} + (\(\frac{1}{2}\) BD)^{2}

[âˆµ OA = OC and OB = OD]

â‡’ AB^{2} = 7 AC^{2} + 7BD^{2}

â‡’ 4AB^{2} = AC^{2} + BD^{2} ……………(1)

Similarly, we have

4BC^{2} = AC^{2} + BD^{2} ……………….(2)

4CD^{2} = AC^{2} + BD^{2} ……………….(3)

4AD^{2} = AC^{2} + BD^{2} ……………….(4)

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

4(AB^{2} + BC^{2} + CD^{2} + AD^{2}) = 4(AC^{2} + BD^{2})

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

Hence Proved.

Question 8.

In given figure, O is a point in the interior of a triangle ABC, OD âŠ¥ BC, OE âŠ¥ AC and OF âŠ¥ 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 âˆ†ABC in which OD âŠ¥BC, OE âŠ¥ AC, OF âŠ¥ AB.

To Prove :

(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}

Construction : Join OB, OC and OA.

Proof :

(i) By Pythagoras theorem for each of the right triangles namely âˆ†OFA, âˆ†ODB, and âˆ†OEC, we get

OA^{2} = OF^{2} + AF^{2} …………….(1)

OB^{2} = OD^{2} + BD^{2} ……………(2)

OC^{2} = OE^{2} + CE^{2} ……………..(3)

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

OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2} + CE^{2} = AF^{2} + BD^{2} + CE^{2}

â‡’ OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

Hence Proved.

(ii) In right triangles ODB and ODC, we have

OB^{2} = OD^{2} + BD^{2} …………….(4) [By Pythagoras theorem]

OC^{2} = OD^{2} + CD^{2} ……………..(5)

Subtracting (5) from (4), we get

OB^{2} – OC^{2} = BD^{2} – CD^{2} ………….(6)

Similarly OC^{2} – OA^{2} = CE^{2} – AE^{2} ……………(7)

and OA^{2} – OB^{2} = AF^{2} – BF^{2} ……………(8)

Adding (6), (7) and (8), we get

OB^{2} – OC^{2} + OC^{2} – OA^{2} + OA^{2} – OB^{2} = BD^{2} – CD^{2} + CE^{2} – AE^{2} + AF^{2} – BF^{2}

â‡’ 0 = BD^{2} – CD^{2} + CE^{2} – AE^{2} + AF^{2} – BF^{2}

â‡’ AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

Hence Proved.

Question 9.

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

Solution :

Let AC be ladder and AB be wall and BC be the distance of the ladder from wall.

âˆ´ AB = 8 m

AC = 10 m

In right âˆ†ABC, AC^{2} = BC^{2} + AB^{2}

â‡’ 10^{2} = BC^{2} + 8^{2}

â‡’ 100 = BC^{2} + 64

â‡’ BC^{2} = 100 – 64

â‡’ BC^{2} = 36

â‡’ BC = 6m.

Hence, distance of the foot of the ladder from base of the wall = 6m.

Question 10.

A guy wire attached to a vertical pole of height 18m is 24m 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 AC be guy wire and AB be vertical pole and BC be distance of the stake from the base of the pole.

âˆ´ AB = 18m, AC = 24m,

In right âˆ†ABC,

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

â‡’ (24)^{2} = BC^{2} + (18)^{2}

â‡’ 576 = BC^{2} + 324

â‡’ BC^{2} = 576 – 324

â‡’ BC^{2} = 252

â‡’ BC = âˆš252

â‡’ BC = 6âˆš7 m

Hence, distance of stake from the base of the pole = 6âˆš7 m.

Question 11.

An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1\(\frac{1}{2}\)hours?

Solution:

Let point B represents the position of airport. Then distance covered by first plane in the north direction (AB) = Speed Ã— Time

= 1000 Ã— 1\(\frac{1}{2}\)

= 1000 Ã— \(\frac{3}{2}\)

= 1500 km

and distance covered by second plane in the west direction (BC) = Speed Ã— Time

= 1200 Ã— \(\frac{1}{2}\)

= 1200 Ã— \(\frac{3}{2}\) = 1800 km.

Now, in right âˆ†ABC,

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

[By Pythagoras theorem]

â‡’ AC^{2} = (1800)^{2} + (1500)^{2}

â‡’ AC^{2} = 3240000 + 2250000

â‡’ AC^{2} = 5490000

â‡’ AC = \(\sqrt{5490000}\)

â‡’ AC = 300âˆš61 km

Hence, distance between two planes = 300âˆš61 km.

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:

Let AB and CD be two poles, BD be distance between the feet of two poles and AC be distance between their tops.

âˆ´ AB = 11 m, BD = 12 m, CD = 6 m

âˆ B = âˆ D = 90Â°

CE || BD

âˆ AEC = 90Â°, CE = 12m, AE = 11 – 6 = 5 m

Now, in right âˆ†AEC,

âˆ´ AC^{2} = CE^{2} + AE^{2} [By Pythagoras theorem]

â‡’ AC^{2} = 12^{2} + 5^{2}

â‡’ AC^{2} = 144 + 25

â‡’ AC^{2} = 169

â‡’ AC = âˆš169

â‡’ AC = 13 m.

Hence, distance between tops of two poles = 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}.

Solution :

Given : A triangle ABC in which âˆ C = 90Â°, D and E are points on the sides CA and CB respectively.

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

Proof: In right âˆ†ACE,

AE^{2} = AC^{2} + CE^{2} ………………(1)

[By Pythagoras theorem]

In right âˆ†DCB,

BD^{2} = BC^{2} + DC^{2} ………………(2)

In right âˆ†ABC,

AB^{2} = BC^{2} + AC^{2} ……………….(3)

In right âˆ†DCE,

DE^{2} = EC^{2} + CD^{2} ………………..(4)

Adding (1) and (2), we get

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

AE^{2} + BD^{2} = AB^{2} + DE^{2} [Using (3) and (4)]

Hence Proved.

Question 14.

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

Solution :

Given : A âˆ†ABC such that AD âŠ¥ BC and DB = 3CD

To Prove: 2AB^{2} = 2AC^{2} + BC^{2}

Proof: DB = 3CD (given)

Now BC = DB + CD

â‡’ BC = 3CD + CD [. DB = 3CD]

â‡’ BC = 4CD

â‡’ CD = \(\frac{1}{4}\)BC

â‡’ and BD = 3CD = \(\frac{3}{4}\)BC

In right âˆ†ADB ,

AB^{2} = BD^{2} + AD^{2} ………………..(1)

In right âˆ†ADC,

AC^{2} = CD^{2} + AD^{2} …………………(2)

Subtracting (2) from (1), we get

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

â‡’ AB^{2} – AC^{2} = (\(\frac{3}{4}\)BC)^{2} – (\(\frac{1}{4}\)BC)^{2}

[âˆµ BD = \(\frac{3}{4}\)BC and CD = \(\frac{1}{4}\)BC]

â‡’ AB^{2} – AC^{2} = \(\frac{9}{16}\) BC^{2} – \(\frac{9}{16}\) BC^{2}

â‡’ AB^{2} – AC^{2} = \(\frac{8}{16}\)BC^{2}

â‡’ AB^{2} – AC^{2} = \(\frac{1}{2}\)BC^{2}

â‡’ 2AB^{2} – 2AC^{2} = BC^{2}

â‡’ 2AB^{2} = 2AC^{2} + BC^{2}

Hence Proved.

Question 15.

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

Solution :

Given : An equilateral âˆ†ABC such that BD = \(\frac{1}{3}\)BC.

To Prove: 9AD^{2} = 7AB^{2}

Construction : Draw AE âŠ¥ BC and Join AD.

Proof: Let each side of equilateral triangle be x units.

BD = \(\frac{1}{3}\)BC

â‡’ BD = \(\frac{x}{3}\)

We know that in equi1ateral perpendicular bisects its corresponding opposite side

BE = EC = \(\frac{x}{2}\)

DE = BE – BD

DE = \(\frac{x}{2}\) – \(\frac{x}{3}\)

DE = \(\frac{x}{6}\)

In right âˆ†AED, AD^{2} = AE^{2} + DE^{2} ……………..(1)

[By Pythagoras theorem]

In right âˆ†AEB, AB^{2} = BE^{2} + AE^{2}

[By Pythagoras theorem]

AE^{2} = AB^{2} – BE^{2}

Put AE^{2} = AB^{2} – BE^{2} in (1), we get

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

AD^{2} = x^{2} – (\(\frac{x}{2}\))^{2} + (\(\frac{x}{6}\))^{2}

â‡’ AD^{2} = x^{2} – \(\frac{x^2}{4}+\frac{x^2}{36}\)

â‡’ AD^{2} = \(\frac{36 x^2-9 x^2+x^2}{36}\)

â‡’ AD^{2} = \(\frac{28 x^2}{36}\)

â‡’ AD^{2} = \(\frac{7}{9}\) x^{2}

â‡’ AD^{2} = \(\frac{7}{9}\) AB^{2}

â‡’ 9AD^{2} = 7AB^{2}

Hence Proved.

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 aftitudes.

Solution :

Given : A triangle ABC in which AB = BC = CA and AD âŠ¥ BC.

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

Proof: In right âˆ†ADB and right âˆ†ADC.

AB = AC (Given)

âˆ ADB = âˆ ADC (Each is 90Â°)

AD = AD (Common)

âˆ´ âˆ†ADB â‰… âˆ†ADC (By RHS congruence criterion)

â‡’ BD = CD (CPCT)

In right âˆ†ADB, AB^{2} = AD^{2} + BD^{2} [By Pythagoras theorem]

â‡’ AB^{2} = AD^{2} + (\(\frac{1}{2}\)AB)^{2}

[âˆµ BD = CD

âˆ´ BD = \(\frac{1}{2}\)AB]

â‡’ AB^{2} = 4AD^{2} + \(\frac{1}{4}\)AB^{2}

â‡’ 4AB^{2} = 4AD^{2} + AB^{2}

â‡’ 3AB^{2} = 4 AD^{2}

Hence Proved.

Question 17.

Tick the correct answer and justify. In âˆ†ABC, AB = 6âˆš3 cm, AC = 12 cm and BC = 6 cm. The angle B is :

(A) 120Â°

(B) 60Â°

(C) 90Â°

(D) 45Â°

Solution :

In âˆ†ABC,

AC^{2} = 12^{2}

â‡’ AC^{2} = 144

BC^{2} + AB^{2} = (6)^{2} + (6âˆš3)^{2}

â‡’ BC^{2} + AB^{2} = 36 + 108

â‡’ BC^{2} + AB^{2} = 144

âˆ´ AC^{2} = BC^{2} + AB^{2}

By converse of Pythagoras theorem,

âˆ B = 90Â°

Correct answer = (C)

Solution is also justified.