Gujarat Board GSEB Solutions Class 10 Maths Chapter 6 Triangle Ex 6.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 6 Triangle Ex 6.5

Question 1.

Sides of triangles are given below. Determine which of these are 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) 7 cm, 24 cm, 25 cm.

Here 72 = 49cm

24^{2} = 576 cm

25^{2} = 625 cm

we observed that

49 + 576 = 625

7^{2} + 24^{2} = 252

The given triangle is right angled triangle.

Hypotenuse = 25 cm

(ii) 3 cm, 8 cm, 6 cm

Here 32 = 9cm

82 = 64 cm

62 = 36 cm

we observed that

3^{2} + 6^{2 }+ 8^{2}

The given triangle is not right angled triangle.

(iii) 50 cm, 80 cm, 100 cm.

Here 50^{2} = 2500 cm

80^{2} = 6400 cm

100^{2} = 10,000 cm

We observed that

50^{2} + 80^{2} ≠ 100^{2}

The given triangle is not right angled.

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

Here 13^{2} = 169 cm

12^{2} = 144 cm

5^{2} = 25 cm

we observed that

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

∴ The given triangle is not right angled.

Hypotenuse = 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:

In ∆QMP and ∆QPR

∠QMP = ∠QPR (each 90°)

∠PQM = ∠PQR (Common)

∴ ∆PQM – ∆QPR ………(1) (by AA similarity)

Again in ∆PMR and ∆QPR

∠PMR = ∠QPR (each 900)

∠PRM = ∠QRP (Common)

∴ ∆PMR – ∆DQPR ………(2)

From (1) and (2)

∆QMP ~ ∆PMR

\(\frac{Q M}{P M}=\frac{P M}{R M}\)

(The corresponding sides of similar triangles are proportional)

PM^{2} = QM.RM

Question 3.

In the 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 (CBSE 2010)

Solution:

(i) In ∆ABC and ∆BAD

∠ACB = ∠BAD (each 900)

∠ABC = ∠ABD (common)

∆BAC ~ ∆BDA (AA similarity)

\(\frac{A B}{D B}=\frac{B C}{B A}\)

(Corresponding sides are proportional)

⇒ AB^{2} = BC.BD

(ii) In ∆ABC

∠BAC + ∠CBA = 90° ………(1) (by angle sum property)

In ∆ABD

∠BDA + ∠CBA = 90° …….(2)

From eqn (1) and (2)

∠BAC + ∠CBA = ∠BDA + ∠CBA

∠BAC = ∠BDA

In ∆ACB and ∆DCA

∠ACB = ∠DCA (each 900)

∠BAC = ∠BDA

(proved above from eqn (3))

∴ ∆ACB ~ ∆DCA (by AA similarity)

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

(corresponding sides of two similar triangle are proportional)

⇒ AC^{2} = BC.DC

(iii) In ∆ADB and ∆CDA

∠DAB = ∠DCA (each 900)

∠ADB = ∠CDA (Common)

∴ ∆ADB ~ ∆CDA (by AA similarity)

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

(Corresponding sides of two similar triangles are proportional)

⇒ AD^{2} = BD.CD

Question 4.

ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}. (CBSE 2012)

Solution:

We have ∆ABC is right angled such that

∠C = 90° and AC = AB

Applying Pythagoras theorem

– AC^{2} + BC^{2}

= AC^{2} + AC^{2} (BC = AC given)

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

Question 5.

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

Solution:

We have ∆ABC such that AB = AC

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

AB^{2} = BC^{2} +AC^{2}(BC =AC given)

∠ACB = 90° (by converse of Pythagoras theorem)

i.e. ∆ABC is a right angle triangle

Question 6.

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

Solution:

It is given that LABC is an equilateral triangle in which

AB = AC = CB = 2a

Let us draw AD ⊥ BC

Now in ∆ABD and ∆ACD

AB = AC (Side of equilateral triangle)

AD = AD (Common)

∠ADB = ∠ADC (each 90°)

∴ ∆ABD ≅ ∆ACD (by RHS)

∴ BD = CD (by CPCT)

i.e. D is a mid point of BC

Hence BD = \(\frac {2a}{2}\) = a

Now in right ∆ABD

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

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

4a^{2} – a^{2} = AD^{2}

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

AD = \(\sqrt { 3 } \)

Similarly BE = CF = \(\sqrt { 3 } \)

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. (CBSE 2012) of its diagonals. (CBSE 2012)

Solution:

Given: ABCD is a rhombus i.e. AB = BC = CD =

DA whose diagonals AC and BD intersect at O.

To prove: AC^{2} + BD^{2}

= AB^{2} + BC^{2} + CD^{2} + DA^{2}

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

∴ OA = OC and OB = OD

Now in right ∆AOB

AB^{2} = OA^{2} + OB^{2} (by Pythagoras theorem)

In right ∆BOC

BC^{2} = OB^{2} + OC^{2}……..(2)

Similarly in ∆COD

CD^{2} = OC^{2} + OD^{2}

and in right ∆AOD

AD^{2} – OA^{2} + OD^{2}

Now adding eqn (1), (2), (3) and (4) we get

AB^{2} + BC^{2} + CD^{2} + DA^{2}

= OA^{2} + OB^{2} + OC^{2} + OB^{2} + OD^{2} + OC^{2} + OD^{2} + OA^{2}

= 20A^{2} + 20B^{2} + 20C^{2} + 20D^{2}

= 20A^{2} + 20B^{2} + 20A^{2} + 20B^{2}

= 40A^{2} + 40B^{2}

= (20A)^{2} + (20B)^{2} + AB^{2} + BC^{2} + CD^{2} + DA^{2}

= AC^{2} + BD^{2}

(20A = AC and 20B = BD)

Question 8.

In the 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} – DE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

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

Solution:

In the given figure, 0 is in the interior of ∆ABC,

OD ⊥ BC, OF ⊥AB, OE ⊥AC

To prove:

(i) OA^{2} + OB^{2} + OC^{2} + 0D^{2} + 0E^{2} + OF^{2} = AF^{2} + BD^{2} + CE^{2}

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

Construction: Join OA, OB and OC

Proof: In right ∆AOF

OA^{2} = AF^{2} + OF^{2} (by Pythagoras theorem)

AF^{2} = OA^{2} + OF^{2} ……….(1)

In right ∆OBD

OB^{2} + BD^{2} + OD^{2}

BD^{2} = OB^{2} + OD^{2} ………..(2)

In right ∆OCE

OC^{2} = CE^{2} + OE^{2}

CE^{2} = OC^{2} + OE^{2}

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

AF^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} + OD^{2} + OE^{2} – OF^{2} ………..(3)

(ii) In right ∆OBD

OB^{2} = OD^{2} + BD^{2}

(by Pythagoras theorem)

⇒ BD^{2} = OB^{2} – OD^{2} ……….(1)

In right triangle ∆OCD

OC^{2} = OD^{2} + CD^{2}

(by Pythagoras theorem)

⇒ CD^{2} = OC^{2} – OD^{2} ……….(2)

Subtracting eqn (2) from eqn (1)

BD^{2} – CD^{2} = OB^{2} – OD^{2} – OC^{2} – OD^{2}

BD^{2} – CD^{2} = OB^{2} – OC^{2} ……….(3)

Similarly in right triangle OCE and OAE we get

CE^{2} = OC^{2} – OE^{2} ………(4)

(by Pythagoras theorem)

AE^{2} = OA^{2} – OE^{2} ………(5)

Subtracting eqn (5) from eqn (4)

CE^{2} – AE^{2} = OC^{2} – OE^{2} – OA^{2} + OE^{2}

⇒ CE^{2} – AE^{2} = OC^{2} – OA^{2} ……….(6)

And in right triangle OAF and OBF

AF^{2} = OA^{2} – OF^{2} ……….(7)

BF^{2} = OB^{2} – OF^{2} ………..(8)

Subtracting eqn (8) form (7)

AF^{2} – BF^{2} = AO^{2} – OF^{2} – OB^{2} + OF^{2}

2 – BF^{2} = OA^{2} – OB^{2} ………..(9)

Adding eqn (3), (6) and eqn (9) we get

BD^{2} + CE^{2} + AF^{2} – CD^{2} – AE^{2} – BF^{2}

⇒ OB^{2} + OC^{2} + OA^{2} – OC^{2} – OA^{2} – 0B^{2}

BD^{2} + CE^{2} + AF^{2} – CD^{2} – AE^{2} – BF^{2} = O

BD^{2} + CE^{2} + AF^{2} = CD^{2} + AE^{2} + BF^{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:

Let AB is a ladder and A is the foot of the ladder BC is the wall.

Now in right triangle ABC

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

(by Pythagoras theorem)

(10)^{2} + AC^{2} + 8^{2}

⇒ 100 – 64 = AC^{2}

⇒ AC^{2} = 36

⇒ AC = 6m

Hence, the distance of the foot of the ladder from the base of 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 state 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 the wire and BC is the vertical pole. The point A is the stake.

Now in right triangle ABC

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

24^{2} = AC^{2} + 18^{2}

AC^{2} = 24^{2} – 18^{2}

AC^{2} = (24 – 18)(24 + 18)

AC^{2} = 6 x 42

AC = \(\sqrt{6 \times 42}\)

AC = \(\sqrt{6 \times 6 \times 7}\)

⇒ AC = 6\(\sqrt{6 \times 42}\)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 plane after 1 \(\frac{1}{2}\) hours?

Solution:

Let the point A represent the airport. First plane flies due North with speed of 1000 km/h.

Distance of plane after 1\(\frac{1}{2}\) hour from airport = speed x time

Distance due North = 1000 x 1\(\frac{1}{2}\)

= 1000 x \(\frac{3}{2}\)

= 1500 km

Similarly, distance of plane after 11\(\frac{1}{2}\) hour From airport due west

= 1200 x 11\(\frac{1}{2}\) = 1200 x \(\frac{3}{2}\)

= 1800 km

Hence in right triangle ABC

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

BC^{2} = (1500)^{2} + (1800)^{2}

BC^{2} = 22500 + 32400

BC^{2} = 54900

BC = \(\sqrt{54900}\) km

BC = \(\sqrt{61 x 90000}\)

BC = 300 \(\sqrt{61}\)m.

Question 12.

Two poles of height 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 the two poles. Distance between their feet is 12 m. Height of pole 1st

AB = 11 m and height of pole lind CD = 6 m

AB = 11m

CD = 6m

BB = AB – AE (AE = CD)

⇒BE =11 – 6

⇒BE = 5m

In ∆BED

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

= 12^{2} + 5^{2} = 144 + 25

⇒ BD^{2} = 169

⇒ BD = 13m

Thus the distance between tops is 13 metres.

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: ∆ABC in which ∠C = 90°, D and E are the mid prints on the sides AC and BC.

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

Proof: In right angle AACB

AB^{2} = AC^{2} + BC^{2} (by Pythagoras theorem)

In DCE

DE^{2} = CD^{2} + CE^{2} ………(2)

(by Pythagoras theorem)

Adding eqn (1) and (2)

We get AB^{2} + DE^{2} = AC^{2} + BC^{2} + CD^{2} + CE^{2}

AB^{2} + DE^{2} = (AC^{2} + CE^{2}) +(BC^{2} + CD^{2}) …….(3)

Now in ∆ACE

AE^{2} = AC^{2} + CE^{2} ……. (4)

In ∆BCD

BC^{2} + CD^{2} = BD^{2} ……(5)

from eqn (3), (4) and (5)

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

Question 14.

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

Solution:

Given: ∆ABC such that AD ⊥ BC. The position of D is such that BD = 3 CD.

In right ∆ABD, we have

AB^{2} = AD^{2} + BD^{2} …….(1) (by Pythagoras theorem)

Similarly from right IACD,

AC^{2} = AD^{2} + CD^{2} ……(2)

Note that, BC = DB + CD

BC = 3CD + CD (BD = CD given)

BC = 4 CD

Subtracting eqn (2) from eqn (1), we get

⇒ AB^{2} – AC^{2} = BD^{2} – CD^{2}

⇒ AB^{2} – AC^{2} = (BD + CD) (BD – CD)

⇒ AB^{2} – AC^{2} = BC (3CD – CD)

⇒ AB^{2} – AC^{2} = BC x 2CD

⇒ AB^{2} – AC^{2} = BC x \(\frac{2 \times \mathrm{BC}}{4}{r}\) CD = \(\frac{BC}{4}\)

⇒ AB^{2} – AC^{2} = \(\frac{\mathrm{BC} \times \mathrm{BC}}{2}\)

⇒ 2AB^{2} – 2AC^{2} = BC^{2}

⇒ AB^{2} = 2AC^{2} + BC^{2}

Question 15.

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

Solution:

Given: AABC is an equilateral triangle.

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

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

Construction: Draw AP⊥BC

Proof: In right AAPB, we have

AB^{2} =AP^{2} + BP^{2} …….(1) (by Pythagoras theorem))

Now, in right triangle APD, we have

AD^{2} = AP^{2} + DP^{2}

(by Pythagoras theorem)

AP^{2} =AD^{2} – DP^{2}

From eqn (1) and (2) we get

AB^{2} =AD^{2} – DP^{2} + BP^{2}

AB^{2} =AD^{2} – DP^{2} + (BP = BC)

AD^{2} = AB^{2} + DP^{2} – 4

Note that, DP = BP – BD

DP = \(\frac{1}{2}\)BC – \(\frac{1}{3}\)BC = \(\frac{1}{6}\)BC in (3),gives

Substituting value of DP = \(\frac{1}{6}\)BC

AD^{2} = AB^{2} + (\(\frac{1}{6}\)BC^{2})^{2} – \(\frac{\mathrm{BC}^{2}}{4}\)

⇒ AD^{2} = AB^{2} + \(\frac{\mathrm{BC}^{2}}{36}\) – \(\frac{\mathrm{BC}^{2}}{4}\)

⇒ AD^{2} = AB^{2} + \(\frac{\mathrm{BC}^{2}-9 \mathrm{BC}^{2}}{36}\)

⇒ AD^{2} = AB^{2} – \(\frac{8 \mathrm{BC}^{2}}{36}\)

⇒ AD^{2} = AB^{2} – \(\frac{2 \mathrm{BC}^{2}}{9}\)

⇒ AD^{2} = AB^{2} – \(\frac{2 \mathrm{2AB}^{2}}{9}\) (BC = AB = AC)

⇒ 9AD^{2} = 9AB^{2} – 2AB^{2}

⇒ 9AD^{2} – 7AB^{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 altitude.

Solution:

Given: ABC is an equilateral triangle,

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

Construction: Draw AD ⊥BC

Proof In right ∆ADB

BD = DC = \(\frac{BC}{2}\) ………(1) (D is the midpoint)

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

AB^{2} = AD^{2} + \(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)[(Using (1)]

AB^{2} = AD^{2} + \(\frac{\mathrm{BC}^{2}}{4}\) (∆ABC is an equilateral)

= AB^{2} = AD^{2} + \(\frac{\mathrm{AB}^{2}}{4}\) (asAB = BC)

4AB^{2} = 4AD^{2} + AB^{2}

3AB^{2} = 4AD^{2}

Question 17.

Tick the correct answer and justify: In AB = 6\(\sqrt {3} \) cm, AC = 12 cm, and BC = 6 cm. The angle B is

(a) 120°

(b) 60°

(c) 90°

(d) 45°

Solution:

Wehave AB = 6\(\sqrt {3} \) cm,AC = 12cm, and BC = 6 cm

AB^{2} = (6\(\sqrt {3} \))^{2} = 36 x 3

AB^{2} = 108

AC^{2} = 122 = 144

BC^{2} = ^{2} = 36

Since 144 = 108 + 36

i.e. AC^{2} = AB^{2} + BC^{2}

Hence the given triangle is right angle triangle at B

∠B = 90°

∴ The correct answer is (c).