Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 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:

1. 7 cm, 24 cm, 25 cm

2. 3 cm, 8 cm, 6 cm

3. 50 cm, 80 cm, 100 cm

4. 13 cm, 12 cm, 5 cm

Solution :

1. 7 cm, 24 cm, 25 cm.

Here, the longest side is 25 cm.

25² = 625 and 7² + 24² = 49 + 576 = 625

∴ 25² = 7² + 24²

Here, the square of the longest side equals the sum of squares of the other two sides. Hence, the triangle with sides 7 cm, 24 cm and 25 cm is a right triangle and the length of its hypotenuse, is 25 cm.

2. 3 cm, 8 cm, 6 cm

Here, the longest side is 8 cm.

8² = 64 and 3² + 6² = 9 + 36 = 45

∴ 8² ≠ 3² + 6²

Hence, the triangle with sides 3 cm, 8 cm and 6 cm is not a right triangle.

3. 50 cm, 80 cm, 100 cm

Here, the longest side is 100 cm.

100² = 10000 and

50² + 80² = 2500 + 6400 = 8900

∴ 100² ≠ 50² + 80²

Hence, the triangle with sides 50 cm, 80 cm and 100 cm is not a right triangle.

4. 13 cm, 12 cm, 5 cm

Here, the longest side is 13 cm.

13² = 169 and 12² + 5² = 144 + 25 = 169

∴ 13² = 12² + 5²

Here, the square of the longest side equals the sum of squares of the other two sides. Hence, the triangle with sides 13 cm. 12 cm and 5 cm is a right triangle and the length of its hypotenuse 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 PM² = QM. MR.

Solution :

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

∴ ΔRMP ~ ΔPMQ ~ ΔRPQ (Theorem 6.7)

Now, ΔRMP ~ ΔPMQ

∴ \(\frac{PM}{QM}=\frac{RM}{PM}\)

∴ PM² = QM. MR

Question 3.

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

1. AB² = BC.BD

2. AC² = BC.DC

3. AD² = BD.CD

Solution :

ABD is a triangle right angled at A and AC ⊥ BD.

∴ ΔBCA ~ ΔACD ~ ΔBAD (Theorem 6.7)

1. ΔBCA ~ ΔBAD

∴ \(\frac{AB}{DB}=\frac{CB}{AB}\)

∴ AB² = BC. BD

2. ΔBCA ~ ΔACD

∴ \(\frac{AC}{DC}=\frac{BC}{AC}\)

∴ AC² = BC. DC

3. ΔACD ~ ΔBAD

∴ \(\frac{AD}{BD}=\frac{CD}{AD}\)

∴ AD² = BD . CD

Question 4.

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

Solution :

ABC is an isosceles triangle right angled at C.

Hence, AB is the hypotenuse and the other two sides are equal, i.e., BC = AC

In ΔABC, ∠C = 90°

∴ By Pythagoras theorem,

AB² = BC² + AC²

∴ AB² = AC² + AC² (∵ BC = AC)

∴ AB² = 2AC²

Question 5.

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

Solution :

In ΔABC, AC = BC and AB² = 2AC²

AB² = 2AC²

∴ AB² = AC² + AC²

∴ AB² = AC² + BC² (∵ AC = BC)

Hence, by the converse of Pythagoras theorem, ΔABC is right triangle in which ∠C is a right angle.

Question 6.

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

Solution :

In ΔABC, AB = BC = CA = 2a.

Let AD be its altitude

∴ ∠ADB = ∠ADC = 90°

In ΔADB and ΔADC,

∠ADB = ∠ADC = 90°

AB = AC

AD = AD

∴ By RHS criterion,

= ΔADC

∴ BD = CD

But, BD + CD = BC

∴ BD = CD = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)(2a) = a

Now, in ΔADB, ∠D = 90°

∴ By Pythagoras theorem,

AB² = AD² + BD²

∴ (2a)² = AD² + (a)²

∴ 4a² – a² = AD²

∴ AD² = 3a²

∴ AD = \(\sqrt{3}\)a

All the altitudes of an equilateral triangle are equal.

Hence, each of the altitudes of equilateral ΔABC with side 2a is \(\sqrt{3}\)a.

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.

Solution :

Given: ABCD is a rhombus.

To prove : AB² + BC² + CD² + DA² = AC² + BD²

Proof: ABCD is a rhombus.

∴ AB = BC = CD = DA ……………(1)

Let its diagonals AC and BD intersect at M.

Then, MA = MC = \(\frac{1}{2}\)AC.

MB = MD = \(\frac{1}{2}\)BD and

∠AMB = ∠BMC = ∠CMD = ∠DMA = 90°

In ΔAMB, ∠AMB = 90°

∴ AB² = MA² + MB² (Pythagoras theorem)

∴ AB² = (\(\frac{AC}{2}\))² + (\(\frac{BD}{2}\))²

∴ AB² = (\(\frac{AC}{2}\))² + (\(\frac{BD}{2}\))²

∴ 4AB² = \(\frac{\mathrm{AC}^2}{4}+\frac{\mathrm{BD}^2}{4}\)

∴ 4AB² = AC² + BD²

∴ AB² + AB² + AB² + AB² = AC² + BD²

∴ AB² + BC² + CD² + DA² = AC² + BD²

Question 8.

In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:

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

2. AF² + BD² + CE² = AE² + CD² + BF².

Solution :

Join OA, OB and OC.

Here, in ΔOFA and ΔOFB, ∠F = 90°, in ΔODB and ΔODC, ∠D = 90° and in ΔOEC and ΔOEA. ∠E = 90°.

Then, Pythagoras theorem is applicable in all the triangles.

1. In ΔOFA, ∠F = 90°

∴ OA² = OF² + AF²

∴ AF² = OA² – OF² …………..(1)

In ΔODB, ∠D = 90°

∴ OB² = OD² + BD²

∴ BD² = OB² – OD² …………..(2)

In ΔOEC, OE² + CE²

∴ CE² = OC² – OE² …………..(3)

Adding (1), (2) and (3).

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

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

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

∴ AF² + BD² + CE² = (OA² – OE²) + (OB² – OF²) + (OC² – OD²)

∴ AF² + BD² + CE² = AE² + BF² + CD² (∵ ΔOAE, ΔOBF and ΔOCD are right triangles)

∴ AF² + BD² + CE² = AE² + CD² + BF²

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 :

Here, AB is the wall with window at point A and AC is the ladder.

Then, AC = 10m and AB = 8 m.

In ΔABC, ∠B = 90°.

∴ AC² = AB² + BC² (Pythagoras theorem)

∴ 10² = 8² + BC²

∴ BC² = 10² – 8²

∴ BC² = 100 – 64

∴ BC² = 36

∴ BC = 6 m

Thus, 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?

Solution :

Here, AB is the vertical pole in which the guy wire is attached at point A and AC is the guy wire and 18 m the stake is attached to its end C.

Then, AC = 24 m and AB = 18 m.

In ΔABC, ∠B = 90°

∴ AC² = AB² + BC² (Pythagoras theorem)

∴ 24² = 18² + BC²

∴ BC² = 576 – 324

∴ BC² = 252

∴ BC² = 4 × 9 × 7

∴ BC = 2 × 3 × \(\sqrt{7}\)

∴ BC = 6\(\sqrt{7}\) m

Thus, the stake should be driven 6\(\sqrt{7}\)m far from the base of the pole, so as to make the wire taut.

Question 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 of 1200 km per hour. How far apart will be the two planes after 1\(\frac{1}{2}\) hours?

Solution :

Here, A is the airport, B is the position of the first plane flying due north after 1\(\frac{1}{2}\) hours and C is the position of the second c- plane flying due west after 1\(\frac{1}{2}\) hours.

[Note: For the sake of simplicity, we consider that both the planes are flying at the same height and point A representing the airport is also imagined to be at the same height.]

Then, AB = distance covered by the first plane in 1\(\frac{1}{2}\) hours

= Speed × Time

= 1000 × \(\frac{3}{2}\)

= 1500 km

Similarly, AC = distance covered by the second plane in 1\(\frac{1}{2}\) hours

= Speed × Time

= 1200 × \(\frac{3}{2}\)

= 1800 km

Also, ∠BAC is the angle formed by north direction and west direction.

Hence ∠BAC = 90°

Now, in ΔABC, ∠A = 90°

∴ BC² = AB² + AC² (Pythagoras theorem)

∴ BC² = (1500)² + (1800)²

∴ BC² = 22500 + 32400

∴ BC² = 54900

∴ BC = \(\sqrt{100 \times 9 \times 61}\)

∴ BC = 300\(\sqrt{61}\) km

Thus, the two planes will be 300\(\sqrt{61}\) km apart from each other after 1\(\frac{1}{2}\) hours.

Question 12.

Two poles of heights 6m 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 :

Here, AB and CD are two erect poles of height 6 m and 11 m respectively.

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

Then, AB = 6 m, BD = 12 m, CD = 11 m, ∠B = 90° and ∠D = 90°.

Draw AE || BC.

Then, in quadrilateral ABDE.

∠B = ∠D = ∠E = ∠A = 90°.

Hence, ABDE is a rectangle.

∴ ED = AB = 6m and AE = BD = 12 m.

Then, CE = CD – DE = 11 – 6 = 5m

Now, in ΔAEC, ∠E = 90°.

∴ AC² = AE² + CE² (Pythagoras theorem)

∴ AC² = 12² + 5²

∴ AC² = 144 + 25

∴ AC² = 169

∴ AC = 13 m

Thus, the distance between the tops of the vertical poles 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 AE² + BD² = AB² + DE²

Solution :

In ΔABC, ∠C is a right angle, point D lies on CA and point E lies on CB.

Then, all the four triangles BCD, BCA, ECD and ECA are right triangles and in each of them C is a right angle.

Hence, Pythagoras theorem is applicable in all the four triangles.

In ΔECA, AE² = EC² + CA² ……………..(1)

In ΔBCD, BD² = BC² + CD² ……………..(2)

In ΔBCA, AB² = BC² + CA² ……………..(3)

In ΔECD, DE² = EC² + CD² ……………..(4)

Adding (1) and (2).

AE² + BD² = EC² + CA² + BC² + CD²

= (BC² + CA²) + (EC² + CD²)

= AB² + DE² [By (3) and (4)]

Thus, AE² + BD² = AB² + DE²

Question 14.

The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3CD (see the given figure). Prove that 2AB² = 2AC² + BC².

Solution :

DB = 3CD

∴ BC = DB + CD = 3CD + CD

∴ BC = 4CD …………..(1)

In ΔADB, ∠D = 90°

∴ AB² = AD² + DB² (Pythagoras theorem) …………..(2)

In ΔADC, ∠D = 90°

∴ AC² = AD² + CD² (Pythagoras theorem) …………..(3)

Subtracting (3) from (2),

AB² – AC² = (AD² + DB²) – (AD² + CD²)

∴ AB² – AC² = DB² – CD²

∴ AB² – AC² = (DB + CD) (DB – CD)

∴ AB² – AC² = (BC) (3CD – CD)

∴ AB² – AC² = (BC) (2CD)

Multiplying the equation by 2, we get

2AB² – 2AC² = (BC) (4CD)

∴ 2AB² – 2AC² = (BC) (BC)

∴ 2AB² = 2AC² + BC² [By (1)]

Question 15.

In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD² = 7AB².

Solution :

Given: In equilateral ΔABC, D is a point on BC such that BD = \(\frac{1}{3}\)BC.

To prove: 9AD² = 7AB²

Construction: Draw AM ⊥ BC, such that M lies on BC.

Proof: ΔABC is an equilateral triangle. Suppose, AB = BC = AC = a

In equilateral ΔABC, AM is an altitude.

∴ AM is a median.

∴ BM = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)a

∴ BD = \(\frac{1}{3}\)BC. Hence, DC = \(\frac{2}{3}\)BC

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

DM = BM – BD = \(\frac{1}{2}\)a – \(\frac{1}{3}\)a = \(\frac{1}{6}\)a

In ΔAMB, ∠M = 90°

∴ AB² = AM² + BM²

∴ a² = AM² + \(\frac{1}{4}\)a²

∴ AM² = \(\frac{3}{4}\)a²

In ΔAMD, ∠M = 90°

∴ AD² = AM² + DM²

∴ 9AD² = 7a²

∴ 9AD2 = 7AB² (∵ AB = a)

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.

Solution :

ABC is an equilateral triangle in which AD is an altitude.

Let AB = BC CA- a units.

In an equilateral triangle, an altitude is a median also.

∴ AD is a median.

∴ BD = \(\frac{1}{2}\)BC = \(\frac{a}{2}\)units

In ΔADB, ∠D = 90°

∴ AB² = AD² + BD²

∴ (a)² = AD² + (\(\frac{a}{2}\))²

∴ a² = AD² + \(\frac{a^2}{4}\)

∴ \(\frac{3}{4}\)a² = AD²

∴ 3a² = 4AD²

∴ 3 (side)² = 4 (altitude)²

Question 17.

Tick the correct answer and justify: In ΔABC, 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 :

In ΔABC, AB = 6\(\sqrt{3}\) cm = 10.38 cm (approx),

AC = 12 cm and BC = 6 cm

Here, AC is the longest side.

Then, 12² = 144 and

(6\(\sqrt{3}\))² + (6)² – 108 + 36 = 144

Thus, 12² = (6\(\sqrt{3}\))² + (6)²

Hence, by the converse of Pythagoras theorem, ΔABC is a right triangle in which the longest side AC is the hypotenuse and its opposite angle ∠B is a right angle.

Hence, the correct answer is (C) 90°.