JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5

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.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5

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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 1
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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 2
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°
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 3
∴ By Pythagoras theorem,
AB² = BC² + AC²
∴ AB² = AC² + AC² (∵ BC = AC)
∴ AB² = 2AC²

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5

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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 5
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
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 6
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

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5

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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 7
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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 8
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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 9
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.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5

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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 10
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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 11
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.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5

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.
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 12
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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 13
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
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 14
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
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 15
∴ 9AD² = 7a²
∴ 9AD2 = 7AB² (∵ AB = a)

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5

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 :
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 - 16
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°.

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