RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

RBSE 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.
(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) Let in △ABC,
AB = 7 cm, BC = 24 cm, AC = 25 cm
AB² + BC² = (7)² + (24)² = 49 + 576 = 625
AC² = (25)² = 625
Now AB² + BC² = AC²
ABC is a right triangle and length of hypotenuse = 25 cm.

(ii) Let in △PQR,
PQ = 3 cm, QR = 8 cm, PR = 6 cm
PQ² + PR² = (3)² + (6)²= 9 + 36 = 45
QR² = (8)² = 64
Here PQ² + PR² ≠ QR²
∴ △PQR is not a right triangle.

(iii) Let in △MNP,
MN = 50 cm, NP = 80 cm, MP = 100 cm
MN² + NP² = (50)² + (80)²
= 2500 + 6400 = 8900
MP² = (100)² = 10000
Here MP² ≠ MN² + NP²
∴ △MNP is not a right triangle.

(iv) Let in △ABC,
AB = 13 cm, BC = 12 cm, AC = 5 cm
BC² + AC² = (12)² + (5)² = 144 + 25 = 169
AB² = (13)² = 169
∴ AB² = BC² + AC²
∴ △ACB is 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 ⊥ QR. Show that PM² = QM . MR.
Solution:
Given : In right △PQR angle P is a right angle. M is a position QR such that PM ⊥ QR.
To prove : PM² = QM . MR

Proof: ∵ ∠P = 90° (Given)
∴ ∠1 + ∠2 = 90° …(i)
∠M = 90°
In △PMQ, ∠1 + ∠3 + 90° = 180°
∠1 + ∠3 = 90° ….(ii)
(∠M = 90°)
From (i) and (ii), ∠1 + ∠2 = ∠1 + ∠3
∠2 = ∠3
In △QPM and △RPM, ∠3 = ∠2
(Proved above)
∠5 = ∠6 (Each 90°)
∴ △QMP ~ △PMR (AA similarity)
\(\frac{a r(\Delta \mathrm{QMP})}{a r(\Delta \mathrm{PMR})}\) = \(\frac{\mathrm{PM}^{2}}{\mathrm{MR}^{2}}\)
[∵ If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides ]

Question 3.
In Fig., ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD . CD

Solution:
In △DAB and △DCA,
∠D = ∠D (Common angle)
∠A = ∠C (Each 90°)
∴ △DAB ~ △DCA …(i)
(By AA similarity criterion)
Now in △DAB and △ACB,
∠B = ∠B (Common angle)
∠A = ∠C (Each 90°)
∴ △DAB ~ △ACB ….(ii)
From (i) and (ii),
∴ △DAB ~ △ACB ~ △DCA

(i) In △ABC ~ △ADC,
\(\frac{A B}{B D}\) = \(\frac{B C}{A B}\)
⇒ AB² = BC × BD (Hence Proved)

(ii) In △ABC ~ △ADC,
\(\frac{A C}{B C}\) = \(\frac{D C}{A C}\)
AC DC BC ” AC
⇒ AC² = BC × DC (Hence Proved)

(iii) In △ACD ~ △ABD,
\(\frac{A D}{C D}\) = \(\frac{B D}{A D}\)
⇒ AD² = BD × CD (Hence Proved)

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Given : △ABC is an isosceles triangle whose angle C is a right angle.
To prove : AB² = 2AC²

Proof: In △ACB,
∠C = 90°
AC = BC (Given)
AB² = AC² + BC²
(By Pythagoras Theorem)
= AC² + AC² (∵ BC = AC)
AB² = 2AC² (Hence Proved)

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.
Solution
Given : △ABC is an isosceles triangle in which AC = BC
and AB² = 2AC²
To prove : △ABC is a right triangle.

Proof : AB² = 2AC² (Given)
AB² = AC² + AC²
AB2 = AC² + BC² [∵AC = BC]
∴ By converse of Pythagoras Theorem, △ABC is a right triangle. (Hence Proved)

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
△ABC is an equilateral triangle whose side is 2a
AD ⊥ BC
AB = AC = BC = 2a
△ADB = △ADC (By RHS congruence)
∴ BD = DC = a

Now in right △ADB,
AB² = AD² + BD²
(2a)² = AD² + (a)²
4a² – a² = AD²
AD² = 3a²
AD = \(\sqrt {3}\)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 : Diagonals AC and BD of rhombus ABCD intersect each other at O.

To prove:
AB² + BC² + CD² + AD² = AC² + BD²
Proof : ∵ Diagonals of a rhombus bisect each other at right angle
∴ AO = CO, BO = DO
∴ Angle at O are right angles.
In △AOB, ∠AOB =90°
∴ AB² = AO² + BO² ….(i)
Similarly, BC² = CO² + BO² …..(ii)
CD² = CO² + DO² ….(iii)
and DA² = DO² + AO² . ….(iv)
Adding (i), (ii), (iii) and (iv) we obtain
AB² + BC² + CD² + DA² = 2AO² + 2CO² + 2BO² + 2DO²
= 4AO² + 4BO²
(∵ AO = CO and BO = DO)
= (2AO)² + (2BO)² = AC² + BD²
∴ AB² + BC² + CD² + AD² =AC ² + BD²
(Hence Proved)

Question 8.
In Fig., 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
(ii) AF² + BD² + CE² = AE² + CD² + BF²
Solution:
Given : In △ABC in which OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

To prove:
(i) OA² + OB² + OC² – OD² – OE² – OF²
= AF² + BD² + CE²

(ii) AF² + BD² + CE² = AE² + CD² + BF²
Construction : Join OA, OB and OC.
Proof: (i) In right △AFO,
OA² = OF² + AF² (By Pythagoras Theorem)
or AF² = OA² – OF² ….(i)
In right △BDO,
OB² = BD² + OD²
(By Pythagoras Theorem)
⇒ BD² = OB² – OD² …..(ii)
In right △CEO,
OC² = CE² + OE²
(By Pythagoras Theorem)
⇒ CE² = OC² – OE² ….(iii)
Adding (i), (ii), (iii)
AF² + BD² + CE² = OA² – OF² + OB² – OD² + OC² – OE²
= OA² + OB² + OC² – OD² – OE² – OF²
(Hence Proved)

(ii) Again,
AF² + BD² + CE² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)
= AE² + CD² + BF²
AE² = AO² – OE²
∵ CD² = OC² – OD²
BF² = OB² – OF²
(Hence Proved)

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:
According to the question, height of the window from ground (PQ) = 8 m,
Length of ladder (PR) = 10 m

Distance between the foot of the ladder and the base of the wall QR = ?
In right △PQR,
PQ² + QR² = PR²
(By Pythagoras Theorem)
(8)² + (QR)² = (10)²
64 + QR² = 100
QR² = 100 – 64
QR = \(\sqrt {36}\)
QR =6m.
Distance of the foot of the ladder from the base of the wall = 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:

Let the height of the pole AB = 18 m
Length of wire AC = 24 m
C is the position of the stake. Its distance from the base of the pole BC = ?
In right △ABC,
AB² + BC² = AC²
(18)² + (BC)² = (24)²
324 + (BC)² = 576
BC² = 576 – 324
BC = \(\sqrt {252}\) = 6\(\sqrt {7}\) m
Hence the distance of the stake from the base of the pole = 6\(\sqrt {7}\) m or 15.87 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 planes after 1\(\frac{1}{2}\) hours?
Solution:
Speed of the first aeroplane = 1000 km/hr.

Distance covered by first aeroplane in \(\frac{1}{2}\) hours towards north OA = 1000 × \(\frac{3}{2}\)
⇒ OA = 1500 km
Speed of the another aeroplane = 1200 km/hr
Distance covered by another aeroplane in 1\(\frac{1}{2}\) hours towards west
OB = 1200 × \(\frac{3}{2}\)
⇒ OB = 1800 km.
In right △AOB,
AB² = AO² + OB²
AB² = (1500)² + (1800)²
AB = \(\sqrt{2250000+3240000}\)
= \(\sqrt{5490000}\) = 100\(\sqrt{549}\)
= 100\(\sqrt{9 \times 61}\) = 100 × 3 \(\sqrt{61}\)
AB = 300\(\sqrt{61}\) km.
Distance between the two aeroplanes
= 300\(\sqrt{61}\) 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:
According to the question,
height of the first pole, AB = 11 m
height of the second pole (CD) = 6 m
Distance between the poles = 12 m

Draw CE ⊥ AB from C.
BE = DC = 6 m
AE = AB – BE
= (11 – 6) m = 5 m
In right △AEC,
AC² = AE² + EC²
AC = \(\sqrt{(5)^{2}+(12)^{2}}\)
= \(\sqrt{25+144}\) = \(\sqrt{169}\) = 13
∴ The distance between the tops of the 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 AE² + BD² = AB² + DE².
Solution:
Given : △ABC is a right triangle in which right angle is at C, D and E are points on the sides CA and CB respectively, i.e.,

CD = AD = \(\frac{1}{2}\)AC
BE =EC= \(\frac{1}{2}\)BC
To prove : AE² + BD² = AB² + DE
Proof: In right △BCA,
AB² = BC² + CA²
(By Pythagoras Theorem) (i)
In right △ECD,
DE² = EC² + DC²
(By Pythagoras Theorem) ….(ii)
In right △ACE,
AE² = AC² + CE² ….(iii)
In right △BCD,
BD² = BC² + CD² ….(iv)
Adding (iii) and (iv),
AE² + BD² = AC² + CE² + BC² + CD² = [AC² + CB²] + [CE² + DC²]
= AB² + DE²
or AE² + BD² = AB² + DE² (Hence Proved)

Question 14.
The perpendicular from A on side BC of a △ABC intersects BC at D such that DB = 3 CD (see Fig.). Prove that 2AB² = 2 AC² + BC².

Solution:
Given : In △ABC, AD ⊥ BC and BD = 3CD
To prove: 2AB² = 2AC² + BC²
Proof: In right triangles ADB and ADC,
AB² = AD² + BD²
AC² = AD² + DC²
∴ AB² – AC² = BD² – DC² = 9CD² – CD²
[∵ BD = 3CD]
= 8CD² = 8\(\left(\frac{\mathrm{BC}}{4}\right)^{2}\)
[∵ BC = DB + CD
= 3CD + CD
= 4CD
∴ CD = \(\frac{1}{4}\)BC]
∴ AB² – AC² = \(\frac{\mathrm{BC}^{2}}{2}\)
or 2(AB² – AC²) = BC²
or 2AB² – 2AC² = BC²
∴ 2AB² = 2AC² + BC² (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 9AD² = 7AB².
Solution:
Given : D is a point on the side BC of an equilateral triangle ABC such that BD = \(\frac{1}{3}\)BC
To prove : 9AD² = 7AB²
Construction : Draw perpendicular AM from A on BC, i.e., AM ⊥ BC
Proof: △AMB = △AMC
[By RHS Rule as AM = AM and AB = AC]

∴ BM = MC = \(\frac{1}{2}\)BC
Again, BD = \(\frac{1}{3}\)BC and DC = (1 – \(\frac{1}{3}\))BC = \(\frac{2}{3}\)BC
(∵ BC is divided into three parts at D)
Now in △ADC, ∠C is an acute angle
∴ AD² = AC² + DC² – 2DC × MC
= AC² + (\(\frac{2}{3}\)BC)² – 2(\(\frac{2}{3}\)BC)\(\frac{1}{2}\)BC
[∵ DC = \(\frac{2}{3}\)BC and MC = \(\frac{2}{3}\)BC]
= AB² + \(\frac{4}{9}\)AB² – \(\frac{2}{3}\)AB²
[∵ AC = BC = AB]
= \(\left(1+\frac{4}{9}-\frac{2}{3}\right)\)AB
= \(\left(\frac{9+4-6}{9}\right)\)AB = \(\frac{7}{9}\)AB
∴ AD² = \(\frac{7}{39}\)AB²
or 9AD² = 7AB² (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 altitudes.
Solution:
Given : △ABC is an equilateral △ in which AB = BC = AC
and AD ⊥BC.

To prove : 3AB² = 4AD²
Proof: In △ABC, for an equilateral triangle
AB = BC = AC – 2a (say)
∵ AD ⊥ BC
∴ BD = DC = \(\frac{1}{2}\)BC = a
In right triangle ADB,
AB² = AD² + BD²
(2a)² = AD² + (a)²
4a² = AD² + a² ⇒ 4a² – a² = AD²
[∵ AB = 2a
a = \(\frac{\mathrm{AB}}{2}\)]
⇒ AD² = 3a² = 3\(\left[\frac{\mathrm{AB}}{2}\right]^{2}\)
⇒ AD² = \(\frac{3 \mathrm{AB}^{2}}{4}\)
3AB² = 4AD² (Hence proved)

Question 17.
Tick die correct answer and justify : In △ABC, 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:
∵ AC = 12 cm, AB = 6\(\sqrt {3}\) cm,
and BC = 6 cm
AC² = (12)² = 144 cm
AB² + BC² = (6\(\sqrt {3}\))² + (6)² = 108 + 36
AB² + BC² = 144 = (12)² = AC²
∴ AB² + BC² = AC²
By converse of Pythagoras Theorem, in △ABC there is right angle at 8.
∴ ∠B =90°
∴ Correct choice =(C)