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RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

April 18, 2022 by Safia Leave a Comment

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.

RBSE Solutions for Class 10 Maths 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:
Given : In right △PQR angle P is a right angle. M is a position QR such that PM ⊥ QR.
To prove : PM2 = QM . MR
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 1
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 ]
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 2

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
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 3
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)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 4
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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 5
Proof : AB2 = 2AC2 (Given)
AB2 = AC2 + AC2
AB2 = AC2 + BC2 [∵AC = BC]
∴ By converse of Pythagoras Theorem, △ABC is a right triangle. (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 6
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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 7
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)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Question 8.
In Fig., O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 8
(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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 9
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)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 10
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:
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 11
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

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 12
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
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 13
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.

RBSE Solutions for Class 10 Maths 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:
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.,
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 14
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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 15
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)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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]
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 16
∴ 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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 17
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)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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)

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