In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 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) The sides are : 7 an, 24 cm, 25 cm
Here, (7 cm)2 = 49 cm2
(24 cm)2 = 576 cm2
(25 cm)2 = 625 cm2
∵ (49 + 576)cm2 = 625 cm2
∴ It is a right triangle.
Hypotenuse = 25 cm.

(ii) The sides are: 3 cm, 8 cm, 6 cm
Here, (3 cm)2 = 9 cm2
(8 cm)2 = 64 cm2
(6 cm)2 = 36 cm2
∵ (9 + 36) ≠ 64 cm2
∴ It is not a right triangle.

(iii) The sides are : 50 cm, 80 cm, 100 cm
Here, (50 cm)2 = 2500 cm2
(80 cm)2 = 6400 cm2
(100 cm)2 = 10000 cm2
∵ (2500 + 6400) cm2 ≠ 10000 cm2
∴ It is not a right triangle.

(iv) The sides are : 13 cm, 12 cm, 5 cm
Here, (13 cm)2 = 169 cm2
(12 cm)2 = 144 cm2
(5 cm)2 = 25 cm2
∵ (144 + 25)cm2 = 169 cm2
∴ It is a right triangle.
Hypotenuse = 13 cm.

MP Board Class 10th 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:
In ∆QMP and ∆QPR,
∠QMP = ∠QPR [Each 90°]
∠Q = ∠Q [Common]
⇒ ∆QMP ~ ∆QPR …. (1) [AA similarity]
Again, in ∆PMR and ∆QPR,
∠PMR = ∠QPR [Each = 90°]
∠R = ∠R [Common]
⇒ ∆PMR ~ ∆QPR …… (2) [AA similarity]
From (1) and (2), we have
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 1

Question 3.
In the figure, ABD is a triangle, right angled at A and AC ⊥ BD.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 2
Show that
(i) AB2 = BC.BD
(ii) AC2 = BC.DC
(iii) AD2 = BD.CD
Solution:
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 3
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 4

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
We have, right ∆ABC such that ∠C = 90° and AC = BC.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 5
∴ By Pythagoras theorem, we have AB2 = AC2 + BC2 = AC2 + AC2 = 2AC2
[∵ BC = AC (given)]
Thus, AB2 = 2AC2

Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Solution:
We have, an isosceles AABC such that BC = AC.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 6
Also, AB2 = 2AC2
∴ AB2 = AC2 + AC2
But AC =BC
∴ AB2 = AC2 + BC2
∴ Using the converse of Pythagoras theorem, ∠ACB = 90°
f.e., ∆ABC is a right angled triangle.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
In equilateral triangle, altitude bisects the base.
⇒ AD = DB
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 7
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 8

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:
Let us have a rhombus ABCD.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 9
∵ Diagonal of a rhombus bisect each other at right angles.
∴ OA = OC and OB = OD
Also, ∠AOB = ∠BOC [Each = 90°]
And ∠COD = ∠DOA [Each = 90°]
In right ∆AOB, we have,
AB2 = OA2 + OB2 …… (1)
[Using Pythagoras theorem]
Similarly, in right ∆BOC,
BC2 = OB2 + OC2 …… (2)
In right ∆COD,
CD2 = OC2 + OD2 …… (3)
In right ∆AOD,
DA2 = OD2 + OA2 ……. (4)
Adding (1), (2), (3) and (4)
AB2 + BC2 + CD2 + DA2
= [OA2 + OB2] + [OB2 + OC2] + [OC2 + OD2] + [OD2 + OA2]
= 2OA2 + 2OB2 + 2 OC2 + 2OD2 = 2[OA2 + OB2 + OC2 + OD2]
= 2[OA2 + OB2 + OA2 + OB2]
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 10
Thus, sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 11
Solution:
We have a point in the interior of a ∆ABC such that
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
(i) Let us join OA, OB and OC.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 12
In right ∆OAF, by Pythagoras theorem
OA2 = OF2 + AF2 …(1)
Similarly, from right triangle ODB and OEC, we have
OB2 = BD2 + OD2, …(2)
and OC2 = CE2 + OE2 …(3)
Adding (1), (2) and (3), we get OA2 + OB2 + OC2
= (AF2 + OF2) + (BD2 + OD2) + (CE2 + OE2)
⇒ OA2 + OB2 + OC2
= AF2 + BD2 + CE2 + (OF2 + OD2 + OE2)
⇒ OA2 + OB2 + OC2 – (OD2 + OE2 + OF2)
= AF2 + BD2 +CE2
⇒ OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= AF2 + BD2 + CE2

(ii) In right triangle OBD and triangle OCD, by Pythagoras theorem:
OB2 = OD2 + BD2 and OC2 = OD2 + CD2
⇒ OB2 – OC2 = OD2 + BD2 – OD2 – CD2
⇒ OB2 – OC2 = BD2 – CD2 ….. (1)
Similarly, we have
OC2 – OA2 = CE2 – AE2 …… (2)
and OA2 – OB2 = AF2 – BF2 ….. (3)
Adding (1), (2) and (3), we get (OB2 – OC2) + (OC2 – OA2) + (OA2 – OB2) = (BD2 – CD2) + (CE2 – AE2) + (AF2 – BF2)
⇒ 0 = BD2 + CE2 + AF2 – (CD2 + AE2 + BF2)
⇒ BD2 + CE2 + AF2 = CD2 + AE2 + BF2
or AF2 + BD2 + CE2 = AE2 + BF2 + CD2

MP Board Class 10th Maths Solutions 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:
Let PQ be the ladder and PR be the wall
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 13
⇒ PQ = 10 m, PR = 8 m
Now, in the right ∆PQR, PQ2 = PR2 + QR2
⇒ 102 = 82 + QR2
[using Pythagoras theorem]
⇒ QR2 = 102 – 82 = (10 + 8)(10 – 8)
= 18 × 2 = 36
QR = \(\sqrt{36}\) = 6m
Thus, the distance of the foot of the ladder from the base to 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:
Let AB is the wire and BC is the vertical pole. The point A is the stake.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 14
Now, in the right AABC, using Pythagoras Theorem, we have
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 15
Thus, the stake is required to be taken at \(6 \sqrt{7}\)m from the base of the pole to make the wire taut.

MP Board Class 10th Maths Solutions 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:
Let the point A represent the airport. Plane-I fly towards North,
∴ Distance of the plane-I from the airport after \(1 \frac{1}{2}\) hours = speed × time
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 16
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 17

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:
Let the two poles AB and CD are such that the distance between their feet AC = 12m.
∵ Height of pole-1, AB = 11 m
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 18

MP Board Class 10th 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:
We have a right ∆ABC such that ∠C = 90°.
Also, D and E are points on CA and CB respectively.
rain We have a right ∆ABC such that ∠C = 90°.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 19
Let us join AE and BD.
In right ∆ACB, using Pythagoras theorem
AB2 = AC2 + BC2 …… (1)
In right ∆DCE, using Pythagoras theorem,
DE2 = CD2 + CE2 …… (2)
Adding (1) and (2), we get
AB2 + DE2 = [AC2 + BC2] + [CD2 + CE2]
= AC2 + BC2 + CD2 + CE2 = [AC2 + CE2] + [BC2 + CD2] …. (3)
In right ∆ACE,
AC2 + CE2 = AE2 …… (4)
In right ∆BCD,
BC2 + CD2 = BD2 …….. (5)
From (3), (4) and (5), we have AB2 + DE2 = AE2 + BD2 or AE2 + BD2 = AB2 + DE2

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB2 = 2AC2 + BC2.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 20
Solution:
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 21
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 22

MP Board Class 10th Maths Solutions 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:
We have an equilateral ∆ABC; in which D is a point on BC such that BD = \(\frac{1}{3}\) BC.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 23
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 24

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:
We have an equilateral ∆ABC, in which AD ⊥ BC.
Since, an altitude in an equilateral A, bisects the corresponding side.
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 25
∴ D is the mid-point
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 26

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 17.
Tick the 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:
(C): We have, AB = \(6 \sqrt{3}\) cm, AC = 12 cm, and BC = 6 cm
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5 27

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5