NCERT Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.5. These solutions are based on new NCERT Syllabus.

NCERT 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.

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

Question 3.
In the figure, 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:

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.

∴ 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.

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.

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

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.

∵ 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]

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

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.

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

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

⇒ 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.

Now, in the right AABC, using Pythagoras Theorem, we have

Thus, the stake is required to be taken at \(6 \sqrt{7}\)m from the base of the pole to make the wire taut.

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

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

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°.

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.

Solution:

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.

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.

∴ D is the mid-point

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