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 cm^{2}

(24 cm)^{2} = 576 cm^{2}

(25 cm)^{2} = 625 cm^{2}

∵ (49 + 576)cm^{2} = 625 cm^{2}

∴ It is a right triangle.

Hypotenuse = 25 cm.

(ii) The sides are: 3 cm, 8 cm, 6 cm

Here, (3 cm)^{2} = 9 cm^{2}

(8 cm)^{2} = 64 cm^{2}

(6 cm)^{2} = 36 cm^{2}

∵ (9 + 36) ≠ 64 cm^{2}

∴ It is not a right triangle.

(iii) The sides are : 50 cm, 80 cm, 100 cm

Here, (50 cm)^{2} = 2500 cm^{2}

(80 cm)^{2} = 6400 cm^{2}

(100 cm)^{2} = 10000 cm^{2}

∵ (2500 + 6400) cm^{2} ≠ 10000 cm^{2}

∴ It is not a right triangle.

(iv) The sides are : 13 cm, 12 cm, 5 cm

Here, (13 cm)^{2} = 169 cm^{2}

(12 cm)^{2} = 144 cm^{2}

(5 cm)^{2} = 25 cm^{2}

∵ (144 + 25)cm^{2} = 169 cm^{2}

∴ 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 PM^{2} = 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) AB^{2} = BC.BD

(ii) AC^{2} = BC.DC

(iii) AD^{2} = BD.CD

Solution:

Question 4.

ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}.

Solution:

We have, right ∆ABC such that ∠C = 90° and AC = BC.

∴ By Pythagoras theorem, we have AB^{2} = AC^{2} + BC^{2} = AC^{2} + AC^{2} = 2AC^{2}

[∵ BC = AC (given)]

Thus, AB^{2} = 2AC^{2}

Question 5.

ABC is an isosceles triangle with AC = BC. If AB^{2} = 2AC^{2}, prove that ABC is a right triangle.

Solution:

We have, an isosceles AABC such that BC = AC.

Also, AB^{2} = 2AC^{2}

∴ AB^{2} = AC^{2} + AC^{2}

But AC =BC

∴ AB^{2} = AC^{2} + BC^{2}

∴ 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,

AB^{2} = OA^{2} + OB^{2} …… (1)

[Using Pythagoras theorem]

Similarly, in right ∆BOC,

BC^{2} = OB^{2} + OC^{2} …… (2)

In right ∆COD,

CD^{2} = OC^{2} + OD^{2} …… (3)

In right ∆AOD,

DA^{2} = OD^{2} + OA^{2} ……. (4)

Adding (1), (2), (3) and (4)

AB^{2} + BC^{2} + CD^{2} + DA^{2}

= [OA^{2} + OB^{2}] + [OB^{2} + OC^{2}] + [OC^{2} + OD^{2}] + [OD^{2} + OA^{2}]

= 2OA^{2} + 2OB^{2} + 2 OC^{2} + 2OD^{2} = 2[OA^{2} + OB^{2} + OC^{2} + OD^{2}]

= 2[OA^{2} + OB^{2} + OA^{2} + OB^{2}]

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) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

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

OA^{2} = OF^{2} + AF^{2} …(1)

Similarly, from right triangle ODB and OEC, we have

OB^{2} = BD^{2} + OD^{2}, …(2)

and OC^{2} = CE^{2} + OE^{2} …(3)

Adding (1), (2) and (3), we get OA^{2} + OB^{2} + OC^{2}

= (AF^{2} + OF^{2}) + (BD^{2} + OD^{2}) + (CE^{2} + OE^{2})

⇒ OA^{2} + OB^{2} + OC^{2}

= AF^{2} + BD^{2} + CE^{2} + (OF^{2} + OD^{2} + OE^{2})

⇒ OA^{2} + OB^{2} + OC^{2} – (OD^{2} + OE^{2} + OF^{2})

= AF^{2} + BD^{2} +CE^{2}

⇒ OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

= AF^{2} + BD^{2} + CE^{2}

(ii) In right triangle OBD and triangle OCD, by Pythagoras theorem:

OB^{2} = OD^{2} + BD^{2} and OC^{2} = OD^{2} + CD^{2}

⇒ OB^{2} – OC^{2} = OD^{2} + BD^{2} – OD^{2} – CD^{2}

⇒ OB^{2} – OC^{2} = BD^{2} – CD^{2} ….. (1)

Similarly, we have

OC^{2} – OA^{2} = CE^{2} – AE^{2} …… (2)

and OA^{2} – OB^{2} = AF^{2} – BF^{2} ….. (3)

Adding (1), (2) and (3), we get (OB^{2} – OC^{2}) + (OC^{2} – OA^{2}) + (OA^{2} – OB^{2}) = (BD^{2} – CD^{2}) + (CE^{2} – AE^{2}) + (AF^{2} – BF^{2})

⇒ 0 = BD^{2} + CE^{2} + AF^{2} – (CD^{2} + AE^{2} + BF^{2})

⇒ BD^{2} + CE^{2} + AF^{2} = CD^{2} + AE^{2} + BF^{2}

or AF^{2} + BD^{2} + CE^{2} = AE^{2} + BF^{2} + CD^{2}

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, PQ^{2} = PR^{2} + QR^{2}

⇒ 10^{2} = 8^{2} + QR^{2}

[using Pythagoras theorem]

⇒ QR^{2} = 10^{2} – 8^{2} = (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 AE^{2} + BD^{2} = AB^{2} + DE^{2}.

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

AB^{2} = AC^{2} + BC^{2} …… (1)

In right ∆DCE, using Pythagoras theorem,

DE^{2} = CD^{2} + CE^{2} …… (2)

Adding (1) and (2), we get

AB^{2} + DE^{2} = [AC^{2} + BC^{2}] + [CD^{2} + CE^{2}]

= AC^{2} + BC^{2} + CD^{2} + CE^{2} = [AC^{2} + CE^{2}] + [BC^{2} + CD^{2}] …. (3)

In right ∆ACE,

AC^{2} + CE^{2} = AE^{2} …… (4)

In right ∆BCD,

BC^{2} + CD^{2} = BD^{2} …….. (5)

From (3), (4) and (5), we have AB^{2} + DE^{2} = AE^{2} + BD^{2} or AE^{2} + BD^{2} = AB^{2} + DE^{2}

Question 14.

The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB^{2} = 2AC^{2} + BC^{2}.

Solution:

Question 15.

In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD^{2} = 7AB^{2}.

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