Pythagoras Theorem Class-9th Concise Selina ICSE Maths Solutions Chapter-13. We provide step by step Solutions of Exercise / lesson-13 Pythagoras Theorem for ICSE Class-9 Concise Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-13 A and Exe-13 B, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

## Pythagoras Theorem Class-9th Concise Selina ICSE Maths Solutions Chapter-13

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Exercise 13 A,

Exercise-13 B,

### Exercise – 13 A, Pythagoras Theorem Class-9th Concise Selina ICSE Maths Solutions

#### Question 1

A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

The pictorial representation of the given problem is given below, Therefore, the distance of the other end of the ladder from the ground is 12m

#### Question 2

A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.

Here , we need to measure the distance AB as shown in the figure below, Therefore the required distance is 64.03 m.

#### Question 3

In the figure: PSQ = 90o, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

…………….

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the ΔPQS and applying Pythagoras theorem we get,

#### Question 4

The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ABD = BCD = 90o. Calculate the length of AB.

…………………

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔBCD and applying Pythagoras theorem we get, The length of AB is 4 cm.

#### Question 5

AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

Since ABC is an equilateral triangle therefore, all the sides of the triangle are of same measure and the perpendicular AD will divide BC in two equal parts.. Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, we consider the and applying Pythagoras theorem we get #### Question 6

In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC= 3 cm. Calculate the length of OC.

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the and applying Pythagoras theorem we get, #### Question 7

In triangle ABC,

AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm2. Find x.

Here, the diagram will be, We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since, ABC is an isosceles triangle, therefore perpendicular from vertex will cut the base in two equal Therefore, x is 13cm

#### Question 8

If the sides of triangle are in the ratio 1 : : 1, show that is a right-angled triangle. Here, in (i) it is shown that, square of one side of the given triangle is equal to the addition of square of other two sides. This is nothing but Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Therefore, the given triangle is a right angled triangle.

#### Question 9

Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m; find the distance between their tips.

The diagram of the given problem is given below, #### Question 10

In the given figure, AB//CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.

……………….

Take M be the point on CD such that AB = DM.

So DM = 7cm and MC = 10 cm

Join points B and M to form the line segment BM. #### Question 11

In the given figure, ∠B = 90°, XY || BC, AB = 12cm, AY = 8cm and AX: XB = 1: 2 = AY: YC. Find the lengths of AC and BC.

………………

Given that AX: XB = 1: 2 = AY: YC.

Let x be the common multiple for which this proportion gets satisfied.

So, AX = 1x and XB = 2x

AX + XB = 1x + 2x = 3x

⇒ AB = 3x .….(A – X – B)

⇒ 12 = 3x

⇒ x = 4

AX = 1x = 4 and XB = 2x = 2 × 4 = 8

Similarly,

AY = 1y and YC = 2y

AY = 8…(given)

⇒ 8 = y

∴ YC = 2y = 2 × 8 = 16

∴ AC = AY + YC = 8 + 16 = 24 cm

∆ABC is a right angled triangle. …. Given

∴ By Pythagoras Theorem, we get

⇒ AB2 + BC2 = AC2

⇒ BC= AC2 – AB2

⇒ BC= (24)2 – (12)2

⇒ BC= 576 – 144

⇒ BC= 432 #### Question 12

In ΔABC, …….. Find the sides of the triangle, if:

(i) AB = (x – 3) cm, BC = (x + 4) cm and AC = (x + 6) cm

(ii) AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm  ### Pythagoras Theorem Class-9th Concise Selina ICSE Maths Solutions  Exercise – 13 B

#### Question 1

In the figure, given below, AD  BC. Prove that: c2 = a2 + b2 – 2ax.

………….

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔACD  and applying Pythagoras theorem we get, #### Question 2

In equilateral Δ ABC, AD….  BC and BC = x cm. Find, in terms of x, the length of AD. #### Question 3

ABC is a triangle, right-angled at B. M is a point on BC. Prove that:

AM2 + BC2 = AC2 + BM2. 