# Class 9 SELINA Solutions Maths Chapter 13 - Pythagoras Theorem [Proof and Simple Applications with Converse]

## Pythagoras Theorem [Proof and Simple Applications with Converse] Exercise Ex. 13(A)

### Solution 1

The pictorial representation of the given problem is given below,

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.

(i)Here, AB is the hypotenuse. Therefore applying the Pythagoras theorem we get,

_{}

Therefore, the distance of the other end of the ladder from the ground is 12m

### Solution 2

Here , we need to measure the distance AB as shown in the figure below,

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.

Therefore , in this case

_{}

Therefore the required distance is 64.03 m.

### Solution 3

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 _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

The length of PR_{}

### Solution 4

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

The length of AB is 4 cm.

### Solution 5

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.

Here, we consider the _{}and applying Pythagoras theorem we get,

_{}Therefore, the length of AD is 8.7 cm

### Solution 6

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,

_{}

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, from (i) and(ii),

_{}

_{}

### Solution 7

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 segments.

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

_{}

Therefore, x is 13cm

### Solution 8

Let, the sides of the triangle be, _{}

Now, _{}

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.

### Solution 9

The diagram of the given problem is given below,

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.

_{}

### Solution 10

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.

So BM || AD also BM = AD.

_{}

### Solution 11

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

⇒ AB^{2} + BC^{2} = AC^{2}

⇒ BC^{2 }= AC^{2} - AB^{2}

⇒ BC^{2 }= (24)^{2} - (12)^{2}

⇒ BC^{2 }= 576 - 144

⇒ BC^{2 }= 432

⇒ BC= cm

∴ AC = 24 cm and BC =cm

### Solution 12

_{}

_{}

## Pythagoras Theorem [Proof and Simple Applications with Converse] Exercise Ex. 13(B)

### Solution 1

First, we consider the _{} and applying
Pythagoras theorem we get,

_{}

First,
we consider the _{} and applying
Pythagoras theorem we get,

_{}

From
(*i*) and (*ii*) we get,

_{}

Hence Proved.

### Solution 2

In
equilateral Δ ABC, AD _{} BC.

Therefore, BD = DC = x/2 cm.

_{}

### Solution 3

The pictorial form of the given problem is as follows,

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

From (*i*) and (*ii*) we get,

_{}

### Solution 4

We draw , PM,MN,NR

Since, M andN are the mid-points of the sides QR and PQ respectively, therefore, PN=NQ,QM=RM

(i)

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras

theorem we get,

_{}

Adding (*i*) and (*ii*) we get,

_{}

(ii)

We consider the _{}and applying Pythagoras theorem we get,

_{}

_{}^{.}

(iii)

We consider the _{}and applying Pythagoras theorem we get,

_{}

(iv)

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

Adding (*i*) and (*ii*) we get,

_{}

### Solution 5

In triangle ABC,_{ }B = 90^{o} and D is the mid-point of BC. Join AD. Therefore, BD=DC

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Similarly, we get from rt. angle triangles ABC we get,

_{}

From (i) and (ii) ,

_{}

### Solution 6

Since, ABCD is a rectangle angles A,B,C and D are rt. angles.

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Similarly, we get from rt. angle triangle BDC we get,

_{}

Adding (i) and (ii) ,

_{}

### Solution 7

_{}

### Solution 8

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Using Pythagorean theorem we have from the above diagram:

OA^{2 }= AH^{2 }+ OH^{2 }= AH^{2 }+ AE^{2 }

OC^{2 }= CG^{2 }+ OG^{2 }= EB^{2 }+ HD^{2 }

OB^{2 }= EO^{2 }+ BE^{2 }= AH^{2 }+ BE^{2 }

OD^{2 }= HD^{2 }+ OH^{2 }= HD^{2 }+ AE^{2 }

Adding these equalities we get:

OA^{2 }+ OC^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2 }

OB^{2 }+ OD^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2 }

From which we prove that for any point within the rectangle there is the relation

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

Hence Proved.

### Solution 9

Here, we first need to join OA, OB, and OC after which the figure becomes as follows,

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 _{}and applying Pythagoras theorem we get,

_{}

_{Similarly, from triangles, BPO,COQ,AOQ,CPO and BRO we get the following results,}

_{}

Adding (*i*), (*ii*) and (*iii*),we get

_{}

Adding (*iv*), (*v*) and (*vi*),we get ,

_{}

From (*vii*) and (*viii*), we get,

AR^{2} + BP^{2} + CQ^{2} = AQ^{2} + CP^{2} + BR^{2}

Hence proved.

### Solution 10

Diagonals of the rhombus are perpendicular to each other.

_{}

### Solution 11

We consider the _{}and applying Pythagoras theorem we get,

_{}

### Solution 12

In an isosceles triangle ABC; AB = AC and D is point on BC produced. Construct AE perpendicular BC.

We consider the rt. angled _{}and applying Pythagoras theorem we get,

_{}

_{}

### Solution 13

We consider the rt. angled _{}and applying Pythagoras theorem we get,

_{}

Similarly, in _{,}

_{}

Putting, _{}from (ii) in (i) we get,

_{}

Hence Proved.

### Solution 14

_{}

### Solution 15

Here,

_{}