# 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(a)

Correct option: (iii) right-angled triangle

Let the three sides of a triangle be 5x, 12x and 13x respectively.

Now,

(5x)^{2} + (12x)^{2} = 25x^{2}
+ 144x^{2} = 169x^{2} = (13x)^{2}

Since, the sum of the squares of two sides is equal to the square of the largest side, the given triangle is the right-angled triangle.

### Solution 1(b)

Correct option: (ii) 8 cm and 6 cm

In right-angled triangle

Hypotenuse = 10 cm

Let the other two sides be 3x and 4x respectively.

Now,

(3x)^{2} + (4x)^{2} = (10)^{2}

⇒
9x^{2} + 16x^{2} = 100

⇒
25x^{2} = 100

⇒
x^{2} = 4

⇒ x = 2

Therefore,

3x = 3(2) = 6 cm

4x = 4(2) = 8 cm

### Solution 1(c)

Correct option: (i)

Construction: Draw median AD.

BD = CD (AD is the median)

Now, median of an isosceles triangle is also perpendicular to the base.

∴ AD ⊥ BC

In right-angled ΔADC,

Therefore,

### Solution 1(d)

Correct option: (iv) 100 cm

ABCD is a rhombus.

Diagonal AC = 40 cm

Diagonal BD = 30 cm

Now, all sides of a rhombus are equal and diagonals of a rhombus bisect each other at right angles.

So, in right-angled triangle AOB,

Therefore, the perimeter of rhombus ABCD

= AB + BC + CD + AD

= (25 + 25 + 25 + 25) cm

= 100 cm

### Solution 1(e)

Correct option: (i) 4 cm

In right-angled triangle BCD,

In right-angled triangle ABD,

### Solution 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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.

_{}

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

### Solution 1(a)

Correct option: (ii) 2AC^{2}

In right-angled triangle ACB,

### Solution 1(b)

Correct option:
(ii) AB^{2} + DE^{2}

In right-angled ΔACE,

(I)

In right-angled ΔDCB,

(II)

Adding equations (I) and (II),

### Solution 1(c)

Correct option: (i) AC × BC

(I)

*Back answer is option (ii).

### Solution 1(d)

Correct option: (iv) AB^{2}

ΔABC is an isosceles triangle right-angled at C.

Therefore,

### Solution 2

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 3

In equilateral Δ ABC, AD _{} BC.

Therefore, BD = DC = x/2 cm.

_{}

### Solution 4

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 5

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 6

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 7

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 8

_{}

### Solution 9

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 10

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.

## Pythagoras Theorem [Proof and Simple Applications with Converse] Exercise Test Yourself

### Solution 1

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 2

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 3

_{}

_{}

### Solution 4

PQRS is a rhombus with side length 10 cm

Diagonal QS = 16 cm

Now, all sides of a rhombus are equal and diagonals of a rhombus bisect each other at right angles.

Then, SO = QO = 8 cm

So, in right-angled triangle POQ,

Therefore, the length of other diagonal of rhombus is 12 cm.

### Solution 5

In right-angled ΔAOB,

(I)

In right-angled ΔCOD,

(II)

In right-angled ΔAOD,

(III)

In right-angled ΔBOC,

(IV)

Adding equations (I) and (II),

### Solution 6

Diagonals of the rhombus are perpendicular to each other.

_{}

### Solution 7

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

_{}

### Solution 8

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 9

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

_{}

Similarly, in _{,}

_{}

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

_{}

Hence Proved.

### Solution 10

_{}

### Solution 11

Here,

_{}

### Solution 12

Construction: Draw median AP.

In ΔABC, AB = AC = x and BC = 10 cm

So, ABC is an isosceles triangle.

BP = CP (AP is the median)

Now, median of an isosceles triangle is also perpendicular to the base.

∴ AP ⊥ BC

Also,

A(ΔABC) = 60 cm^{2}

Now, in right-angled ΔAPC,

### Solution 13

In right-angled DADB,

In right-angled ΔADC,

From equations (I) and (II),

Hence, the length of CD is cm.

### Solution 14

ABCD is a
quadrilateral in which ∠A
+ ∠D
= 90^{o}

Construction: Extend AB and DC and let them intersect at point M.

In ΔAMD, ∠A + ∠D = 90^{o}

⇒ ∠M = 90^{o}

Now,

In right-angled ΔAMC,

In right-angled ΔBMD,

In right-angled ΔAMD,

In right-angled ΔBMC,

Adding equations (I) and (II),