Question

The breadth of a rectangle is 5 cm and its diagonal is 13 cm. Find it's Perimeter.

Answer 1

Answer:

• The perimeter of the rectangle is 34 cm.

Step-by-step explanation:

Given :-

• Breadth of the rectangle = 5 cm.
• Diagonal of the rectangle = 13 cm.

To Find :-

• Perimeter of the rectangle.

Basic Terms :-

• Length : Length is a measure of how long an object is or the distance between two points.
• Breadth : Breadth is the distance from the right side to the left side of a shape.
• Diagonal : A diagonal is a straight line connecting the opposite corners of a polygon through its vertex.

Formula Used :-

⦾ To find the perimeter we know that,

✦ Perimeter of rectangle = 2(l + b) ✦

where,

• l = length.
• b = breadth.

Solution :-

➊ To find the perimeter of rectangle it is important to have it's length. So let the length of the rectangle be l. So firstly, by using Pythagoras Theorem we will find out length.

Given :

• Breadth = 5 cm.
• Diagonal = 13 cm.

According to the question by using the formula we get,

↦ (Length)² + (Breadth)² = (Diagonal)²

↦ (l)² + (5)² = (13)²

↦ (l)² + 25 = 169

↦ (l)² = 169 - 25

↦ (l)² = 144

↦ l = √144

➦ l = 12

∴ Hence, the length of the rectangle is 12 cm.

➋ Now we have the length. So, now we will find out the perimeter of the rectangle.

Given :

• Length = 12 cm.
• Breadth = 5 cm.

According to the question by using the formula we get,

↦ Perimeter of rectangle = 2(l + b)

↦ Perimeter of rectangle = 2(12 + 5)

↦ Perimeter of rectangle = 2(17)

↦ Perimeter of rectangle = 2 × 17

➦ Perimeter of rectangle = 34 cm

∴ Hence, perimeter of rectangle is 34 cm.

❖ Verification ❖

➲ We have the perimeter of rectangle as 34 cm. So, now we will verify both LHS and RHS.

↦ Perimeter of rectangle = 2(l + b)

Putting perimeter = 34 we get,

↦ 34 = 2(12 + 5)

↦ 34 = 2(17)

↦ 34 = 2 × 17

↦ 34 = 34

➦ LHS = RHS

∴ Hence, Verified ✔

Answer 2

Given:

• The breadth of a rectangle is 5 cm
• its diagonal is 13 cm

To Find:

• It's perimeter

Solution:

★ Here we have the measure of the diagonal and the breadth of the rectangle using pythagoras therom let's find the measure of the length of the rectangle so, that we can find the perimeter of the rectangle

Pythagoras Therom ;

${\large }\bf{(d)}^{2} = {(b)}^{2} + {(l)}^{2}$

Here;

• Diagonal ( D ) = 13cm
• Breadth ( B ) = 5cm
• Length ( L ) = x

Applying concept ;

${ : \implies} \rm \: {(AB)}^{2} + {(BD)}^{2} = {(AD)}^{2} \\ \\ \\ { : \implies} \rm {(5cm)}^{2} + {(x)}^{2} = {(13cm)}^{2} \\ \\ \\ { : \implies} \rm 25 {cm}^{2} + {x}^{2} = 169 {cm}^{2} \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm {x}^{2} = 169 {cm}^{2} - 25 {cm}^{2} \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm {x}^{2} = 144 {cm}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\ \\ \\ { : \implies} \rm x = \sqrt{144 {cm}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm x = 12cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Henceforth, the length of the rectangle is 12cm

★ Now let's find the perimeter

${ : \implies} \bf \: perimeter = 2(l + b) \: \: \: \\ \\ \\ { : \implies} \bf \: perimeter = 2(5 + 12) \\ \\ \\ { : \implies} \bf \: perimeter = 2(17) \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \bf \: perimeter = 34cm \star \: \: \: \: \:$

• Henceforth the perimeter of the rectangle is 36cm