Question

find the perimeter of a rectangle whose one side measures 20m and the diagonal is 29m

answer it quick and no spamming or you'll regret it​

Answer 1

heyy ARMY !

Let the length (AB) be xm.

Breadth

(BC) = 20m. (Given)

Diagonal (AC) = 29m (given)

(29)^2 = (x)^2 + (20)^2

ATQ

(AC)^2 = (AB)^2 + (BC)^2

841 = x^2 + 400

841 - 400 = x^2

V441 = x

21 = x

Hence length of the rectangle be 21m.

Perimeter of rectangle = 2(1 + b)

= 2(20 + 21)

= 2*41

= 81m

Answer 2

Answer :

›»› The perimeter of a rectangle is 82 m.

Given :

• Breadth of a rectangle = 20 m.
• Diagonal of a rectangle = 29 m.

To Find :

• Perimeter of a rectangle = ?

Solution :

Let us assume that, the length of a rectangle is "x m".

From pythagorean theorem

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

→ (d)² = (x)² + (b)²

→ (29)² = x² + (20)²

→ 841 = x² + (20)²

→ 841 = x² + 400

→ x² = 841 - 400

→ x² = 441

→ x = √441

→ x = 21

The length of a rectangle is 21 m.

Now,

As we know that

→ Perimeter of rectangle = 2(length * breadth)

→ Perimeter of rectangle = 2(21 + 20)

→ Perimeter of rectangle = (2 * 21) + (2 * 20)

→ Perimeter of rectangle = 42 + (2 * 20)

→ Perimeter of rectangle = 42 + 40

→ Perimeter of rectangle = 82

Hence, the perimeter of a rectangle is 82 m.