Question

a rectangle and a square have the same perimeter of 80m. If the lenght of the rectangles is 6m more than the breadth,find area of the rectangle​

Answer 1

Answer:

Area of Rectangle = 391 m²

Step-by-step explanation:

Correct Question :-

A Rectangle has Perimeter 80 m . If the Length of the Rectangle is 6m more than the Breadth , find Area of the Rectangle​ ?

We Have :-

Rectangle and a Square have the same perimeter of 80m

Length of the rectangle is 6m more than the breadth

To Find :-

Area Of Rectangle

Formula Used :-

Perimeter of Square = 4 * side

Perimeter of Rectangle = 2 ( Length + Breadth )

Area of Rectangle = Length * Breadth

Solution :-

Let the Length be x + 6

Breadth = x

Perimeter of Rectangle = 2 ( x + 6 + x )

80  = 2 ( 6 + 2x )

40 = 6 + 2x

2x = 34

x = 17

Length = 23 m

Breadth = 17 m

Area of Rectangle = 23 * 17

                              = 391 m²

Area of Rectangle = 391 m²

Answer 2

Answer:

${\boxed{\bold\red{Area\:of\:rectangle=391\:{m}^{2} }}}$

Step-by-step explanation:

Given:-

• Perimeter of rectangle = Perimeter of square = 80 m
• Length = Breadth + 6
• Area of rectangle = ?

${\boxed{\bold\green{Perimeter\:of\:rectangle=2(l+b)}}}$

${\boxed{\bold\green{Area\:of\:rectangle=l \times b }}}$

$\rule{300}{1}$

Let breadth of rectangle be x m.So the length of rectangle be (x + 6) m

→ $\tt 80 = 2(x + 6 + x)$

→ $\tt 80 = 2(2x + 6)$

→ $\tt 80 = 4x + 12$

→ $\tt 4x = 80 - 12$

→ $\tt 4x = 68$

→ $\tt x =\dfrac{68}{4}$

→ ${\boxed{\tt{x = 17\:m}}}$

$\rule{300}{1}$

Dimensions:-

→ $\tt Breadth = x =17 \: m$

→ $\tt Length = x + 6 = 17+6 = 23\:m$

$\tt Area = Length \times Breadth$

→ $\tt Area = (23\times 17)\:{m}^{2}$

→ ${\boxed{\tt\green{Area = 391\:{m}^{2} }}}$

$\therefore\bold{\underline{Area\:of\:rectangle=391\:{m}^{2} }}$