Question

If the perimeter of rectangle is 52metere and area is 160m² thae lenghth of rectangle is

Answer 1

Given :-

• Area of the rectangle = 160 m²

• Perimeter of the rectangle = 52 m

To Find :-

• Length of the rectangle

Solution :-

Perimeter of the rectangle :-

$:\implies\sf{Perimeter = 2(l+b)}$

$:\implies\sf{52 = 2(l+b)}$

$:\implies\sf{52/2 = (l+b)}$

$:\implies\sf{26 = (l+b)}$

$:\implies\sf{b=26-l}...(1)$

Area of the rectangle :-

$:\implies\sf{Area = l \times b}$

$:\implies\sf{160 = l \times (26-l)}$

$:\implies\sf{160 = 26l-l^2}$

$:\implies\sf{ 26l-l^2-160=0}$

$:\implies\sf{ -l^2+26l-160=0}$

$:\implies\sf{ l^2-26l+160=0}$

$:\implies\sf{ l^2-10l-16l+160=0}$

$:\implies\sf{ l(l-10)-16(l-10)=0}$

$:\implies\sf{ (l-10)(l-16)=0}$

$:\implies\sf{ (l=16)\:or\:(l=10)}$

So,

Length of the rectangle = 16 m

Breadth of the rectangle = 10 m

To check ↓

➥  Area of the rectangle = l × b

➥  Area = 16 × 10

➥  Area = 160 m²

Hence Verified!

Answer 2

Answer:

Let the length and breadth be = l and b

2 ( l + b ) = 52

=> l + b = 26

=> l = 26 - b ... (1)

lb = 160 ... (2)

(1) in (2) :

b^2 - 26b + 160 = 0

When you will take out the zeroes, you'll get:

b = 10 m , 16 m

l = 16 m , 10 m

Both are the possible values of the breadth and length