Question

rectangle has twice the area of square the width of rectangle is 8cm and length is 8cm more than the sides of square find area of square

Answer 1

Define x:

Let the length of the square be x

Find the area of the square in term of x:

Area = length x length

Area = x²

Find the area of the rectangle in term of x:

Area = Length x Breadth

Area =8 ( x + 8) = 8x + 64

Since the area of the rectangle is twice the area of the square:

2x² = 8x + 64

2x² - 8x - 64 = 0

2(x - 8) (x + 4) = 0

(x - 8) (x + 4) = 0

x  = 8 or x = -4 (rejected, because length cannot be negative)

Find the area of the square:

Area = length x Length

Are = 8 x 8 = 64 cm²

Answer: The area is 64 cm²

Answer 2

Given that : -

A rectangle has twice the area of a square.

width of rectangle = 8 cm

length of rectangle = 8 cm more than the sides of square.


Let the side of the square be x.

So,

length of rectangle = (8 + x) cm

Area of the rectangle = length × breadth = 8 × (x + 8) cm²

Area of the square = side² = x²

Now,

area of rectangle = 2 × area of square

=> 8(x + 8) = 2x²

=> 4(x + 8) = x²

=> 4x + 32 = x²

=> x² - 4x - 32 = 0

=> x² - 8x + 4x - 32 = 0

=> x(x - 8) + 4(x - 8) = 0

=> (x - 8)(x + 4) = 0

So,

x + 4 = 0 => x = - 4
But side of square cannot be negative.

So,

x - 8 = 0 => x = 8

Therefore , Measure of each side of the square = 8 cm

Area of the square = (8 × 8) cm² = 64 cm²