Question

A square poster has sides measuring 2 feet less than the sides of a square sign. If the difference between their areas is 40 square​ feet, find the le

Answer 1

Let us assume side length of a square sign = x feet.

Side of the square poster is 2 feet less than the sides of a square sign.

Therefore, side of the square poster in terms of x is =  (x-2) feet.

Given the difference between their areas is 40 square​ feet.

We know, area of a square = (side)^2.

Therefore, we can setup an equation,

(Square sign side length)^2 -  ( square poster side length)^2 = 40.

(x)^2 - (x-2)^2 = 40.

Expanding (x-2)^2 = x^2 +(2)^2 -2(x)(2) = x^2 + 4 - 4x, we get

x^2- (x^2 + 4 - 4x ) = 40.

Distributing minus sign over parenthesis, we get

x^2 -x^2 -4 +4x =40.

Combining like terms, x^2-x^2=0, we get

0-4+4x=40.

-4 +4x =40.

Adding 4 on both sides, we get

-4 +4 +4x =40+4.

4x = 44.

Dividing both sides by 4.

4x/4 = 44/4.

x= 11.

Therefore,  side length of a square sign = 11 feet.

The square poster is 2 feet less than the sides of a square sign.

2 less than 11 is 11-2 = 9 feet.

Therefore, the square poster is 9 feet.