Question

You have two rectangles, rectangle ABCD and EFGH. The length of AB is x meters. The length of CD is 12 meters. The length of EF is 16 meters. The length of FG is x - 2 meters. You want to find the value of x so that the rectangles have the same area.

Answer 1

Let the area of the rectangle ABCD be denoted by $A_1$.

Therefore, $A_1=x\times 12=12x$........(Equation 1) [Reason: Area of any rectangle is the product of it's length and breadth]

Likewise, let the area of the rectangle EFGH be denoted by $A_2$

Therefore, $A_2=16 \times (x-2)=16(x-2)=16x-32$....(Equation 2) [Reason: Area of any rectangle is the product of it's length and breadth]

Since the area of the two rectangles are equal, we will have:

$A_1=A_2$

Therefore, equating the two equations (Equation 1 and Equation 2) we get:

$12x=16x-32$

$12x-16x=-32$

$-4x=-32$

$x=\frac{-32}{-4}=8$

Thus, the value of x such that the rectangles have the same area is 8 meters.