Question

The perimeter of the rectangle is 80 cm. The area of the rectangle is acm2 . (i) show that x 2 40x + a = 0.​

Answer 1

Answer:

Proved

Step-by-step explanation:

Given

$p = 80$ ---- perimeter

$a \to area$

Required

$x^2 + 40x + a = 0$

The perimeter is calculated as:

$p = 2(x + y)$

Where

$x,y \to$ the rectangle dimension

So, we have:

$2(x + y) = 80$

Divide by 2

$x + y = 40$

Make y the subject

$y = 40 - x$

The area of the rectangle is:

$a = xy$

$a = x * (40 -x)$

$a = 40x -x^2$

Equate to 0:

$x^2 - 40x + a = 0$