Question

Will give brainliest!!!

The length of a rectangle is 2 inches more than the width. If the diagonal is radical 20 inches, find the area of the rectangle.

A) 4 in^2
B) 8 in^2
C) 16 in^2
D) 32 in^2

Answer 1

Answer:

B.  8 in^2.

Step-by-step explanation:

Let the width be x inches, then the length = x + 2 inches.

So by The Pythagoras theorem:

(√20)^2 =  x^2 + (x + 2)^2

x^2 + x^2 + 4x + 4 = 20

2x^2 + 4x - 16 = 0

x^2 + 2x - 8 = 0

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

x = 2, -4    (we ignore the negative so x = 2)

So the area = 2 * (2 + 2)

= 8 in^2.

Answer 2

Answer:

B

Step-by-step explanation:

L = W + 2

$a^{2}$ + $b^{2}$ = $c^{2}$

(W + 2)2 + W2 = $\sqrt{20} ^{2$  

2$w^{2}$ + 4W + 4 = 20

2$w^{2}$ + 4W − 16 = 0

W = 2 or W = −4

And,

L = W + 2

L = 2 + 2

L = 4

Therefore,

A = LW = (4)(2) = 8