Question

13. A rectangle has a width that is twice as long as its length and an area of 722 square inches. Find the length of the diagonal, rounded to the nearest tenth.

Answer 1

42.5 inches

Explanation:

l = length

w = 2l

A = 722

A = l*w

722 = l * 2l

722 = 2l^2

Divide each side by 2

361 = l^2

Take the square root of each side

sqrt(361) = l

w = 2 * sqrt(361)

We want to find the diagonal so we can use the pythagorean theorem

a^2+ b^2 = c^2 where c is the length of the diagonal

l^2 + w^2 = c^2

(sqrt(361)) ^2 + (2 sqrt(361))^2 = c^2

361+1444 = c^2

1805 = c^2

Take the square root of each side

sqrt(1805) = sqrt(c^2)

42.48529 = c

To the nearest tenth

42.5 =c

Answer 2

The diagonal of the rectangle is approximately 42.5 inches.

Explanation:

The area of a rectangle is given by the following formula:

area = width*length

In this case the width = 2*length, therefore we have:

area = 2*length²

722 = 2*length²

2*length² = 722

length² = 361

length = sqrt(361) = 19 inches

width = 2*length = 2*19 = 38 inches

The diagonal forms a right triangle with the sides of the rectangle, where it is the hypotenuse. Therefore we can use Pytagora's theorem:

diagonal = sqrt(length² + width²)

diagonal = sqrt(19² + 38²) = 42.485 inches

The diagonal of the rectangle is approximately 42.5 inches.