A rectangle has a width that is twice as long as its lengths and an area of 722 square inches, how do you find the length of the diagonal?

Question

Nathan Axel · Accepted Answer

See below.

If #x# is the length then the width is #2x#. Area is #x# x #2x# = #2x^2# = #722 => #x = #sqrt ((722)/2) ## => x = 19# Length is 19, width is 38. By Pythagoras theorem: Diagonal is #sqrt(19^2 + 38^2) = 42.49# # color(blue)(2.d.p.)#

Hazel Anglin · Answer

undefined To find the length of the diagonal of a rectangle, you can use the Pythagorean theorem.

Let the length of the rectangle be x inches.

Given that the width is twice as long as the length, the width would be 2x inches.

The area of the rectangle is given as 722 square inches, so we have the equation:

x * 2x = 722

Simplifying this equation, we get:

2x^2 = 722

Dividing both sides by 2, we have:

x^2 = 361

Taking the square root of both sides, we get:

x = 19

Therefore, the length of the rectangle is 19 inches.

To find the length of the diagonal, we can use the Pythagorean theorem:

Diagonal^2 = Length^2 + Width^2

Substituting the values, we have:

Diagonal^2 = 19^2 + (2*19)^2

Simplifying this equation, we get:

Diagonal^2 = 361 + 1444

Diagonal^2 = 1805

Taking the square root of both sides, we get:

Diagonal ≈ 42.49 inches

Therefore, the length of the diagonal of the rectangle is approximately 42.49 inches.