What is the area of a rectangle if its perimeter is 62 decimeter and its one side is one decimeter less than its diagonal?

Answer 1

x = unknown side;

d = length of diagonal;

d-1 = other side.

You also know that the perimeter, which is the sum of all sides, equals 62. Therefore:

x + (d-1) + x + (d-1) = 62 -->

2x + 2d - 2 = 62 -->

2x + 2d = 64 -->

2d = 64 - 2x -->

d = 32 - x ------- equation (1)

The diagonal of the rectangle and its two sides form a right triagle, where the diagonal is the hypotenuse. So, apply Pythagorean Theorem (in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides) and you have:

d² = x² + (d - 1)² -->

d² = x² + d² - 2d + 1 -->

x² = 2d - 1 -->

x² + 1 = 2d --> d = (x² + 1)/2 ---equation (2)

Substitute equation (2) into equation (1) and solve for x:

d = 32 - x -->

(x² + 1)/2 = 32 - x -->

x² + 1 = 64 - 2x --->

x² + 1 - 64 + 2x = 0 --->

x² + 2x - 63 = 0

Apply the quadratic formula, to find roots x1 and x2:

x1,2 = [-b ± √(b² - 4ac)]/2a, where:

a = 1, b = 2, c = -63, thus:

x1,2 = [-2 ± √((2² - 4*1*(-63))]/2*1 -->

x1,2 = [-2 ± √(4 + 252)]/2 = (-2 ± √256)/2 -->

x1,2 = (-2 ± 16)/2 --> x1 = 14/2 = 7, x2 = -18/2 = -9

Ignore the second root, x2 = -9, as the rectangle's side cannot be negative.

Now, you know that one of the rectangle's sides is 7. Substitute this in equation (1), to find the diagonal, d:

d = 32 - x -->

d = 32 - 7 -->

d = 25

So the other side is (d-1):

d - 1 = 25 - 1 = 24.

Now, you know both sides, so multiply them to find the rectangle's area:

Area = side 1 * side 2 -->

Area = 7 * 24 = 168 dm²