Question

The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 129cm^2. What is the length of the diagonal? Give your answer to 2 decimal places.

Answer 1

Answer: 16.14 cm

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Step-by-step explanation:

• L = x = length of the rectangle
• W = 2x-9 = width of the rectangle, since its 9 less than twice the length

area of rectangle = L*W = 129

L*W = 129

x*(2x-9) = 129

2x^2-9x = 129

2x^2-9x-129 = 0

Apply the quadratic formula. We'll use a = 2, b = -9, c = -129.

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-9)\pm√((-9)^2-4(2)(-129)))/(2(2))\\\\x = (9\pm√(1113))/(4)\\\\x \approx (9\pm33.36165464)/(4)\\\\x \approx (9+33.36165464)/(4)\ \text{ or } \ x \approx (9-33.36165464)/(4)\\\\x \approx (42.36165464)/(4)\ \text{ or } \ x \approx (-24.36165462)/(4)\\\\x \approx 10.59041366\ \text{ or } \ x \approx -6.09041364\\\\`

We ignore the negative solution because a negative length makes no sense.

The length is approximately L = 10.5904 cm.

The width is 2L-9 = 2*10.5904-9 = 12.1808 cm approximately.

As a quick check,

L*W = 10.5904*12.1808 = 128.99954432

which isn't too far off from 129. We have rounding error which is why we don't perfectly land on the target area value. If you wanted to get closer to the value 129, then use more decimal digits in the approximations of L and W.

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If you draw a diagonal in the rectangle, then you form two identical or congruent right triangles.

Focusing on one of those triangles, we have

• a = 10.5904
• b = 12.1808
• c = unknown hypotenuse = diagonal length

Apply the pythagorean theorem

a^2+b^2 = c^2

c = sqrt( a^2 + b^2 )

c = sqrt( (10.5904)^2 + (12.1808)^2 )

c = 16.1408940520653

c = 16.14

The diagonal is roughly 16.14 cm long.