An arhitect wants to draw a rectangle with a diagonal of 20 inches. The length of the rectangle is to be 8 inches more than twice the width.what dimensions should she make the rectangle

The diagonal of a rectangle = sqrt(w^2 + l^2)

w = width

l = length

In this problem,

The diagonal = 20 in

w = x

l = 2x + 8

Let's plug our numbers into the formula above.

20in = sqrt((x)^2 + (2x + 8)^2)

Let's simplify the inside of the sqrt

20 in = sqrt(5x^2 + 32x + 64)

Now, let's square both sides.

400 = 5x^2 + 32x + 64

Subtract 400 from both sides.

0 = 5x^2 + 32x - 336

Factor

0 = (5x - 28)(x + 12)

Set both terms equal to zero and solve.

x + 12 = 0

Subtract 12 from both sides.

x = -12

5x - 28 = 0

Add 28 to both sides.

5x = 28

Divide both sides by 5

x = 28/5

The width cant be a negative number so now we know that the only real solution is 28/5

Let's plug 28/5 into our length equation.

Length = 2(28/5) + 8 = 56/5 + 8 = 96/5

In conclusion,

Length = 96/5 inches

Width = 28/5

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