A rectangle is drawn so the width is 28 inches longer than the height. If the rectangle's diagonal measurement is 52 inches, find the height.

Guest
Apr 20, 2017

#1
**+4 **

width = w

height = w + 28

from the Pythagorean theorem:

w^{2} + (w + 28)^{2} = 52^{2}

w^{2} + (w + 28)(w + 28) = 2704

w^{2} + w^{2} + 28w + 28w + 784 = 2704

2w^{2} + 56w - 1920 = 0

from the quadratic formula:

\(w = {-56 \pm \sqrt{56^2-4(2)(-1920)} \over 2(2)} = \frac{-56\pm \sqrt{18496}}{4}=\frac{-56\pm 136}{4}=-14\pm 34\)

Since we are looking for the length of a line, the answer is positive.

w = -14 + 34 = 20

and

height = 20 + 28 = 48 inches

hectictar
Apr 20, 2017

#1
**+4 **

Best Answer

width = w

height = w + 28

from the Pythagorean theorem:

w^{2} + (w + 28)^{2} = 52^{2}

w^{2} + (w + 28)(w + 28) = 2704

w^{2} + w^{2} + 28w + 28w + 784 = 2704

2w^{2} + 56w - 1920 = 0

from the quadratic formula:

\(w = {-56 \pm \sqrt{56^2-4(2)(-1920)} \over 2(2)} = \frac{-56\pm \sqrt{18496}}{4}=\frac{-56\pm 136}{4}=-14\pm 34\)

Since we are looking for the length of a line, the answer is positive.

w = -14 + 34 = 20

and

height = 20 + 28 = 48 inches

hectictar
Apr 20, 2017