Question

a rectangle is drawn so the width is 5 inches longer than the height. if the rectangles diagonal measurement is 38 inches, find the height. give your answer rounded to 1 decimal place. here is a desmos diagram to help. inches question help: message instructor

Explanation:

Step1: Let the height be $x$ inches.

The width is $x + 5$ inches.

Step2: Apply the Pythagorean theorem.

In a rectangle, $height^{2}+width^{2}=diagonal^{2}$. So $x^{2}+(x + 5)^{2}=38^{2}$.
Expanding $(x + 5)^{2}=x^{2}+10x + 25$, the equation becomes $x^{2}+x^{2}+10x + 25=1444$.
Combining like - terms, we get $2x^{2}+10x+25 - 1444 = 0$, or $2x^{2}+10x-1419 = 0$.

Step3: Use the quadratic formula.

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b = 10$, $c=-1419$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(10)^{2}-4\times2\times(-1419)=100 + 11352=11452$.
Then $x=\frac{-10\pm\sqrt{11452}}{4}=\frac{-10\pm106.976}{4}$.
We take the positive root since length cannot be negative. $x=\frac{-10 + 106.976}{4}=\frac{96.976}{4}=24.244\approx24.2$.

Answer:

$24.2$