A rectangle is drawn so the width is 5 inches longer than the height. If the rectangle's diagonal measurement is 52 inches,

Answer 1

Let x=height

Let y=width

The diagonal is the hypotenuse of the rectangle=52 inches

x²+y²= 52² (Pythagoras theorem)

y=x+5

x²+(x+5)² =52²

x²+(x+5)(x+5)=2704

x²+x²+5x+5x+25=2704

2x²+10x+25-2704=0

2x²+10x-2679=0

This has brought us to quadratic equation.

You can solve using General formulae method to get the value of x and then determine y as well.

Answer 2

h=height

h+5=width (or length)

use the Pythagorean Theorem

52^2=h^2+(h+5)^2

2704=h^2+h^2+10h+25

2h^2+10h+25-2704=0

2h^2+10h-2679=0

use the Quadratic Formula

(-10±√(10^2-4*2*[-2679]))/(2*2)

(-10±√21,532)/4

(-10±146.737)/4

(-10+146.737)/4

136.737/4

34.184=height

34.184+5=39.184=length

h=34.184 in

l=39.184 in

now you have the length, width and diagonal of the rectangle and you can find the area or the perimeter

Answer 3

Let w be the width of the rectangle and h be the height.

From the question we know that w = h + 5.

From the pythagorean theorem we know that w^2 + h^2 = 52

So we can plug in to get h^2 + (h+5)^2 = 52.

Once you FOIL out the (h+5)^2 term, and put everything on one side of the equation, you'll see that you have a quadratic equation.

You can use the quadratic formula, or any other method you like, to solve for h and then get w too.