How do you solve this?

Question

The length of a rectangle is 31 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the length and width of the rectangle

 

 

Answer 1

Width of the rectangle say x.

According to given problem length of rectangle = 31+2x

Diagonal of the rectangle would be 2+31+2x = 33+2x

We know that rectangle length,width and it's diagonal forms a right angle triangle.

From the pythageron therom

(33+2x)^2 = (31+2x)^2+x^2

1089+4x^2+132x = 961+4x^2+124x+x^2

Bring all terms to one side.

961-1089+4x^2-4x^2+124x-132x+x^2 = 0

x^2-8x-128 = 0

x^2-16x+8x-128 = 0

x(x-16)+8(x-16) = 0

(x-16)(x+8) = 0

(x-16) = 0 and (x+8) = 0

x = 16 and x = -8

Rectangle dimensions are always positive value.

So x = 16

Width of the rectangle = 16nches.

Length of rectangle = 31+2x = 31+32 = 63inches.