Question

a rectangle is 3cm longer than it is wide. the diagonal is 15cm. find the dimensions of the rectangle

Answer 1

I thought, it was a bit odd. :)

Solve simultaneous equations:
Width + 3 = Height
Width^2 + Height ^2 = 15^2

Answer 2

let the width be =x cm
then length is = (x+3) cm

Withe the diagonal, length and width form a right angled triangle whose hypotenuse is the diagonal

Using Pythagoras theorem

(15)^2 = (x)^2 + (x+3)^2
225 = x^2 + x^2 + 6x + 9
225 = 2x^2 + 6x + 9
or 2x^2 + 6x - 216 = 0
Factorising the quadratic equation
2x^2 + 6x - 216
= 2(x^2 + 3x - 108)
= 2(x^2 + 12x - 9x - 108)
= 2[x(x + 12) -9(x + 12)]
= 2[(x - 9)(x + 12)]

so 2(x-9)(x+12) = 0
=> either x-9 = 0 or x+12 = 0
if x-9 = 0 ==> x = 9
if x+12 = 0 => x = -12

since length can never be negative, so the usable value is x = 9

so width = 9 cm
length = 9+3 = 12 cm