Calculate the diagonal of a rectangle if the base is 14 less than the height and the diagonal is 22 less than twice the height.

We'll note the height of the rectangle as
x.

The base is x - 14.

We'll
apply the Pythagorean theorem into the right angled triangle, whose cathetus are the
base and the height and the hypothenuse is the
diagonal.

We'll note the diagonal as
d.

d^2 = h^2 + b^2

d^2 = x^2 +
(x-14)^2

But, from enunciation, d = 2x -
22.

(2x - 22)^2 = x^2 +
(x-14)^2

We'll expand the squares from both
sides:

4x^2 - 88x + 484 = x^2 + x^2 - 28x +
196

We'll combine like
terms:

4x^2 - 88x + 484 = 2x^2 - 28x +
196

We'll subtract both sides 2x^2 - 28x +
196:

4x^2 - 88x + 484 - 2x^2 + 28x - 196 =
0

We'll combine like
terms:

2x^2 - 60x + 288 =
0

We'll divide by 2:

x^2 - 30x
+ 144 = 0

We'll apply the quadratic
formula:

x1 = [30+sqrt(900 -
576)]/2

x1 = (30+18)/2

x1 =
24

x2 = (30-18)/2

x2 =
6

Since the base is x - 14, the height has to have a value
> 14. For this reason, the only valid value for x is
24.

height = x =
24

base = 24 -
14

base =
10

diagonal = 2x -
22

diagonal = 2*24 -
22

diagonal = 48 -
22

diagonal = d =
26