# 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.

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### 4 Answers

let the base be (B) and the height be (H) , and the diagonal be (D)

given B = H - 14 ..........(1)

and D = 2H - 22 ........(2)

But we know that:

D^2 = H^2 + B^2

==> D =sqrt (B^2 +H^2)

Now substiute in (2)

sqrt(B^2 + H^2) = 2H - 22

square both sides:

==> B^2 + H^2 = 4H^2 - 88H + 484

==> B^2 = 3H^2 - 88H + 484.........(3)

From (1), squre both sides:

B^2 = (H-14)^2

B^2 = H^2 - 28H + 196........(4)

From (3) and (4):

3h^2 - 88H + 484 = H^2 - 28H + 196

Combine like terms:

2H^2 - 60H + 288 = 0

Divide by 2:

==> H^2 - 30H + 144 = 0

==> (H - 6)(H-24) = 0

==> H1 = 6 ==> B1 = 6-14 = -8 ( impossible)

**==> H = 24 **

**==> B = 24-14 = 10 **

**==> D = 2H-22 = 48-22 = 26**

For a rectangle, the base and height are at right angles, so the diagonal , base and height form a Pythagorean Triplet.

Lets take the height as H. We are given that base the 14 less than the height or H-14.

The diagonal is 22 less than twice the height or 2H-22

Therefore (2H-22)^2 = H^2 + (H-14)^2

=> 4H^2 + 484 - 88H = H^2 + H^2 + 196 - 28H

=> 2H^2 -88H + 28H + 484 - 196 =0

=> 2H^2 - 60H + 288 =0

=> H^2 - 30H + 144 =0

=> H^2 - 24H- 6H + 144= 0

=> H(H - 24) - 6 ( H-24) =0

=> (H - 24)(H - 6) =0

So H can be 24 or 6.

The base of the rectangle is H-14 and it cannot be negative so H= 6 can be eliminated.

**Therefore the height is 24, base is 10 and the diagonal is 26**

Let heifght of rectangle = h.

Then its base b = h-14 .

the diagonal d = -22+2h

to find d .

We know that b^2+h^2 = d^2 for the rectangle by Pythagoras.

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

h^2-28h+14^2 + h^2 = 22^2-88h +4h^2

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

0 = (484-196) +(28-88)h+(4-2)h^2

2h^2-60h -288 = 0

Divide by 2:

h^2 -30h+144 = 0

(h-6)(h-24) = 0

h = 24 , base = 24 -14 = 10

Therefore diagonal of the rectangle = -22+2x = -22+2*24 = 26

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**