# A right triangle has a perimeter of 24 cm and a hypotenuse of 10 cm. Find the sides x and y, x > y, that make the right angle of the triangle.

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Let the sides of the triangle be x, y and z. Here z is the hypotenuse given as 10. The perimeter of the triangle equal to x+ y + z = 24.

Now as this is a right triangle, we know that x, y and z form a Pythagorean triplet and x^2+ y^2 = z^2.

So we have x^2 + y^2 = 10^2 = 100.

Also, using the perimeter x+ y + z = 24

=> x + y = 24 - 10

=> x+ y = 14

=> y = 14 - x

So (14-x)^2 + x^2 = 100

=> 196 + x^2 - 28x +x^2 = 100

=> 2x^2 - 28x + 96 =0

=> x^2 - 14x + 48 = 0

=> x^2 - 6x - 8x + 48 =0

=> x(x-6) -8(x-6) =0

=> (x-6)*(x-8) =0

Therefore x can be 6 or 8. But as it is given that x> y we take x = 8.

So y = 14 - 8 =6

**So we have the sides of the triangle as x = 8, y = 6 and the hypotenuse equal to 10.**

Given the perimeter = 24 cm

The hypotenuse = 10

Then we know that the perimeter of the triangle is the sum of all three sides:

Let the sides be x , y , and the hypotenuse = 10

==> x + y + 10 = 24

==> x + y = 14

==> x = 14 - y ............(1)

Also we know that:

x^2 + y^2 = 10^2

==> x^2 + y^2 = 100 .............(2)

now we will substitute (1) in (2):

==> (14-y)^2 + y^2 = 100

==> 196 - 28y + y^2 + y^2 = 100

==?> 2y^2 - 28y + 196 = 100

==> 2y^2 - 28y + 96 = 0

Let us divide by 2:

==> y^2 -14y + 48 = 0

==> we will factor:

==> ( y - 8 ) ( y- 6)

==> y1= 8 ==> x1= 6

==> y2 = 8 ==> x2= 6

But given : x> y

**==> x = 8 and y = 6**

We'll write the perimeter of the triangle as the sum of the lengths of the sides:

x + y + 10 = 24 (1)

Since it is a right angle triangle, we'll use the Pythagorean theorem to calculate the hypothenuse:

x^2 + y^2 = 10^2 (2)

We'll write x from (1), with respect to y:

x = 24 - 10 - y

x = 14 - y (3)

We'll substitute (3) in (2):

(14 - y)^2 + y^2 = 100

196 - 28y + y^2 + y^2 - 100 = 0

2y^2 - 28y + 96 = 0

We'll divide by 2:

y^2 - 14y + 48 = 0

We'll apply the quadratic formula:

y1 = [14 + sqrt(196 - 192)]/2

y1 = (14 + 2)/2

y1 = 8

y2 = (14-2)/2

y2 = 6

x1 = 14 - y1 = 14 - 8 = 6

x2 = 14 - y2 = 14 - 6 = 8

**Since form enunciation we have the condition x>y, we'll take x = 8 and y = 6 as the cathetus of the right angle triangle.**

The perimeter of the triangle = 24cm

The length of the hypotenuse = 10.

To find the sides x and y that make the right angle.

x^2+y^2 = square of the hypotenuse = 10^2 by Pytha goras theorem.

Therefore y^2 = 10^2-x^2. Or y = sqrt(100-x^2).

Therefore the perimeter : x+ y+sqrt(x^2+y^2) = 24

x+sqrt(100-x^2) + 10 = 24.

sqrt(100-x^2) = 24-10-x = 14-x.

sqrt(100-x^2) = (14-x).

We square both sides:

100 -x^2 = (14-x)^2 = 196-28x+x^2.

100 -x^2 = 196 -28x+x^2.

Subtract 100 -x^2 from both sides:

0 = 196 -28x+x^2 - 100 +x^2.

Or 2x^2 -28x +96 = 0.

2x^2 - 16x -12x +96 = 0

2x(x-8) - 12(x-8) = 0.

(x-8)(2x-12) = 0

x-8 = 0 , or 2x-12 = 0.

x-8 = 0 gives x= 8.

2x-12 = 0 gives 2x= 12, or x = 12/2 = 6.

Therefore x = 8, or y = 6.

So the right angle making sides of the triangle are :x = 8 and y = 6.