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.

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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

 

 

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