The perimeter is given of right triangle is 24 cm if its hypotenuse is 10 cm find its area
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)
...
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 = 6 ==> x2= 8
But given : x> y
==> x = 8 and y = 6
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)
...
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 = 6 ==> x2= 8
But given : x> y
==> x = 8 and y = 6
