Question

the area of a right angled triangle is 480 CM square if the base of the triangle is 8 cm more than twice the height of the triangle then find the sides of the triangle

Answer 1

So, b is 2*20 + 8 = 48cm.
Third side be y
Then,
y^2 = 48^2 + 20^2
y = 52


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Answer 2

Answer:

The sides of the triangle are 20 cm , 48 cm and 52 cm

Step-by-step explanation:

We are given that the base of the triangle is 8 cm more than twice the height of the triangle

Let height be x

So, Base = 2x+8

Area of triangle = $\frac{1}{2} \times Base \times Height$

                          = $\frac{1}{2} \times 2x+8 \times x$  

The Area of the triangle is 480 sq.cm

So, $480 = \frac{1}{2} \times 2x+8 \times x$  

$480 = \frac{1}{2} \times (2x^2+8x)$  

$480 \times 2 = (2x^2+8x)$  

$960= (2x^2+8x)$  

$2x^2+8x-960=0$  

$x^2+4x-480=0$  

$x^2+24x-20x-480=0$  

$x(x+24)-20(x+24)=0$  

$(x+24)(x-20)=0$  

$x=-24,20$  

Side cannot be negative

So, x = 20

So, The height of triangle is 20 cm

Base = 2x+8=2(20)+8 =48 cm

To Find Hypotenuse :

$Hypotenuse^2= Perpendicular^2+Base^2$

$Hypotenuse^2= 20^2+48^2$

$Hypotenuse= \sqrt{20^2+48^2}$

$Hypotenuse=52$

Hence the sides of the triangle are 20 cm , 48 cm and 52 cm