Question

1.The sum of the lengths of the two sides containing the right-angled triangle is 15cm. If the area of the triangle is 28sq cm, find the lengths of these two sides.

2.The lengths of the sides of a right-angled triangle are (2x+1)cm, 2x cm and (x-1)cm. Find x

Answer 1

One side is x cm
Then other side will be (15-x) cm.
Area of triangle=1/2*base*height
28=1/2*x*(15-x)
28*2=x*(15-x)
56=15x-x²
x²-15x+56=0
x²-8x-7x+56=0
x(x-8)-7(x-8)=0
(x-8)(x-7)=0
x=8,7
Therefore, one side =8cm 
Then , other side =(15-8)cm=7cm

Answer 2

let the sides be x,y
given : x+y = 15  eqn------1
and      xy = 56
hence y =56/x
putting in eqn 1
x+ 56/x  =15
(x^2 + 56) x = 15
x^2 -15x +56 =0
on solving further we get values of x an 7 and 8
and the second question .....
the hypotenuse is 2x+1 cms
the second longer side is 2x
and the shortest side is x-1
hence on applying pythagoras theorem 
(2x+1)^ = (2x)^2 + (x-1)^2...
on solving u gt x = 6