Question

12. In AABC, ZB = 90°. Find the sides of the
triangle, if :
(i) AB = (x - 3) cm, BC = (x + 4) cm and
AC = (x + 6) cm
(ii) AB = x cm, BC = (4x + 4) cm and
AC = (4x + 5) cm​

Answer 1

Answer:

I) (X+6)^2=(x-3)^2+(X+4)^2

x^2-10x-11=0

(x-11)(X+1)=0

X=11or x=-1

three sides of the triangle are 8,15,17cm

ii) (4x+5)^2=(x^2+(4x+4)^2

x^2 -8x -9=0

(x-9)(X+1)

X=9or-1

Three sides of the triangle are 9,40,41cm