Question

The difference between the sides at right angles in a right angled triangle14cm .if the area of the triangle is 120sq.cm then the perimeter of the triangle is

Answer 1

Answer:

let the 1st side be x

second side -x=14 (given)

second side =14 + x

area=1÷2×x× 14+x

=1÷2×(14x+x^2)=120

=14x+x^2=120×2

=x^2+14x=240

=x^2+14x-240=0

=x^2+(24-10)x-240=0

=x^2+24x-10x-240=0

=x (x+24)-10 (x+24)=0

=(x+24)(x-10)=0

x=-24 and x=10

taking x=10

1st side =10

2nd side=14+10

=24

h^2=24^2+10^2

h^2=576+100

h^2=676

h=26

perimeter =26+10+24=60

Answer 2

Answer:

let the triangle be ABC

area =1/2*base*height

120 = 1/2*bc*ab

let one side be x and other be 14-x

so, 240=x *(x-14)

hence 240=x2-14x

0= x2-14x-240

0= x(x-24)+10(x-24)

0=(x-24)(x+10)

hence x =24 , x-14=2-14=10 cm

now using phythagoras theorem

(24)sq +(10)sq =a sq

576 + 100= a sq

676=a sq

26 = a

now perimeter of triangle= sum of all the sides

= (24+10+26)cm

= 60cm