Question

The difference between sides at right angles in a right angled triangle is 14 cm. The area of the triangle is 120 cm2
Find the length of any side of right angled triangle..​

Answer 1

Answer:

let one side at right angle be x

the other must be x-14

area of a triangle =(1/2) × (height) (base)

120 = (1/2)(x)(x-14)

120 = (x^2 -14x)/2

240 = x^2 -14x

x^2 -14x -240 = 0

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

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

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

now, find x

x-24 =0

x = 24

again

x+10 =0

x = -10

as the side couldn't be -v so the value of x as a side must be 24cm

so the measure of one side we got is x =24cm and the other side = x-14 = 24 -14 =10cm

apply pythagoras theorem for third side as it is a right angled triangle

[let hypotaneous be a]

(24)^2 + (10^2) = a^2

576 +100 = a^2

676 = a^2

a = 26 cm

so the hypataneous = 26cm

perimeter of a triangle = sum of all sides

= (24 + 10 + 26 ) cm

= 60cm

Answer 2

Answer:

plz gives 30 thanks...........

Step-by-step explanation:

Let the sides at right angles in the given right-angled triangle be a and b.

Without loss of generality, we let a>b.

The by the problem a−b=14⇒a=(b+14)........(1).

Now the are of the right-angled triangle be

2

1

×a×b=

2

b(b+14)

cm

2

. [Using (1)]

According to the problem

2

b(b+14)

=120

b

2

+14b−240=0

(b−10)(b+24)=0

This gives b=10⇒a=24.

Now the hypotenuse of the right-angled triangle be

10

2

+24

2

=26 cm.

Now the perimeter of the triangle be (10+24+26)=60 cm.