Question

Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is larger than the other by 5 cm, find the lengths of the other two sides.

Answer 1

GIVEN:
Hypotenuse of right ∆(H) = AC = 25 cm
Let the one side of a right ∆(B) = BC = x cm
Other side(P) = AB = ( x+5) cm

In right ∆ ABC,
H² = B² + P²
AC² = BC² + AB²

[By Pythagoras theorem]
(25)² = x² + (x+5)²
625 = x² + x² + 5² + 2×x×5

[(a+b)² = a²+b²+2ab]

625 = 2x² +25 +10x
2x² +25 +10x - 625= 0
2x² +10x - 600 = 0
2(x² +5x -300)= 0
x² +5x -300 = 0
x² +20x - 15x -300 = 0   [ by factorization]
x(x +20) -15(x + 20) = 0
(x +20) (x - 15) = 0
(x +20) = 0 or  (x - 15) = 0

x = -20 or x= 15

Side Can't be negative, so x = 15

One side (x) = 15 cm
Other side (x+5) = 15 +5 = 20 cm

Hence, the lengths of  the other two sides be 15 cm & 20 cm.

HOPE THIS WILL HELP YOU...

Answer 2

Hey...!!

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Then other side is (x+5) cm

x^2 + (x+5)^2 = (25)^2

x^2 + x^2 +10x + 25 = 625

2x^2 + 10x - 600 = 0

x^2 + 5x - 300 = 0

x^2 + 20x - 15x - 300 = 0

x(x+20) - 15(x+20) = 0

(x+20) (x-15) = 0

x = -20     x = 15

Side cannot be negative 

So, therefore one side is 15cm
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I Hope it's help you...!!!



other side is (x+5) = (15+5) = 20cm