One leg of a right angled triangle exceeds the other leg by 2 inches. The hypotenuse is 10 inches. Find the length of the shorter leg of the triangle.

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Solution

The correct option is B
6

Let the length of the legs of the triangle be $x a n d x + 2 i n c h e s .$

Applying Pythagoras theorem,

$x^{2} + (x + 2)^{2} = (10)^{2}$

$x^{2} + x^{2} + 4 x + 4 = 100$

$2 x^{2} + 4 x - 96 = 0$

$x^{2} + 2 x - 48 = 0$

$x^{2} + 8 x - 6 x - 48 = 0$

$x (x + 8) - 6 (x + 8) = 0$

$(x + 8) (x - 6) = 0$

Solving we get x = -8 and 6.

Length can't be negative, hence
$x = 6 i n c h e s$

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One leg of a right angled triangle exceeds the other leg by 2 inches. The hypotenuse is 10 inches. Find the length of the shorter leg of the triangle.

The correct option is B 6 Let the length of the legs of the triangle be x and x+2 inches. Applying Pythagoras theorem, x2+(x+2)2=(10)2 x2+x2+4x+4=100 2x2+4x−96=0 x2+2x−48=0 x2+8x−6x−48=0 x(x+8)−6(x+8)=0 (x+8)(x−6)=0 Solving we get x = -8 and 6. Length can't be negative, hence x=6 inches