The sum of the sides of a right-angled triang | vedclass

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(C) Let the sides of the right-angled triangle be $b$,$c$ (legs) and $x$ (hypotenuse).Given $b + c + x = 42$  --- $(1)$In a right-angled triangle,the

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$63$

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(C) Let the sides of the right-angled triangle be $b$,$c$ (legs) and $x$ (hypotenuse).Given $b + c + x = 42$  --- $(1)$In a right-angled triangle,the median to the hypotenuse is half the hypotenuse,so $M = \frac{x}{2}$.The altitude $h$ to the hypotenuse is given by $h = \frac{bc}{x}$.Given $M - h = 2$,so $\frac{x}{2} - \frac{bc}{x} = 2 \Rightarrow x^2 - 2bc = 4x$ --- $(2)$From $(1)$,$b + c = 42 - x$. Squaring both sides: $b^2 + c^2 + 2bc = (42 - x)^2$.Since $b^2 + c^2 = x^2$,we have $x^2 + 2bc = (42 - x)^2$ --- $(3)$Adding $(2)$ and $(3)$: $2x^2 = (42 - x)^2 + 4x$.$2x^2 = 1764 - 84x + x^2 + 4x \Rightarrow x^2 + 80x - 1764 = 0$.Solving the quadratic equation: $(x + 98)(x - 18) = 0$. Since $x > 0$,$x = 18$.Substitute $x = 18$ into $(2)$: $18^2 - 2bc = 4(18)$ $\Rightarrow 324 - 2bc = 72$ $\Rightarrow 2bc = 252$ $\Rightarrow bc = 126$.The area of the triangle is $\frac{1}{2}bc = \frac{1}{2} 	imes 126 = 63$.

The sum of the sides of a right-angled triangle is $42$,and the difference between the median and altitude drawn from the vertex at the right angle is $2$. The area of the triangle is

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