If the hypotenuse of a right angled triangle is $41 c m$ and the area of the triangle is $180 s q c m$ , then the difference between the lengths of the triangle must be

22 c m

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25 c m

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27 c m

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31 c m

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Solution

The correct option is D $31 c m$
$L e t b a s e o f r i g h t a n g l e d t r i a n g l e △ A B C b e x a n d h e i g h t b e y H y p o t e n u s e = A B = 41 I n r i g n t a n g l e d △ A B C 41^{2} = x^{2} + y^{2} \Rightarrow 1681 = x^{2} + y^{2} . . . (1) A r e a o f △ A B C = \frac{1}{2} x y . . . . [A r e a o f T r i a n g l e = \frac{1}{2} b h] \Rightarrow \frac{1}{2} x y = 180 \Rightarrow x y = 360..... (2) D i f f e r e n c e b e t w e e n t w o l e n g t h s = x - y {(x - y)}^{2} = x^{2} + y^{2} - 2 x y \Rightarrow {(x - y)}^{2} = 1681 - 2 \times 360 = 1681 - 720 = 961 \Rightarrow x - y = \sqrt{961} \Rightarrow x - y = 31 H e n c e o p t i o n (D) i s t h e c o r r e c t a n s w e r$

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If the hypotenuse of a right angled triangle is 41cm and the area of the triangle is 180sqcm, then the difference between the lengths of the triangle must be

If the hypotenuse of a right angled triangle is $41 c m$ and the area of the triangle is $180 s q c m$ , then the difference between the lengths of the triangle must be