The perimeter of a right triangle is 12 , cm. | vedclass

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(A) Let the two sides of the right triangle be $a$ and $b$. The hypotenuse is $5 \, cm$.Perimeter $= a + b + 5 = 12 \, cm \implies a + b = 7 \, cm$.By

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$3 \, cm, 4 \, cm$ and $6 \, cm^2$

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(A) Let the two sides of the right triangle be $a$ and $b$. The hypotenuse is $5 \, cm$.Perimeter $= a + b + 5 = 12 \, cm \implies a + b = 7 \, cm$.By Pythagoras theorem,$a^2 + b^2 = 5^2 = 25$.We know that $(a + b)^2 = a^2 + b^2 + 2ab$.Substituting the values: $(7)^2 = 25 + 2ab$.$49 = 25 + 2ab \implies 2ab = 24 \implies ab = 12$.Now,$(a - b)^2 = (a + b)^2 - 4ab = (7)^2 - 4(12) = 49 - 48 = 1$.So,$a - b = 1$ (assuming $a > b$).Solving $a + b = 7$ and $a - b = 1$,we get $2a = 8 \implies a = 4 \, cm$ and $b = 3 \, cm$.Area of the triangle $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes a 	imes b = \frac{1}{2} 	imes 4 	imes 3 = 6 \, cm^2$.

The perimeter of a right triangle is $12 \, cm$. The hypotenuse is $5 \, cm$. What are the other two sides and the area of the triangle?

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