Question

the perimeter of a right triangle is 12m. its hypotenuse is 5m. then the area of the triangle is​

Answer 1

Answer:

In a right angled triangle the hypotenuse is always greater.

Step-by-step explanation:

So, H^2=p^2+b^2

P^2=h^2-b^2

b^2=h^2-p^2

Hope it helps to u

Answer 2

Given: The perimeter of a right ∆ is 12m and it's hypotenuse is 5m

❍ Let's say, that the sides of the triangle be a and b.

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❍ We know that the perimeter of a ∆ is sum of all three sides of the ∆.

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$\sf : \implies \:a + b + 5 = 12$

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$\sf : \implies \: a + b = 7$

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$\sf : \implies \: a = 7 - b \: \: \bold{ (eq)i}$

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❍ Now, using the Pythagoras theorem we'll find other two sides of the right ∆.

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$\sf : \implies \: (Hypotenuse) {}^{2} =( Perpendicular) {}^{2} +( Base) {}^{2}$

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$\sf : \implies(5) {}^{2} = (a) {}^{2} + (b) {}^{2}$

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$\sf : \implies \: 25 = (7 - b) {}^{2} + (b) {}^{2}$

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$\sf : \implies \: 25 = \:(7) {}^{2} + (b) {}^{2} - 2 \times 7 \times b \: \: + b {}^{2}$

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$\sf : \implies \: 25 = 49 + b {}^{2} - 14b + b {}^{2}$

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$\sf : \implies \: 25 = 49 + 2b {}^{2} - 14b$

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$\sf : \implies \: b {}^{2} - 7b + 12 = 0$

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$\sf : \implies \: (b - 4)(b - 3) = 0$

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$\sf : \implies \: b = 4,3$

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❍ Now, we'll find the value of a by using eq i)

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$\sf : \implies \: a = 7 - b$

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$\sf : \implies \: a = 7 - 4$

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$\sf : \implies \: a = 3$

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❍ We know that the area of a right ∆ is :

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$\sf : \implies \: Area_{(right ∆)} = \dfrac{1}{2} \times base \times perpendicular$

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$\sf : \implies \: Area_{(right ∆)} = \dfrac{1}{2} \times 4 \times 3$

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$\sf : \implies \: Area_{(right ∆)} = 2 \times 3$

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$\sf : \implies \: Area_{(right ∆)} = \bold{6m} {}^{2}$

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$\sf \therefore \underline{Hence ,\: the \: area \: of \: the \: right \triangle \: is \: \bold{6m {}^{2}} }$