Question

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The hypotenuse of a right isosceles triangle is "a" .Find it's area .​

Answer 1

Step-by-step explanation:

Isosceles right angled triangle :

Two sides of right angle are equal of which hypotenuse is a.

Let two equal sides of the right angled triangle be x.

According to pythagorus theorem ,

${x}^{2} + {x}^{2} = {a}^{2} \\ \\ \\ 2 {x}^{2} = {a}^{2} \\ \\ \\ {x}^{2} = \frac{ {a}^{2} }{2} \\ \\ \\ x = \frac{a}{ \sqrt{2} } \\ \\ \\ area \: of \: right \: angled \: triangle \: = \frac{1}{2} \times \frac{a}{ \sqrt{2} } \times \frac{a}{ \sqrt{2} } \\ \\ \\ \frac{ {a}^{2} }{4}$

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Answer 2

Answer:

{x}^{2} + {x}^{2} = {a}^{2} \\ \\ \\ 2 {x}^{2} = {a}^{2} \\ \\ \\ {x}^{2} = \frac{ {a}^{2} }{2} \\ \\ \\ x = \frac{a}{ \sqrt{2} } \\ \\ \\ area \: of \: right \: angled \: triangle \: = \frac{1}{2} \times \frac{a}{ \sqrt{2} } \times \frac{a}{ \sqrt{2} } \\ \\ \\ \frac{ {a}^{2} }{4}

Step-by-step explanation:

Isosceles right angled triangle :

Two sides of right angle are equal of which hypotenuse is a.

Let two equal sides of the right angled triangle be x.

According to pythagorus theorem ,

{x}^{2} + {x}^{2} = {a}^{2} \\ \\ \\ 2 {x}^{2} = {a}^{2} \\ \\ \\ {x}^{2} = \frac{ {a}^{2} }{2} \\ \\ \\ x = \frac{a}{ \sqrt{2} } \\ \\ \\ area \: of \: right \: angled \: triangle \: = \frac{1}{2} \times \frac{a}{ \sqrt{2} } \times \frac{a}{ \sqrt{2} } \\ \\ \\ \frac{ {a}^{2} }{4}