Question

The hypotenuse of a right angled isosceles triangle is 16 root 2 find the area​

Answer 1

let others two sides be x units

By pathagorus Theorem

${x}^{2} + {x}^{2} = {(16 \sqrt{2}) }^{2} \\ \\ \sqrt{2 {x}^{2} } = 16 \sqrt{2} \\ \\ x \sqrt{2} = 16 \sqrt{2} \\ \\ x = 16$

Area of ∆ = 1/2 × b × h

$ar = \frac{1}{2} \times 16 \times 16 \sqrt{2} \\ \\ ar = 8 \times 16 \sqrt{2} \\ \\ ar = 128 \sqrt{2} \: {unit}^{2}$

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Answer 2

Question:

The hypotenuse of a right angled isosceles triangle is 16 root 2 find the area.

Answer:

let side of ∆ be x

${x}^{2} + {x}^{2} = {(16 \sqrt{2})}^{2} \\ \\ 2 {x}^{2} = 16 \times 16 \times 2 \\ \\ {x}^{2} = 16 \times 16 \\ \\ x = 16 \: unit$

area of ∆

$ar = \frac{1}{2} \times 16 \times 16 \\ \\ ar = 8 \times 16 \\ \\ ar = 128 \: \: {unit}^{2}$