Question

Answer Properly, This is my First Question

Question : What is the area of triangle with sides a, b and √(a² + b²) ?
If possible provide a diagram

Don't Copy. Give answer fast. I will mark as Genius

Answer 1

Area of Triangle whose sides are $\sf{a}$ , $\sf{b}$ and $\sqrt{\sf{a^2+b^2}}$ = $\sf{ab/2}$

We are all aware of formula to calculate Perimeter of Triangle! It's Simple ; Perimeter of Triangle = Sum of All Sides. We just add all the sides of the triangle and the result will be our perimeter!

$\longrightarrow$ Perimeter of Triangle = Sum of All Sides

We are Given that the Sides of Triangle are $\sf{a}$ , $\sf{b}$ and $\sqrt{\sf{a^2+b^2}}$

$\longrightarrow$ Perimeter of Triangle = $\sf{a+b+\sqrt{a^2+b^2}}$

Now we use the Concept of Herons Formula! Also known as Hero's Formula is used to find the area of a triangle when we know the length of all its sides. First for doing it, We need to find Semi Perimeter of Triangle which is exactly half of Perimeter of Triangle

$\longrightarrow$ Semi Perimeter of Triangle = $\sf{Perimeter/2}$

We already found that Perimeter of Triangle = $\sf{a+b+\sqrt{a^2+b^2}}$

$\longrightarrow$ Semi Perimeter of Triangle = $\sf{(a+b+\sqrt{a^2+b^2})/2}$

Heron's Formula is given by $\sf{\sqrt{s(s-a)(s-b)(s-c)}}$ where s - Semi Perimeter, a, b and c are Sides of the Triangle. We have s = $\sf{(a+b+\sqrt{a^2+b^2})/2}$ and a = a, b = b, c = $\sf{\sqrt{a^2+b^2}}$. Substitute them in $\sf{\sqrt{s(s-a)(s-b)(s-c)}}$ Formula to find the area of the Triangle!

$\longrightarrow$ Area of Triangle = $\sf{\sqrt{s(s-a)(s-b)(s-c)}}$

$\longrightarrow$ Area of Triangle = $\sf{\sqrt{s(s-a)(s-b)(s-\sqrt{a^2+b^2})}}$

$=\sf{\sqrt{\frac{a+b+\sqrt{a^2+b^2}}{2}\left(\frac{a+b+\sqrt{a^2+b^2}}{2}-a\right)\left(\frac{a+b+\sqrt{a^2+b^2}}{2}-b\right)\left(\frac{a+b+\sqrt{a^2+b^2}}{2}-\sqrt{a^2+b^2}\right)}}$

$\sf{=\sqrt{\frac{a+b+\sqrt{a^2+b^2}}{2}\cdot \frac{b+\sqrt{a^2+b^2}-a}{2}\left(-b+\frac{a+b+\sqrt{a^2+b^2}}{2}\right)\left(\frac{a+b+\sqrt{a^2+b^2}}{2}-\sqrt{a^2+b^2}\right)}}$

$\sf{=\sqrt{\frac{a+b+\sqrt{a^2+b^2}}{2}\cdot \frac{b+\sqrt{a^2+b^2}-a}{2}\cdot \frac{a+\sqrt{a^2+b^2}-b}{2}\left(\frac{a+b+\sqrt{a^2+b^2}}{2}-\sqrt{a^2+b^2}\right)}}$

$\sf{ =\sqrt{\frac{a+b+\sqrt{a^2+b^2}}{2}\cdot \frac{a+b-\sqrt{a^2+b^2}}{2}\cdot \frac{b+\sqrt{a^2+b^2}-a}{2}\cdot \frac{a+\sqrt{a^2+b^2}-b}{2}}}$

$\sf{=\sqrt{\frac{\left(a+b+\sqrt{a^2+b^2}\right)\left(-a+\sqrt{a^2+b^2}+b\right)\left(-b+\sqrt{a^2+b^2}+a\right)\left(a+b-\sqrt{a^2+b^2}\right)}{2\cdot \:2\cdot \:2\cdot \:2}}}$

$\sf{=\sqrt{\frac{\left(a+b+\sqrt{a^2+b^2}\right)\left(-a+\sqrt{a^2+b^2}+b\right)\left(-b+\sqrt{a^2+b^2}+a\right)\left(a+b-\sqrt{a^2+b^2}\right)}{16}}}$

$\longrightarrow$ Area of Triangle = $\sf{\sqrt{4a^2b^2/16}}$

$\longrightarrow$ Area of Triangle = $\sf{\sqrt{a^2b^2/4}}$

$\large\boxed{\boxed{\longrightarrow\textbf{Area of Triangle = ab/2}}}$