Question

prove that the area of isosceles triangle with length of equal side a and base b is equal to root 1/2 x b a²-b²/4​

Answer 1

$\:\:\:\: \: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{!! QUESTION!!}}}}}}}$

prove that the area of isosceles triangle with length of equal side a and base b is equal to root

$\frac{b}{2} \sqrt{a {}^{2} - \frac{b {}^{2} }{4} }$

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$\:\:\:\: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{!! ANSWER!!}}}}}}}$

$\:\: \underline{\underline{\bf{\large\mathfrak{formula \:used}}}}$

the area of a triangle with sides a,b,c......is

$= \sqrt{s(s - a)(s - b)(s - c)}$

$where...s = half \: of \: perimeter \: \\ = \frac{a + b + c}{2}$

$\:\: \underline{\underline{\bf{\large\mathfrak{given}}}}$

the triangle is isosceles and having equal sides is a and the base b....

$\:\: \underline{\underline{\bf{\large\mathfrak{~~solution~~}}}}$

the half of perimeter is

$s = \frac{a + a + b}{2} = \frac{2a + b}{2}$

therefore area....

$= \sqrt{\frac{2a + b}{2} ( \frac{2a + b}{2} - a) {}^{2}( \frac{2a + b}{2} - b) } \\ = \sqrt{ \frac{2a + b}{2}( \frac{b}{2} ) {}^{2}( \frac{2a - b}{2} ) } \\ = \sqrt{ \frac{b {}^{2} }{4} \times \frac{1}{4} \times (2a + b)(2a - b)} \\ = \sqrt{ \frac{b {}^{2} }{16} (4a {}^{2} - b {}^{2}) } \\ = \sqrt{ \frac{1}{4}a {}^{2}b {}^{2} - \frac{1}{16}b {}^{4} } \\ = \sqrt{ \frac{b {}^{2} }{4}(a {}^{2} - \frac{b {}^{2} }{4} ) } \\ = \frac{b}{2} \sqrt{a {}^{2} - \frac{b {}^{2} }{4} } \: \: (proved)$

$\:\: \underline{\underline{\bf{\large\mathfrak{hope\: this \:helps\: you......Khushi..}}}}$