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Question Number 15893 by Tinkutara last updated on 15/Jun/17

In Δ A B C, sides a, b, c are roots of the

equation x^{3} - {p x}^{2} + q x - r = 0 . Prove

that area of Δ A B C is

\frac{1}{4} \sqrt{p (4 p q - p^{3} - 8 r)}

Answered by RasheedSoomro last updated on 15/Jun/17

It can be shown in above case that

{\begin{cases} a + b + c = - (- p) = p \\ ab + bc + ca = q \\ abc = - (- r) = r \end{cases}

∴ s = \frac{a + b + c}{2} = \frac{p}{2}

We know that the area of triangle

▴ = \sqrt{s (s - a) (s - b) (s - c)}

where a, b, c are sides and s is semiperimeter

of the triangle .

▴ = \sqrt{\frac{p}{2} (\frac{p}{2} - a) (\frac{p}{2} - b) (\frac{p}{2} - c)}

▴ = \sqrt{\frac{p}{2} (\frac{p - 2 a}{2}) (\frac{p - 2 b}{2}) (\frac{p - 2 c}{2})}

▴ = \frac{1}{4} \sqrt{p (p - 2 a) (p - 2 b) (p - 2 c)}

▴ = \frac{1}{4} \sqrt{p {p^{3} - {2 ap}^{2} - {2 bp}^{2} - {2 cp}^{2} + 4 abp + 4 bcp + 4 cap - 8 abc}}

▴ = \frac{1}{4} \sqrt{p {p^{3} - {2 p}^{2} (a + b + c) + 4 p (ab + bc + ca) - 8 (abc)}}

▴ = \frac{1}{4} \sqrt{p {p^{3} - {2 p}^{2} (p) + 4 p (q) - 8 (r)}}

▴ = \frac{1}{4} \sqrt{p {p^{3} - {2 p}^{3} + 4 pq - 8 r}}

▴ = \frac{1}{4} \sqrt{p {- p^{3} + 4 pq - 8 r}}

▴ = \frac{1}{4} \sqrt{p (4 pq - p^{3} - 8 r)}

Commented by Tinkutara last updated on 15/Jun/17

Thanks Sir!