if-the-sides-of-a-triangle-are-a-2-b-2-a-2-4b-2-4a-2-b-2-respectively-what-is-its-area-

Question Number 187205 by mr W last updated on 14/Feb/23

i f t h e s i d e s o f a t r i a n g l e a r e

\sqrt{a^{2} + b^{2}}, \sqrt{a^{2} + 4 b^{2}}, \sqrt{4 a^{2} + b^{2}} r e s p e c t i v e l y,

w h a t i s i t s a r e a ?

Answered by ajfour last updated on 14/Feb/23

A = 4 a b - \frac{1}{2} (a) (b) - \frac{1}{2} (a) (2 b)

- \frac{1}{2} (b) (2 a)

A = \frac{a b}{2} (8 - 1 - 2 - 2) = \frac{3 a b}{2}

Commented by anurup last updated on 14/Feb/23

Is it possible to solve this by using {Heron}^{'} s formula ?

Commented by ajfour last updated on 14/Feb/23

I wouldnt have dared..!

Commented by anurup last updated on 14/Feb/23

It will be too cumbersome I guess .

Commented by mr W last updated on 15/Feb/23

n i c e s o l u t i o n!

Answered by ajfour last updated on 14/Feb/23

Answered by Frix last updated on 14/Feb/23

{Heron}^{'} s Formula for a triangle with sides

p, q, r is

\frac{\sqrt{2 (p^{2} q^{2} + p^{2} r^{2} + q^{2} r^{2}) - (p^{4} + q^{4} + r^{4})}}{4}

\Rightarrow for \sqrt{p}, \sqrt{q}, \sqrt{r} it is

\frac{\sqrt{2 (p q + p r + q r) - (p^{2} + q^{2} + r^{2})}}{4}

p = u + v \land q = u + 4 v \land r = 4 u + v

\frac{3 \sqrt{u v}}{2} = \frac{3 a b}{2}

Commented by mr W last updated on 15/Feb/23

n i c e s o l u t i o n!

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