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

$$ifthesidesofatriangleare$$$$\sqrt{a^2+b^2},\sqrt{a^2+4b^2},\sqrt{4a^2+b^2}respectively,$$$$whatisitsarea?$$

Answered by ajfour last updated on 14/Feb/23

$$A=4ab-\frac{1}{2}\left(a\right)\left(b\right)-\frac{1}{2}\left(a\right)\left(2b\right)$$$$-\frac{1}{2}\left(b\right)\left(2a\right)$$$$A=\frac{ab}{2}(8-1-2-2)=\frac{3ab}{2}$$

Commented by anurup last updated on 14/Feb/23

$$\mathrm{Is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{solve}\mathrm{this}\mathrm{by}\mathrm{using}{\mathrm{Heron}}^{\prime }\mathrm{s}\mathrm{formula}?$$

Commented by ajfour last updated on 14/Feb/23

I wouldnt have dared..!

Commented by anurup last updated on 14/Feb/23

$$\mathrm{It}\mathrm{will}\mathrm{be}\mathrm{too}\mathrm{cumbersome}\mathrm{I}\mathrm{guess}.$$

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

$$nicesolution!$$

Answered by ajfour last updated on 14/Feb/23

Answered by Frix last updated on 14/Feb/23

$${\mathrm{Heron}}^{\prime }\mathrm{s}\mathrm{Formula}\mathrm{for}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{sides}$$$$p,q,r\mathrm{is}$$$$\frac{\sqrt{2(p^2{q}^2+p^2{r}^2+q^2{r}^2)-(p^4+q^4+r^4)}}{4}$$$$\Rightarrow \mathrm{for}\sqrt{p},\sqrt{q},\sqrt{r}\mathrm{it}\mathrm{is}$$$$\frac{\sqrt{2(pq+pr+qr)-(p^2+q^2+r^2)}}{4}$$$$p=u+v\wedge q=u+4v\wedge r=4u+v$$$$\frac{3\sqrt{uv}}{2}=\frac{3ab}{2}$$

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

$$nicesolution!$$

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