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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
if the sides of a triangle are  (√(a^2 +b^2 )), (√(a^2 +4b^2 )), (√(4a^2 +b^2 )) respectively,  what is its area?
$${if}\:{the}\:{sides}\:{of}\:{a}\:{triangle}\:{are} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\:\sqrt{{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} },\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{respectively}, \\ $$$${what}\:{is}\:{its}\:{area}? \\ $$
Answered by ajfour last updated on 14/Feb/23
A=4ab−(1/2)(a)(b)−(1/2)(a)(2b)            −(1/2)(b)(2a)   A=((ab)/2)(8−1−2−2)=((3ab)/2)
$${A}=\mathrm{4}{ab}−\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left({b}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left(\mathrm{2}{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left({b}\right)\left(\mathrm{2}{a}\right)\: \\ $$$${A}=\frac{{ab}}{\mathrm{2}}\left(\mathrm{8}−\mathrm{1}−\mathrm{2}−\mathrm{2}\right)=\frac{\mathrm{3}{ab}}{\mathrm{2}} \\ $$
Commented by anurup last updated on 14/Feb/23
Is it possible to solve this by using Heron′s formula?
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{by}\:\mathrm{using}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{formula}? \\ $$
Commented by ajfour last updated on 14/Feb/23
I wouldnt have dared..!
Commented by anurup last updated on 14/Feb/23
It will be too cumbersome I guess.
$$\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{too}\:\mathrm{cumbersome}\:\mathrm{I}\:\mathrm{guess}. \\ $$
Commented by mr W last updated on 15/Feb/23
nice solution!
$${nice}\:{solution}! \\ $$
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  ((√(2(p^2 q^2 +p^2 r^2 +q^2 r^2 )−(p^4 +q^4 +r^4 )))/4)  ⇒ for (√p), (√q), (√r) it is  ((√(2(pq+pr+qr)−(p^2 +q^2 +r^2 )))/4)  p=u+v∧q=u+4v∧r=4u+v  ((3(√(uv)))/2)=((3ab)/2)
$$\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula}\:\mathrm{for}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides} \\ $$$${p},\:{q},\:{r}\:\mathrm{is} \\ $$$$\frac{\sqrt{\mathrm{2}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} {r}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} \right)−\left({p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} \right)}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{for}\:\sqrt{{p}},\:\sqrt{{q}},\:\sqrt{{r}}\:\mathrm{it}\:\mathrm{is} \\ $$$$\frac{\sqrt{\mathrm{2}\left({pq}+{pr}+{qr}\right)−\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)}}{\mathrm{4}} \\ $$$${p}={u}+{v}\wedge{q}={u}+\mathrm{4}{v}\wedge{r}=\mathrm{4}{u}+{v} \\ $$$$\frac{\mathrm{3}\sqrt{{uv}}}{\mathrm{2}}=\frac{\mathrm{3}{ab}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 15/Feb/23
nice solution!
$${nice}\:{solution}! \\ $$

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