Question Number 199672 by Mingma last updated on 07/Nov/23
Answered by mr W last updated on 07/Nov/23
Commented by mr W last updated on 07/Nov/23
$$\left({a}+\mathrm{4}\right)^{\mathrm{2}} +\left({a}+\mathrm{8}\right)^{\mathrm{2}} =\left(\mathrm{2}{a}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{32}=\mathrm{0} \\ $$$$\left({a}+\mathrm{4}\right)\left({a}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{8} \\ $$$${area}\:{ABC}=\frac{\left(\mathrm{8}+\mathrm{4}\right)\left(\mathrm{8}+\mathrm{8}\right)}{\mathrm{2}}=\mathrm{96} \\ $$
Commented by necx122 last updated on 07/Nov/23
Mr. w, please, I'll love to know how (a+4) came about, sir
Commented by cortano12 last updated on 07/Nov/23
$$\:\:\underline{\underbrace{\:}} ^{} \\ $$
Commented by mr W last updated on 07/Nov/23
$${to}\:{necx}\mathrm{122}\:{sir}: \\ $$$${Say}\:{O}={center}\:{of}\:{circle} \\ $$$${CF}={EO}=\mathrm{4} \\ $$$${CE}={FO}=\mathrm{4} \\ $$$${OF}=\mathrm{4}={CF} \\ $$$$\angle{OBF}={x}=\angle{FAC} \\ $$$$\Rightarrow\Delta{ACF}\equiv\Delta{BFO} \\ $$$$\Rightarrow{BF}={AC}={a}+\mathrm{4} \\ $$$${similarly} \\ $$$${BP}={AC}={a}+\mathrm{4} \\ $$
Commented by necx122 last updated on 07/Nov/23
Oh that's true. They are similar triangles. Thank you sir. It's clear now.
Commented by Mingma last updated on 07/Nov/23
Nice solution, sir!
Commented by mr W last updated on 07/Nov/23
$${they}\:{are}\:{not}\:{only}\:{similar},\:{but}\:{also} \\ $$$${congruent}! \\ $$
Commented by necx122 last updated on 07/Nov/23
Yes, that's true. Thank you.