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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

(a+4)^2 +(a+8)^2 =(2a+4)^2 a^2 −4a−32=0 (a+4)(a−8)=0 ⇒a=8 area ABC=(((8+4)(8+8))/2)=96

$$\left({a}+\mathrm{4}\right)^{\mathrm{2}} +\left({a}+\mathrm{8}\right)^{\mathrm{2}} =\left(\mathrm{2}{a}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{32}=\mathrm{0} \\ $$$$\left({a}+\mathrm{4}\right)\left({a}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{8} \\ $$$${area}\:{ABC}=\frac{\left(\mathrm{8}+\mathrm{4}\right)\left(\mathrm{8}+\mathrm{8}\right)}{\mathrm{2}}=\mathrm{96} \\ $$

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

(a+y)^2 = (a+4)^2 +(4+y)^2 a^2 +2ay+y^2 = a^2 +8a+16+16+8y+y^2 2ay = 8a+8y+32 ay= 4a+4y+16 ay−4y = 4a+16 y = ((4a+16)/(a−4)) (2) tan x=tan x (4/(a+4)) = (4/y) ⇒y=a+4 then a^2 −16 = 4a+16 a^2 −4a−32=0 (a−8)(a+4)=0 a=8 ⇒y=12

$$\: \\ $$$$\:\left(\mathrm{a}+\mathrm{y}\right)^{\mathrm{2}} =\:\left(\mathrm{a}+\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{y}\right)^{\mathrm{2}} \\ $$$$\:\:\mathrm{a}^{\mathrm{2}} +\mathrm{2ay}+\mathrm{y}^{\mathrm{2}} =\:\mathrm{a}^{\mathrm{2}} +\mathrm{8a}+\mathrm{16}+\mathrm{16}+\mathrm{8y}+\mathrm{y}^{\mathrm{2}} \\ $$$$\:\:\mathrm{2ay}\:=\:\mathrm{8a}+\mathrm{8y}+\mathrm{32}\: \\ $$$$\:\:\:\mathrm{ay}=\:\mathrm{4a}+\mathrm{4y}+\mathrm{16}\: \\ $$$$\:\:\:\:\mathrm{ay}−\mathrm{4y}\:=\:\mathrm{4a}+\mathrm{16} \\ $$$$\:\:\:\:\mathrm{y}\:=\:\frac{\mathrm{4a}+\mathrm{16}}{\mathrm{a}−\mathrm{4}}\: \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{tan}\:\mathrm{x}=\mathrm{tan}\:\mathrm{x} \\ $$$$\:\:\:\frac{\mathrm{4}}{\mathrm{a}+\mathrm{4}}\:=\:\frac{\mathrm{4}}{\mathrm{y}}\:\Rightarrow\mathrm{y}=\mathrm{a}+\mathrm{4} \\ $$$$\:\mathrm{then}\:\mathrm{a}^{\mathrm{2}} −\mathrm{16}\:=\:\mathrm{4a}+\mathrm{16} \\ $$$$\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{4a}−\mathrm{32}=\mathrm{0} \\ $$$$\:\:\left(\mathrm{a}−\mathrm{8}\right)\left(\mathrm{a}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\:\:\:\mathrm{a}=\mathrm{8}\:\Rightarrow\mathrm{y}=\mathrm{12} \\ $$

Commented by mr W last updated on 07/Nov/23

to necx122 sir: Say O=center of circle CF=EO=4 CE=FO=4 OF=4=CF ∠OBF=x=∠FAC ⇒ΔACF≡ΔBFO ⇒BF=AC=a+4 similarly BP=AC=a+4

$${to}\:{necx}\mathrm{122}\:{sir}: \\ $$$${Say}\:{O}={center}\:{of}\:{circle} \\ $$$${CF}={EO}=\mathrm{4} \\ $$$${CE}={FO}=\mathrm{4} \\ $$$${OF}=\mathrm{4}={CF} \\ $$$$\angle{OBF}={x}=\angle{FAC} \\ $$$$\Rightarrow\Delta{ACF}\equiv\Delta{BFO} \\ $$$$\Rightarrow{BF}={AC}={a}+\mathrm{4} \\ $$$${similarly} \\ $$$${BP}={AC}={a}+\mathrm{4} \\ $$

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!

$${they}\:{are}\:{not}\:{only}\:{similar},\:{but}\:{also} \\ $$$${congruent}! \\ $$

Commented by necx122 last updated on 07/Nov/23

Yes, that's true. Thank you.

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