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Question Number 187524 by sonukgindia last updated on 18/Feb/23

Answered by HeferH last updated on 18/Feb/23

Commented by HeferH last updated on 18/Feb/23

Area of triangle ((√(20)),(√(20)),(√8))= A A = (√(s(s−a)(s−b)(s−c))) (Heron) A= (√((2(√5)+(√2))((√2))((√2))(2(√5)−(√2)))) A = (√((20−2)2)) = 6 but also: A = ((2(√5)∙h)/2) = 6 h = (6/( (√5))) = ((6(√5))/5) a^2 = r^2 −h^2 a^2 +(r−h)^2 = x^2 2r^2 −2rh=x^2 2(20)−2(√(20))∙((6(√5))/5) = x^2 40−24=x^2 16 = x^2

$$\:{Area}\:{of}\:{triangle}\:\left(\sqrt{\mathrm{20}},\sqrt{\mathrm{20}},\sqrt{\mathrm{8}}\right)=\:{A} \\ $$$$\:{A}\:=\:\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\:\left({Heron}\right) \\ $$$$\:{A}=\:\sqrt{\left(\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{2}}\right)\left(\mathrm{2}\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}\right)} \\ $$$$\:{A}\:=\:\sqrt{\left(\mathrm{20}−\mathrm{2}\right)\mathrm{2}}\:=\:\mathrm{6}\: \\ $$$$\:{but}\:{also}:\:\:{A}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{5}}\centerdot{h}}{\mathrm{2}}\:=\:\mathrm{6} \\ $$$$\:{h}\:=\:\frac{\mathrm{6}}{\:\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\:{a}^{\mathrm{2}} =\:{r}^{\mathrm{2}} −{h}^{\mathrm{2}} \\ $$$$\:{a}^{\mathrm{2}} +\left({r}−{h}\right)^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \\ $$$$\:\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{rh}={x}^{\mathrm{2}} \\ $$$$\:\mathrm{2}\left(\mathrm{20}\right)−\mathrm{2}\sqrt{\mathrm{20}}\centerdot\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{5}}\:=\:{x}^{\mathrm{2}} \\ $$$$\:\mathrm{40}−\mathrm{24}={x}^{\mathrm{2}} \\ $$$$\:\mathrm{16}\:=\:{x}^{\mathrm{2}} \\ $$

Commented by HeferH last updated on 18/Feb/23

Answered by mr W last updated on 18/Feb/23

Commented by mr W last updated on 18/Feb/23

((√2)(√(20)))^2 =s^2 +((√8))^2 +2s×(√8)×((√2)/2) s^2 +4s−32=0 (s−4)(s+8)=0 ⇒s=4 area of square s^2 =16

$$\left(\sqrt{\mathrm{2}}\sqrt{\mathrm{20}}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} +\left(\sqrt{\mathrm{8}}\right)^{\mathrm{2}} +\mathrm{2}{s}×\sqrt{\mathrm{8}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} +\mathrm{4}{s}−\mathrm{32}=\mathrm{0} \\ $$$$\left({s}−\mathrm{4}\right)\left({s}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{4} \\ $$$${area}\:{of}\:{square}\:{s}^{\mathrm{2}} =\mathrm{16} \\ $$

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