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Question Number 80675 by naka3546 last updated on 05/Feb/20

Commented by naka3546 last updated on 05/Feb/20

Area of triangle ABC = ...

$${Area}\:\:{of}\:\:{triangle}\:\:{ABC}\:\:=\:\:... \\ $$

Answered by mr W last updated on 05/Feb/20

Commented by mr W last updated on 05/Feb/20

((x+48+z)/x)=(((BL)/(PQ)))^2 =(((PQ+ML)/(PQ)))^2 =(1+((ML)/(PQ)))^2 1+((48)/x)+(z/x)=(1+(√(z/x)))^2 =1+2(√(z/x))+(z/x) ((48)/2)=(√(xz)) ...(i) similarly ((32)/2)=(√(zy)) ...(ii) ((96)/2)=(√(yx)) ...(iii) (i)×(ii)×(iii): ((48)/2)×((32)/2)×((96)/2)=xyz ...(iv) (iv)/(i)^2 : y=((((48)/2)×((32)/2)×((96)/2))/((((48)/2))^2 ))=((((32)/2)×((96)/2))/((48)/2))=((32×96)/(2×48))=32 similarly x=((48×96)/(2×32))=72 z=((32×48)/(2×96))=8 Δ_(ABC) =48+32+96+72+32+8=288 cm^2

$$\frac{{x}+\mathrm{48}+{z}}{{x}}=\left(\frac{{BL}}{{PQ}}\right)^{\mathrm{2}} =\left(\frac{{PQ}+{ML}}{{PQ}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\frac{{ML}}{{PQ}}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\frac{\mathrm{48}}{{x}}+\frac{{z}}{{x}}=\left(\mathrm{1}+\sqrt{\frac{{z}}{{x}}}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}\sqrt{\frac{{z}}{{x}}}+\frac{{z}}{{x}} \\ $$$$\frac{\mathrm{48}}{\mathrm{2}}=\sqrt{{xz}}\:\:\:...\left({i}\right) \\ $$$${similarly} \\ $$$$\frac{\mathrm{32}}{\mathrm{2}}=\sqrt{{zy}}\:\:\:...\left({ii}\right) \\ $$$$\frac{\mathrm{96}}{\mathrm{2}}=\sqrt{{yx}}\:\:\:...\left({iii}\right) \\ $$$$\left({i}\right)×\left({ii}\right)×\left({iii}\right): \\ $$$$\frac{\mathrm{48}}{\mathrm{2}}×\frac{\mathrm{32}}{\mathrm{2}}×\frac{\mathrm{96}}{\mathrm{2}}={xyz}\:\:\:...\left({iv}\right) \\ $$$$\left({iv}\right)/\left({i}\right)^{\mathrm{2}} : \\ $$$${y}=\frac{\frac{\mathrm{48}}{\mathrm{2}}×\frac{\mathrm{32}}{\mathrm{2}}×\frac{\mathrm{96}}{\mathrm{2}}}{\left(\frac{\mathrm{48}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\frac{\mathrm{32}}{\mathrm{2}}×\frac{\mathrm{96}}{\mathrm{2}}}{\frac{\mathrm{48}}{\mathrm{2}}}=\frac{\mathrm{32}×\mathrm{96}}{\mathrm{2}×\mathrm{48}}=\mathrm{32} \\ $$$${similarly} \\ $$$${x}=\frac{\mathrm{48}×\mathrm{96}}{\mathrm{2}×\mathrm{32}}=\mathrm{72} \\ $$$${z}=\frac{\mathrm{32}×\mathrm{48}}{\mathrm{2}×\mathrm{96}}=\mathrm{8} \\ $$$$ \\ $$$$\Delta_{{ABC}} =\mathrm{48}+\mathrm{32}+\mathrm{96}+\mathrm{72}+\mathrm{32}+\mathrm{8}=\mathrm{288}\:{cm}^{\mathrm{2}} \\ $$

Answered by mr W last updated on 05/Feb/20

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