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  • the line x+y=a meets x y axes at A and B respectively. A triangle APQ is inscribed in the triangle OAB O being origin

The line x + y = p meets the x- and y-axes at A and B,respectively. A triangle APQ is inscribed in triangle OAR, O being the origin, with right angle at Q. P and Q lie respectively, on OR and AB. If the area of triangle APQ is 3/8th of the area of triangle OAB, then AQ/BQ is equal to

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\begin{array}{l} \text {We have given:}\\\\ \frac{\operatorname{arAQP}}{\operatorname{arAOB}}=\frac{3}{8} \\\\ \text {Let } \frac{\mathrm{AQ}}{\mathrm{BQ}}=\mathrm{k} \\ \text {Therefore, } \frac{\frac{\mathrm{p}^{2} \mathrm{k}}{\mathrm{k}+1)^{2}}}{\frac{\mathrm{p}^{2}}{2}}=\frac{3}{8} \end{array}

\begin{aligned} &\frac{2 \mathrm{k}}{(\mathrm{k}+1)^{2}}=\frac{3}{8}\\ &16 \mathrm{k}=3 \mathrm{k}^{2}+6 \mathrm{k}+3\\ &3 \mathrm{k}^{2}-10 \mathrm{k}+3=0\\ &\mathrm{k}=\frac{1}{3} \text { and } 3\\ &\text { But } 3 \text { is taken since } \frac{1}{3} \text { gives a negative value. } \end{aligned}

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