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Question Number 134097 by bobhans last updated on 27/Feb/21

$$\mathrm{In}\mathrm{a}\mathrm{square}\mathrm{ABCD},\mathrm{a}\mathrm{triangle}$$$$\mathrm{APQ}\mathrm{inscribed}\mathrm{in}\mathrm{it}.\mathrm{AP}=4\mathrm{cm},$$$$\mathrm{PQ}=3\mathrm{cm}\mathrm{and}\mathrm{AQ}=5\mathrm{cm}.\mathrm{Point}$$$$\mathrm{P}\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{side}\mathrm{BC}\mathrm{and}\mathrm{point}\mathrm{Q}$$$$\mathrm{is}\mathrm{on}\mathrm{side}\mathrm{CD}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}$$$$\mathrm{square}\mathrm{ABCD}.$$

Answered by mr W last updated on 27/Feb/21

Commented by mr W last updated on 27/Feb/21

$$BP=\sqrt{4^2-a^2}$$$$\frac{PC}{3}=\frac{a}{4}$$$$\Rightarrow PC=\frac{3a}{4}$$$$\sqrt{4^2-a^2}+\frac{3a}{4}=a$$$$4^2-a^2=\frac{a^2}{16}$$$$a^2=\frac{{16}^2}{17}$$$$\Rightarrow a=\frac{16}{\sqrt{17}}$$$$areaofABCD=a^2=\frac{{16}^2}{17}=\frac{256}{17}\approx 15.05$$

Commented by bobhans last updated on 28/Feb/21

$$\mathrm{thanks}$$

Commented by otchereabdullai@gmail.com last updated on 12/Mar/21

$$\mathrm{Thanks}\mathrm{prof}\mathrm{w}$$

Answered by liberty last updated on 01/Mar/21

Commented by liberty last updated on 01/Mar/21

$${\left(4u\right)}^2+u^2=16$$$$\iff {16\mathrm{u}}^2+{\mathrm{u}}^2=16;{\mathrm{u}}^2=\frac{16}{17}$$$$\mathrm{so}\mathrm{area}\mathrm{of}\mathrm{square}\mathrm{is}{16\mathrm{u}}^2=\frac{256}{17}\approx 15.0588$$

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