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Question Number 193149 by Mingma last updated on 04/Jun/23

Answered by ajfour last updated on 05/Jun/23

$$\frac{\sqrt{3}}{4}{p}^2=A$$$$p=k\sqrt{A}where{k}^2=\frac{4}{\sqrt{3}}$$$$q=k\sqrt{B}$$$$B=9A$$$$P(\frac{p}{2},\frac{p\sqrt{3}}{2})$$$$Q(\frac{2p+q}{2},\frac{q\sqrt{3}}{2})$$$$q=3p$$$$PQ=\sqrt{{\left(\frac{p+q}{2}\right)}^2+3{\left(\frac{q-p}{2}\right)}^2}$$$$PQ=c=\sqrt{7}p$$$$m=\frac{(q-p)\sqrt{3}}{q+p}=\frac{\sqrt{3}}{2}$$$$M(\frac{3p+q}{4},\frac{(p+q)\sqrt{3}}{4})$$$$M(\frac{3p}{2},\sqrt{3}p)$$$$\frac{c\sqrt{3}}{2}=\frac{\sqrt{21}}{2}p$$$$\tan \varphi =\frac{1}{m}=\frac{2}{\sqrt{3}}\Rightarrow \sin \varphi =\frac{2}{\sqrt{7}}$$$$H=y_M+\frac{c\sqrt{3}}{2}\sin \varphi$$$$=\sqrt{3}p+\frac{p\sqrt{7}\sqrt{3}}{2}\times \frac{2}{\sqrt{7}}$$$$=2\sqrt{3}p$$$$p=kasA=1$$$$H=2\sqrt{3}k$$$$W=p+q=4p=4k$$$$Rectanglearea=WH$$$$=\left(4k\right)\left(2\sqrt{3}k\right)=8\sqrt{3}{k}^2$$$$=8\sqrt{3}\times \frac{4}{\sqrt{3}}=32sq.units.$$

Commented by Mingma last updated on 05/Jun/23

Perfect �� I'm excited you corrected yourself!

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