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Question Number 137363 by liberty last updated on 02/Apr/21

$$$$As illustrated, the rectangle has an area of 1, and E is the midpoint of AD. BF is one third of AB. What is the area of the shadow?

Commented by liberty last updated on 02/Apr/21

Answered by mr W last updated on 02/Apr/21

Commented by mr W last updated on 02/Apr/21

$$EJ=\frac{AF}{2}=BF$$$$\Rightarrow BG=GE$$$$EJ=\frac{AF}{2}=\frac{1}{2}\times \frac{2}{3}AB=\frac{AB}{3}=\frac{DC}{3}$$$$\Rightarrow EI=\frac{IC}{3}=\frac{EC}{4}$$$$FK=\frac{AE}{3}=\frac{BC}{6}$$$$\Rightarrow BK=6\times HK=\frac{6}{7}\times BK=\frac{6}{7}\times \frac{BE}{3}=\frac{2}{7}\times BE$$$$\mathrm{\Delta }BHC=\frac{2}{7}\times \mathrm{\Delta }BEC$$$$\mathrm{\Delta }EGI=\frac{1}{4}\times \mathrm{\Delta }EGC=\frac{1}{8}\times \mathrm{\Delta }BEC$$$$shados=(1-\frac{2}{7}-\frac{1}{8})\times \mathrm{\Delta }BEC$$$$=\frac{33}{56}\times \mathrm{\Delta }BEC$$$$=\frac{33}{56}\times \frac{1}{2}=\frac{33}{112}$$

Answered by EDWIN88 last updated on 02/Apr/21

$$\mathrm{we}\mathrm{can}\mathrm{make}\mathrm{simpler}\mathrm{by}\mathrm{considering}\mathrm{this}$$$$\mathrm{a}12\times 12\mathrm{square}\mathrm{with}\mathrm{B}\mathrm{at}\mathrm{origin}.\mathrm{We}\mathrm{have}$$$$\mathrm{to}\mathrm{remember}\mathrm{to}\mathrm{divide}\mathrm{by}{12}^2\mathrm{at}\mathrm{the}\mathrm{end}.$$$$\mathrm{We}\mathrm{get}\mathrm{coordinates}\mathrm{B}(0,0),\mathrm{A}(0,12),\mathrm{F}(0,4)$$$$\mathrm{E}(6,12),\mathrm{D}(12,12),\mathrm{C}(12,0).\mathrm{we}\mathrm{need}\mathrm{to}?$$$$\mathrm{calculate}\mathrm{G},\mathrm{H},\mathrm{I}.\mathrm{G}\mathrm{is}\mathrm{the}\mathrm{meet}\mathrm{if}\mathrm{BE}\mathrm{and}\mathrm{DF}$$$$\Rightarrow \mathrm{G}(3,6);\mathrm{H}(\frac{12}{7},\frac{24}{7});\mathrm{I}(\frac{15}{2},9).$$$$\mathrm{The}\mathrm{shadow}\mathrm{area}\mathrm{is}$$$$=\frac{1}{2}\mid \left|\begin{array}{c}120\\ \frac{12}{7}\frac{24}{7}\end{array}\right|+\left|\begin{array}{c}\frac{12}{7}\frac{24}{7}\\ 36\end{array}\right|+\left|\begin{array}{c}36\\ \frac{15}{2}9\end{array}\right|+\left|\begin{array}{c}\frac{15}{2}9\\ 120\end{array}\right|\mid$$$$=\frac{1}{2}\mid \frac{288}{7}+0-18-108\mid$$$$=\frac{1}{2}.\frac{594}{7}=\frac{297}{7}.\mathrm{Therefore}\mathrm{we}\mathrm{get}\mathrm{the}$$$$\mathrm{shadow}\mathrm{area}\mathrm{is}\frac{297}{7\times 144}=\frac{33}{112}$$$$$$

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