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Question Number 21366 by Glorious Man last updated on 22/Sep/17

Answered by mrW1 last updated on 22/Sep/17

$$\left[\mathrm{ADF}\right]=\left[\mathrm{ABE}\right]=25+70=95$$$$\Rightarrow \left[\mathrm{EHG}\right]=95-25-55=15$$$$\left[\mathrm{HGJB}\right]=2\times 95-15-80=95$$

Commented by mrW1 last updated on 22/Sep/17

$$\mathrm{let}\mathrm{S}=\mathrm{area}\mathrm{of}\mathrm{ABCD}=\mathrm{AB}\times \mathrm{AD}$$$$\left[\mathrm{ABE}\right]=\frac{1}{2}\times \mathrm{AB}\times \mathrm{AE}=\frac{1}{2}\times \mathrm{AB}\times \frac{\mathrm{AD}}{2}=\frac{1}{4}\times \mathrm{AB}\times \mathrm{SD}=\frac{\mathrm{S}}{4}=25+70=95$$$$\Rightarrow \mathrm{S}=4\times 95=380$$$$\left[\mathrm{ADF}\right]=\frac{1}{2}\times \mathrm{AD}\times \mathrm{DF}=\frac{1}{2}\times \mathrm{AD}\times \frac{\mathrm{AB}}{2}=\frac{1}{4}\times \mathrm{AD}\times \mathrm{AB}=\frac{\mathrm{S}}{4}=95$$$$\left[\mathrm{EHG}\right]=\left[\mathrm{ADF}\right]-25-55=95-25-55=15$$$$\left[\mathrm{BEC}\right]=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB}=\frac{\mathrm{S}}{2}=\frac{380}{2}=190$$$$\mathrm{shaded}\mathrm{area}=\left[\mathrm{BEC}\right]-80-15=190-80-15=95$$

Commented by Joel577 last updated on 22/Sep/17

$$plsexplainsir$$$$Ididntunderstandwhyarea\mathrm{\Delta }ABC=\mathrm{\Delta }Ae$$$$andhowdidyouget\left[HGJB\right]=95$$

Commented by Joel577 last updated on 23/Sep/17

$$Understood.Thankyouverymuch$$

Answered by sma3l2996 last updated on 23/Sep/17

$$ifABCDisarectanglesotheareaof\mathrm{\Delta }ABEshouldbeequaletheareaof\mathrm{\Delta }BCF,sowhy25+70\ne 80+20$$

Answered by Tikufly last updated on 08/Oct/17

$$\mathrm{In}\mathrm{rectangle}\mathrm{ABCD},\mathrm{F}\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}$$$$\mathrm{of}\mathrm{DC}.$$$$\therefore \mathrm{area}\left(\mathrm{ABF}\right)=2\mathrm{area}\left(\mathrm{ADF}\right)=2\mathrm{area}\left(\mathrm{BCF}\right)$$$$\Rightarrow 2\mathrm{area}\left(\mathrm{ADF}\right)=2\mathrm{area}\left(\mathrm{BCF}\right)$$$$\Rightarrow \mathrm{area}\left(\mathrm{ADF}\right)=\mathrm{area}\left(\mathrm{BCF}\right)$$$$\Rightarrow 25+\mathrm{area}\left(\mathrm{EGH}\right)+55=20+80$$$$\Rightarrow \mathrm{area}\left(\mathrm{EGH}\right)={20\mathrm{cm}}^2.$$$$$$$$\mathrm{E}\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{AD}.$$$$\therefore \mathrm{area}\left(\mathrm{BEC}\right)=2\mathrm{area}\left(\mathrm{ABE}\right)$$$$\Rightarrow \mathrm{area}\left(\mathrm{EGH}\right)+\mathrm{area}\left(\mathrm{HGJB}\right)+\mathrm{area}\left(\mathrm{JCB}\right)=2\times (25+70)$$$$\Rightarrow 20+\mathrm{area}\left(\mathrm{HGJB}\right)+80=190$$$$\Rightarrow \mathrm{area}\left(\mathrm{HGJB}\right)=90$$$$\mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}\mathrm{is}{90\mathrm{cm}}^2.$$

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