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Question Number 11632 by Joel576 last updated on 29/Mar/17

Answered by sandy_suhendra last updated on 29/Mar/17

Commented by sandy_suhendra last updated on 29/Mar/17

Area ΔAOB + Area Δ BPC =((AB×(1/2)CD)/2) + ((BC×(1/2)CD)/2) =(1/4)AB×CD + (1/4)BC×CD =(1/4)CD×(AB+BC) =(1/4)CD×AC=(1/4)×50×200 =2,500 A ΔAEC=((AC×CD)/2)=((200×50)/2) =5,000 A shaded = A ΔAEC−(A ΔAOB+A ΔBPC) =5,000−2,500=2,500

$$\mathrm{Area}\:\Delta\mathrm{AOB}\:+\:\mathrm{Area}\:\Delta\:\mathrm{BPC} \\ $$$$=\frac{\mathrm{AB}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{CD}}{\mathrm{2}}\:+\:\frac{\mathrm{BC}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{CD}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{AB}×\mathrm{CD}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{BC}×\mathrm{CD} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{CD}×\left(\mathrm{AB}+\mathrm{BC}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{CD}×\mathrm{AC}=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{50}×\mathrm{200} \\ $$$$=\mathrm{2},\mathrm{500} \\ $$$$ \\ $$$$\mathrm{A}\:\Delta\mathrm{AEC}=\frac{\mathrm{AC}×\mathrm{CD}}{\mathrm{2}}=\frac{\mathrm{200}×\mathrm{50}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{5},\mathrm{000} \\ $$$$ \\ $$$$\mathrm{A}\:\mathrm{shaded}\:=\:\mathrm{A}\:\Delta\mathrm{AEC}−\left(\mathrm{A}\:\Delta\mathrm{AOB}+\mathrm{A}\:\Delta\mathrm{BPC}\right)\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{5},\mathrm{000}−\mathrm{2},\mathrm{500}=\mathrm{2},\mathrm{500} \\ $$

Commented by Joel576 last updated on 29/Mar/17

but AC = 200

$$\mathrm{but}\:\mathrm{AC}\:=\:\mathrm{200} \\ $$

Commented by sandy_suhendra last updated on 29/Mar/17

oh sorry, I have fixed it

$$\mathrm{oh}\:\mathrm{sorry},\:\mathrm{I}\:\mathrm{have}\:\mathrm{fixed}\:\mathrm{it} \\ $$

Commented by Joel576 last updated on 29/Mar/17

no problem, thank you very much

$$\mathrm{no}\:\mathrm{problem},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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