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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

[ADF]=[ABE]=25+70=95 ⇒[EHG]=95−25−55=15 [HGJB]=2×95−15−80=95

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

Commented by mrW1 last updated on 22/Sep/17

let S=area of ABCD=AB×AD [ABE]=(1/2)×AB×AE=(1/2)×AB×((AD)/2)=(1/4)×AB×SD=(S/4)=25+70=95 ⇒S=4×95=380 [ADF]=(1/2)×AD×DF=(1/2)×AD×((AB)/2)=(1/4)×AD×AB=(S/4)=95 [EHG]=[ADF]−25−55=95−25−55=15 [BEC]=(1/2)×BC×AB=(S/2)=((380)/2)=190 shaded area=[BEC]−80−15=190−80−15=95

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

Commented by Joel577 last updated on 22/Sep/17

pls explain sir I didnt understand why area ΔABC = ΔAe and how did you get [HGJB] = 95

$${pls}\:{explain}\:{sir} \\ $$$${I}\:{didnt}\:{understand}\:{why}\:{area}\:\Delta{ABC}\:=\:\Delta{Ae} \\ $$$${and}\:{how}\:{did}\:{you}\:{get}\:\left[{HGJB}\right]\:=\:\mathrm{95} \\ $$

Commented by Joel577 last updated on 23/Sep/17

Understood. Thank you very much

$${Understood}.\:{Thank}\:{you}\:{very}\:{much} \\ $$

Answered by sma3l2996 last updated on 23/Sep/17

if ABCD is a rectangle so the area ofΔABE should be equale the area of ΔBCF, so why 25+70≠80+20

$${if}\:{ABCD}\:{is}\:{a}\:{rectangle}\:{so}\:{the}\:{area}\:{of}\Delta{ABE}\:{should}\:{be}\:{equale}\:{the}\:{area}\:{of}\:\Delta{BCF},\:\:\:{so}\:\:{why}\:\:\mathrm{25}+\mathrm{70}\neq\mathrm{80}+\mathrm{20} \\ $$

Answered by Tikufly last updated on 08/Oct/17

In rectangle ABCD, F is the mid point of DC. ∴ area(ABF)=2area(ADF)=2area(BCF) ⇒2area(ADF)=2area(BCF) ⇒area(ADF)=area(BCF) ⇒25+area(EGH)+55=20+80 ⇒area(EGH)=20cm^2 . E is the mid point of AD. ∴area(BEC)=2area(ABE) ⇒area(EGH)+area(HGJB)+area(JCB)=2×(25+70) ⇒20+area(HGJB)+80=190 ⇒area(HGJB)=90 Hence, the area of the shaded region is 90cm^2 .

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

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