Question Number 218890 by Spillover last updated on 17/Apr/25 | ||
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Answered by mr W last updated on 17/Apr/25 | ||
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$$\frac{\mathrm{1}}{\mathrm{1}}×\frac{{BF}}{{FD}}×\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{1}\:\Rightarrow\frac{{BF}}{{FD}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\Delta_{{ADF}} =\frac{\mathrm{3}}{\mathrm{7}}×\Delta_{{ADB}} =\frac{\mathrm{3}}{\mathrm{7}}×\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{27}}{\mathrm{14}} \\ $$$$\left[{CDFE}\right]=\Delta_{{ACF}} −\Delta_{{ADF}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{27}}{\mathrm{14}}=\frac{\mathrm{15}}{\mathrm{14}}\:\checkmark \\ $$ | ||
Commented by Spillover last updated on 17/Apr/25 | ||
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$${correct}.{thanks} \\ $$ | ||
Answered by Spillover last updated on 17/Apr/25 | ||
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Answered by Spillover last updated on 17/Apr/25 | ||
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Answered by Spillover last updated on 17/Apr/25 | ||
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Answered by Spillover last updated on 17/Apr/25 | ||
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