Question Number 215715 by Hanuda354 last updated on 15/Jan/25 | ||
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Commented by Hanuda354 last updated on 15/Jan/25 | ||
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$$\mathrm{Find}\:{x} \\ $$ | ||
Answered by som(math1967) last updated on 16/Jan/25 | ||
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$$\:\frac{{CF}}{{FE}}=\frac{{CD}}{{DE}}\:\:\:\left[{DF}\:{is}\:{bisector}\:{of}\:\angle{D}\right] \\ $$$$\:{CF}=\frac{\mathrm{6}}{\mathrm{9}}×\mathrm{7}.\mathrm{2}=\mathrm{4}.\mathrm{8} \\ $$$${now}\:\:{BF}\parallel{AE} \\ $$$$\therefore\:\frac{{x}}{\mathrm{7}.\mathrm{8}}=\frac{\mathrm{4}.\mathrm{8}}{\mathrm{7}.\mathrm{2}}\:\Rightarrow{x}=\frac{\mathrm{4}.\mathrm{8}×\mathrm{7}.\mathrm{8}}{\mathrm{7}.\mathrm{2}}=\mathrm{5}.\mathrm{2}\:{unit} \\ $$ | ||
Commented by Hanuda354 last updated on 16/Jan/25 | ||
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$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much},\:\mathrm{sir}. \\ $$ | ||