Menu Close

Category: Geometry

Question-206292

Question Number 206292 by cortano21 last updated on 11/Apr/24 Answered by A5T last updated on 11/Apr/24 $${Let}\:{AE}={x};{BE}={y};{BF}={v};{FC}={w} \\ $$$${S}+\mathrm{39}=\frac{\left(\mathrm{2}{v}+{w}\right)\left({x}+{y}\right)}{\mathrm{2}}=\frac{\mathrm{2}{v}\left({x}+{y}\right)}{\mathrm{2}}+\mathrm{15}\Rightarrow{S}={vx} \\ $$$${wx}=\mathrm{54}−{S}=\mathrm{30}−{yw}\Rightarrow{yw}={S}−\mathrm{24} \\ $$$$\frac{\left[{BFD}\right]}{\left[{DFC}\right]}=\frac{{v}}{{w}}\Rightarrow\left[{BFD}\right]=\frac{\mathrm{15}{v}}{{w}} \\ $$$$\frac{\left[{EDB}\right]}{\left[{ADE}\left[\right.\right.}=\frac{{y}}{{x}}\Rightarrow{EDB}=\frac{\mathrm{27}{y}}{{x}}…

Question-206275

Question Number 206275 by cortano21 last updated on 11/Apr/24 Answered by HeferH24 last updated on 11/Apr/24 $$\:{CDEF}\:=\:{m} \\ $$$$\:{ABFE}\:=\:\mathrm{3}{m} \\ $$$$\:\left(\frac{\mathrm{4}}{\mathrm{6}}\right)^{\mathrm{2}} =\:\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{4}{k}}{\mathrm{9}{k}} \\ $$$$\:\mathrm{5}{k}\:=\:\mathrm{4}{m} \\…