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Category: Geometry

Question-121615

Question Number 121615 by ajfour last updated on 10/Nov/20 Commented by ajfour last updated on 10/Nov/20 $${The}\:{blue}\:{triangle}\:{is}\:{right}\:{angled}\:{and} \\ $$$${isosceles}\:{with}\:\:{AF}={BF}=\mathrm{3}. \\ $$$${Radius}\:{of}\:{circle}\:{r}=\mathrm{2}. \\ $$$${Find}\:{maximum}\:{side}\:{length}\:{of} \\ $$$${equilateral}\:\bigtriangleup{DEF}.…

Question-187146

Question Number 187146 by Rupesh123 last updated on 14/Feb/23 Answered by a.lgnaoui last updated on 14/Feb/23 $$\bigtriangleup{SMI}\:\:{et}\:{RNI}\:{semblables}\:\:\: \\ $$$${I}\:{centre}\:{de}\:{SRMN}\:\:{SK}=\frac{{PQ}}{\mathrm{2}}=\mathrm{3} \\ $$$$\bigtriangleup{BRQ}\:\:\:{RIO}\:\:{Semblables} \\ $$$$\frac{{OR}}{{OI}}=\frac{\frac{{RN}}{\mathrm{2}}}{{HQ}}=\frac{{RQ}}{{BQ}}\Rightarrow\:\:\:\frac{{RN}}{\mathrm{2}{HQ}}=\frac{{RQ}}{{BQ}} \\ $$$${HQ}=\mathrm{3}\:\:{RQ}=\mathrm{6}\:\:\:{BQ}={BP}+\mathrm{6}…

Question-187100

Question Number 187100 by Tons last updated on 13/Feb/23 Answered by a.lgnaoui last updated on 14/Feb/23 $$\bigtriangleup{APB}\:\:\:\:{AB}\mathrm{sin}\:{X}={AC}\mathrm{cos}\:{Y}\:\: \\ $$$$ \\ $$$$\:{BC}=\mathrm{2}{AB}\mathrm{cos}\:{Y}\:\:\Rightarrow\begin{cases}{{Y}=\frac{\pi}{\mathrm{2}}−{X}}\\{{AB}={AC}}\end{cases} \\ $$$$\mathrm{sin}\:{X}=\mathrm{cos}\:{Y}\:\:\: \\ $$$$\bigtriangleup{ABCD}\:\:\:\:{Sqart}\left({Care}\right)…

Question-121539

Question Number 121539 by ajfour last updated on 09/Nov/20 Commented by ajfour last updated on 09/Nov/20 $$\:\:\:\:\:\:\:\:{Find}\:\frac{{s}}{{R}}\:\:{for}\:{maximum}\:{blue} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{triangular}\:{area}. \\ $$$$\left({square}\:{remains}\:{within}\:{semicircle}\right) \\ $$ Answered by…