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Category: Geometry

Question-57283

Question Number 57283 by ajfour last updated on 01/Apr/19 Commented by ajfour last updated on 01/Apr/19 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{cut}\:\mathrm{out}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{and} \\ $$$$\mathrm{rectangle}\:\mathrm{from}\:\mathrm{an}\:\mathrm{equilateral}\: \\ $$$$\mathrm{triangular}\:\mathrm{sheet}\:\mathrm{of}\:\mathrm{paper},\:\mathrm{edges}\:\mathrm{a}. \\ $$…

Question-188339

Question Number 188339 by normans last updated on 28/Feb/23 Answered by universe last updated on 28/Feb/23 $${ar}\left(\bigtriangleup{ABC}\right)\:=\:{ar}\left({ABD}\right)\:\:+\:\:{ar}\left({ADC}\right) \\ $$$${AB}.{AC}\mathrm{sin}\:\left(\alpha+\beta\right)\:=\:{AB}.{AD}\mathrm{sin}\left(\alpha\right)\:+\:{AD}.{AC}\mathrm{sin}\left(\beta\right) \\ $$$$\frac{\mathrm{sin}\left(\alpha+\beta\right)\:}{{AD}}\:=\:\frac{\mathrm{sin}\alpha\:}{{AC}}\:+\:\frac{\mathrm{sin}\beta\:}{{AB}}\: \\ $$$$\:\: \\ $$…

if-the-on-ABC-inner-circle-radius-and-outer-circle-radius-r-and-R-show-that-r-4R-sin-A-2-sin-B-2-sin-C-2-

Question Number 188320 by normans last updated on 27/Feb/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{if}}\:\boldsymbol{{the}}\:\boldsymbol{{on}}\:\bigtriangleup\boldsymbol{{ABC}},\: \\ $$$$\:\:\boldsymbol{{inner}}\:\boldsymbol{{circle}}\:\boldsymbol{{radius}}\:\boldsymbol{{and}}\:\boldsymbol{{outer}}\:\boldsymbol{{circle}}\:\boldsymbol{{radius}}\:\boldsymbol{{r}}\:\boldsymbol{{and}}\:\boldsymbol{{R}}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{{show}}\:\boldsymbol{{that}}; \\ $$$$\:\:\:\:\:\:\boldsymbol{{r}}\:=\:\mathrm{4}\boldsymbol{{R}}\:\boldsymbol{{sin}}\:\frac{\boldsymbol{{A}}}{\mathrm{2}}\:\centerdot\:\boldsymbol{{sin}}\:\frac{\boldsymbol{{B}}}{\mathrm{2}}\:\centerdot\:\boldsymbol{{sin}}\:\frac{\boldsymbol{{C}}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by mr…

Question-188296

Question Number 188296 by Rupesh123 last updated on 27/Feb/23 Answered by HeferH last updated on 27/Feb/23 $$\mathrm{2}^{\mathrm{2}} \:=\:\mathrm{1}\centerdot{h}\:\Rightarrow\:{h}\:=\:\mathrm{4}\: \\ $$$$\:{h}^{\mathrm{2}} \:=\:\mathrm{2}{x} \\ $$$$\:\mathrm{16}\:=\:\mathrm{2}{x} \\ $$$$\:{x}\:=\:\mathrm{8}\:…