Question Number 4505 by Yozzii last updated on 03/Feb/16 Commented by Yozzii last updated on 03/Feb/16 $$\bigtriangleup{ACD}\:{and}\:\bigtriangleup{DBE}\:{are}\:{equilateral}. \\ $$$${F}\:{is}\:{the}\:{midpoint}\:{of}\:{AE}\:{and}\:{G}\:{is}\:{the} \\ $$$${midpoint}\:{of}\:{BC}.\:{D}\:{is}\:{a}\:{point}\:{on}\:{the} \\ $$$${line}\:{AB}.\:{Prove}\:{that}\:\bigtriangleup{FGD}\:{is}\: \\ $$$${equilateral}.\:…
Question Number 4496 by FilupSmith last updated on 01/Feb/16 $$\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{question}\:\mathrm{but}\:\mathrm{for}\:\mathrm{some} \\ $$$$\mathrm{silly}\:\mathrm{reason}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{figure}\:\mathrm{it}\:\mathrm{out}… \\ $$$$ \\ $$$$\mathrm{1}. \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:{r}\:\mathrm{and} \\ $$$$\mathrm{area}\:{A},\:\mathrm{and}\:\mathrm{I}\:\mathrm{wish}\:\mathrm{to}\:\mathrm{make}\:\mathrm{a}\:\mathrm{new}\:\mathrm{circle}\:\mathrm{with} \\ $$$${n}\:\mathrm{times}\:\mathrm{the}\:\mathrm{area},\:\mathrm{for}\:\mathrm{what}\:\mathrm{new}\:\mathrm{value} \\ $$$$\mathrm{of}\:{r}\:\mathrm{should}\:\mathrm{be}\:\mathrm{used}? \\…
Question Number 135570 by bemath last updated on 14/Mar/21 $${Geometry} \\ $$If the area of the triangle formed by the positive x-axis, the normal and…
Question Number 135522 by Dwaipayan Shikari last updated on 13/Mar/21 Commented by Dwaipayan Shikari last updated on 13/Mar/21 $$ \\ $$ Two large circles altogether…
Question Number 4437 by Rasheed Soomro last updated on 25/Jan/16 $$\mathrm{An}\:\boldsymbol{\mathrm{ellipse}}\:\mathrm{having}\:\boldsymbol{\mathrm{semi}}-\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axis}}\: \\ $$$$\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\boldsymbol{\mathrm{semi}}-\boldsymbol{\mathrm{minor}}\:\boldsymbol{\mathrm{axis}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{b}} \\ $$$$\mathrm{and}\:\mathrm{a}\:\boldsymbol{\mathrm{circle}}\:\mathrm{having}\:\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{r}}\:\mathrm{have}\:\mathrm{equal} \\ $$$$\boldsymbol{\mathrm{area}}. \\ $$$$\mathrm{Express}\:\boldsymbol{\mathrm{r}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}. \\ $$$$ \\ $$ Answered by…
Question Number 4409 by Rasheed Soomro last updated on 22/Jan/16 $$\mathcal{D}{ivide}\:{a}\:{circle}\:{into}\:{two}\:{congruent} \\ $$$${regions}\:{such}\:{that}\:{they}\:{have}\:{no}\: \\ $$$${straightedge}. \\ $$ Commented by prakash jain last updated on 22/Jan/16…
Question Number 4390 by Rasheed Soomro last updated on 17/Jan/16 Answered by Rasheed Soomro last updated on 20/Jan/16 Commented by Rasheed Soomro last updated on…
Question Number 4387 by Rasheed Soomro last updated on 17/Jan/16 Commented by Rasheed Soomro last updated on 17/Jan/16 $$\mathrm{In}\:\mathrm{the}\:\mathrm{trapezium}\:\mathrm{m}\angle\mathrm{A}=\mathrm{m}\angle\mathrm{B}=\frac{\pi}{\mathrm{2}}\:\mathrm{rad}. \\ $$$$\mathrm{m}\overline {\mathrm{AB}}=\mathrm{m}\overline {\mathrm{AD}}=\mathrm{x}\:\mathrm{units}\:\mathrm{and}\:\mathrm{m}\overline {\mathrm{BC}}=\mathrm{2x}\:\mathrm{units}. \\…
Question Number 135458 by Dwaipayan Shikari last updated on 13/Mar/21 Commented by Dwaipayan Shikari last updated on 13/Mar/21 Commented by Dwaipayan Shikari last updated on…
Question Number 4384 by Rasheed Soomro last updated on 16/Jan/16 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{r}_{\mathrm{1}} \:\mathrm{has}\:\mathrm{been}\:\mathrm{divided} \\ $$$$\mathrm{into}\:\mathrm{two}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{area}, \\ $$$$\mathrm{by}\:\mathrm{an}\:\mathrm{arc}\:\mathrm{having}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{radius}\left(\mathrm{r}_{\mathrm{2}} \right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}. \\ $$ Commented by Rasheed Soomro…