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Category: Geometry

Draw-a-rectangle-of-maximum-perimeter-by-ruler-and-compass-when-area-is-ab-AB-a-CD-b-are-given-

Question Number 3830 by Rasheed Soomro last updated on 21/Dec/15 $$\mathcal{D}{raw}\:{a}\:{rectangle}\:{of}\:{maximum}\:{perimeter}, \\ $$$${by}\:{ruler}\:{and}\:{compass},{when}\:{area}\:{is}\:\boldsymbol{\mathrm{ab}}.\: \\ $$$$\left(\boldsymbol{\mathrm{AB}}\:=\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{CD}}=\boldsymbol{\mathrm{b}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}}.\right) \\ $$ Commented by prakash jain last updated on 22/Dec/15…

Consider-a-triangle-ABC-Let-D-and-E-are-two-points-on-AB-and-AC-respectively-such-that-DE-BC-Now-there-are-two-parts-of-ABC-ADE-and-trapizoid-DBCE-If-these-two-regions-have-same-area-W

Question Number 3823 by Rasheed Soomro last updated on 21/Dec/15 $${Consider}\:{a}\:{triangle}\:\mathrm{ABC}.\:{Let}\:\mathrm{D}\:\:{and}\:\:\mathrm{E} \\ $$$${are}\:{two}\:{points}\:{on}\:\mathrm{AB}\:\:{and}\:\:\mathrm{AC}\:{respectively} \\ $$$${such}\:{that}\:\mathrm{DE}\:\parallel\:\mathrm{BC}.\:{Now}\:{there}\:{are}\:{two} \\ $$$${parts}\:{of}\:\bigtriangleup\mathrm{ABC}\::\:\bigtriangleup\mathrm{ADE}\:\:\:{and}\:\:{trapizoid} \\ $$$$\mathrm{DBCE}.\:{If}\:{these}\:{two}\:{regions}\:{have}\:{same}\:{area} \\ $$$${What}\:{will}\:{be}\:{the}\:{ratio}\:{of}\:{two}\:{distances}\:: \\ $$$$\left({i}\right)\:{distance}\:{of}\:\mathrm{DE}\:{from}\:{point}\:\mathrm{A}\:{and} \\ $$$$\left({ii}\right)\:{distance}\:{between}\:\mathrm{BC}\:{and}\:\mathrm{DE}\:\:?…

A-chord-divides-the-circle-in-two-segments-having-areas-s-1-and-s-2-If-diameter-perpendicular-to-this-chord-is-cut-into-1-3-by-the-chord-what-is-s-1-s-2-

Question Number 3808 by Rasheed Soomro last updated on 21/Dec/15 $${A}\:{chord}\:{divides}\:\:{the}\:{circle}\:{in}\:{two} \\ $$$${segments},{having}\:{areas}\:{s}_{\mathrm{1}} \:{and}\:\:{s}_{\mathrm{2}} . \\ $$$${If}\:{diameter},\:{perpendicular}\:{to}\:{this} \\ $$$${chord}\:{is}\:{cut}\:{into}\:\mathrm{1}:\mathrm{3}\:{by}\:{the}\:{chord}\:,{what}\:{is}\:{s}_{\mathrm{1}} :{s}_{\mathrm{2}} \:? \\ $$$$ \\ $$…

Question-69272

Question Number 69272 by TawaTawa last updated on 22/Sep/19 Answered by $@ty@m123 last updated on 22/Sep/19 $${Let}\:\angle{DPC}=\theta \\ $$$$\Rightarrow\angle{Q}=\theta−\mathrm{30} \\ $$$${Let}\:{CP}={x} \\ $$$${In}\:\bigtriangleup{DCP}, \\ $$$$\mathrm{tan}\:\theta=\frac{\sqrt{\mathrm{3}}}{{x}}\:…\left(\mathrm{1}\right)…

Derive-a-formula-of-volume-of-right-circular-cone-when-the-formula-of-volume-of-cyllinder-is-given-

Question Number 3714 by Rasheed Soomro last updated on 19/Dec/15 $$\mathcal{D}\mathrm{erive}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{volume}\:\mathrm{of}\:\boldsymbol{\mathrm{right}}\:\boldsymbol{\mathrm{circular}}\:\boldsymbol{\mathrm{cone}} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{volume}\:\mathrm{of}\:\boldsymbol{\mathrm{cyllinder}}\:\:\mathrm{is}\:\mathrm{given}. \\ $$ Answered by Filup last updated on 19/Dec/15 $$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{answer}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{given}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{can}\:\mathrm{show}\:\mathrm{it}\:\mathrm{this}\:\mathrm{way}:…

Lets-say-we-have-an-n-gon-All-sides-are-equal-When-n-3-interior-angles-180-3-60-n-4-360-4-90-n-t-180-t-2-t-For-a-circle-essentially-an-gon-n-180lim-t-t

Question Number 3662 by Filup last updated on 18/Dec/15 $$\mathrm{Lets}\:\mathrm{say}\:\mathrm{we}\:\mathrm{have}\:\mathrm{an}\:{n}−\mathrm{gon}. \\ $$$$\mathrm{All}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$ \\ $$$$\mathrm{When}\:{n}=\mathrm{3},\:\mathrm{interior}\:\mathrm{angles}\:\theta=\frac{\mathrm{180}}{\mathrm{3}} \\ $$$$\theta=\mathrm{60}° \\ $$$$ \\ $$$${n}=\mathrm{4},\:\theta=\frac{\mathrm{360}}{\mathrm{4}}=\mathrm{90}° \\ $$$$\vdots \\…

Question-69192

Question Number 69192 by TawaTawa last updated on 21/Sep/19 Answered by mr W last updated on 21/Sep/19 $${let}\:\angle{BAD}=\alpha \\ $$$${side}\:{length}\:={a} \\ $$$$\frac{{a}}{\mathrm{sin}\:\left(\mathrm{60}+\alpha\right)}=\frac{\mathrm{3}}{\mathrm{sin}\:\alpha}\:\:\:\:…\left({i}\right) \\ $$$$\frac{{a}}{\mathrm{sin}\:\left(\mathrm{60}+\mathrm{30}−\alpha\right)}=\frac{{a}}{\mathrm{cos}\:\alpha}=\frac{\mathrm{5}}{\mathrm{sin}\:\left(\mathrm{30}−\alpha\right)}\:\:\:…\left({ii}\right) \\…