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Category: Geometry

How-can-cut-a-right-angeled-triangle-to-make-a-square-how-can-cut-a-equilateral-triangle-to-make-a-rectangle-

Question Number 54788 by behi83417@gmail.com last updated on 11/Feb/19 $${H}\boldsymbol{\mathrm{ow}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{cut}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{right}}\:\boldsymbol{\mathrm{angeled}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{make}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{square}}? \\ $$$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{cut}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{equilateral}}\:\boldsymbol{\mathrm{triangle}}\:\boldsymbol{\mathrm{to}} \\ $$$$\boldsymbol{\mathrm{make}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{rectangle}}? \\ $$ Commented by Kunal12588 last updated on 11/Feb/19…

Question-185828

Question Number 185828 by normans last updated on 28/Jan/23 Commented by mr W last updated on 28/Jan/23 $$\left({x}+{y}+{z}\right)_{{min}} =\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta}{\mathrm{2}}} \\ $$$$\Delta={area}\:{of}\:{triangle}\:{ABC} \\…

how-to-justify-that-sin-x-7pi-2-cos-x-

Question Number 120244 by zahaku last updated on 30/Oct/20 $${how}\:{to}\:{justify}\:\:{that}\:\mathrm{sin}\:\left({x}−\frac{\mathrm{7}\pi}{\mathrm{2}}\:\right)=\:\mathrm{cos}\:{x} \\ $$ Commented by zahaku last updated on 30/Oct/20 $$\mathrm{sin}\:\left({x}−\frac{\mathrm{7}\pi}{\mathrm{2}}\:\right)=−\mathrm{sin}\:\left(\mathrm{2}\pi+\left({x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\:\right)\:\:?\right. \\ $$ Commented by Aziztisffola…

Question-185768

Question Number 185768 by Mingma last updated on 27/Jan/23 Answered by JDamian last updated on 27/Jan/23 $${l}=\frac{\mathrm{2}}{\mathrm{3}}{L} \\ $$$$\mathrm{outer}\:\mathrm{triangle}\:\mathrm{area} \\ $$$$\:\:\:\:\:\:\:{A}={kL}^{\mathrm{2}} \\ $$$$\mathrm{inner}\:\mathrm{triangle}\:\mathrm{area} \\ $$$$\:\:\:\:\:\:\:\:\:{a}={kl}^{\mathrm{2}}…

Question-185770

Question Number 185770 by Mingma last updated on 27/Jan/23 Answered by mahdipoor last updated on 27/Jan/23 $$\Delta{ADB}:\:\:\frac{{AB}}{{sin}\left(\mathrm{180}−\left(\mathrm{20}+{x}\right)\right)}=\frac{{BD}}{{sin}\left({x}\right)} \\ $$$$\Delta{BDC}:\:\:\frac{{BD}}{{sin}\mathrm{30}}=\frac{{BC}}{{sin}\left(\mathrm{180}−\left(\mathrm{50}+\mathrm{30}\right)\right)} \\ $$$$\Delta{ABC}:\:\:\frac{{AB}}{{sin}\mathrm{70}}=\frac{{BC}}{{sin}\left(\mathrm{180}−\left(\mathrm{70}+\mathrm{70}\right)\right)} \\ $$$$\Rightarrow\Rightarrow{BD}=\frac{{sin}\left({x}\right)}{{sin}\left(\mathrm{20}+{x}\right)}{AB}=\frac{{sin}\left(\mathrm{30}\right)}{{sin}\left(\mathrm{80}\right)}{BC}= \\ $$$$\frac{{sin}\left(\mathrm{30}\right)}{{sin}\left(\mathrm{80}\right)}×\frac{{sin}\left(\mathrm{40}\right)}{{sin}\left(\mathrm{70}\right)}{AB}…