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Category: Geometry

Question-213250

Question Number 213250 by ajfour last updated on 01/Nov/24 Answered by a.lgnaoui last updated on 02/Nov/24 $$\mathrm{Calcul}\:\mathrm{de}\:\mathrm{l}\:\mathrm{aire} \\ $$$$ \\ $$$$\mathrm{BC}=\mathrm{a}\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\mathrm{DM}=\mathrm{a}\sqrt{\mathrm{2}}\:+\frac{\mathrm{a}}{\mathrm{2}}. \\ $$$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABDMC}}\right)=\mathrm{S}\left(\boldsymbol{\mathrm{ABC}}\right)+\mathrm{2}\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BDE}}\right)+\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCEF}}\right) \\ $$$$…

Question-213021

Question Number 213021 by Spillover last updated on 28/Oct/24 Commented by Ghisom last updated on 29/Oct/24 $${r}_{\mathrm{1}} =\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx.\mathrm{351153302} \\ $$$${r}_{\mathrm{2}} =\frac{\left(\mathrm{1}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\sqrt{\mathrm{2}}}{\mathrm{2}}\approx.\mathrm{165910681} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\mathrm{2}−\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx\mathrm{2}.\mathrm{11652017}…

Question-213020

Question Number 213020 by Spillover last updated on 28/Oct/24 Commented by Ghisom last updated on 29/Oct/24 $$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{only}\:\mathrm{if}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{special}\:\mathrm{one} \\ $$$$\mathrm{rectangular}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{fit}\:\mathrm{but}\:\mathrm{isosceles}\:\mathrm{fits} \\ $$$${a}={b}=\mathrm{8}\wedge{c}=\frac{\mathrm{48}}{\mathrm{5}} \\ $$$$\mathrm{I}\:\mathrm{used}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula}\:\mathrm{and}\:\mathrm{its}\:\mathrm{further}…