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Category: Geometry

Question-207008

Question Number 207008 by mr W last updated on 03/May/24 Answered by A5T last updated on 03/May/24 Commented by A5T last updated on 03/May/24 $$\frac{\mathrm{5}}{{BO}}=\frac{\mathrm{6}}{{CO}}\Rightarrow\frac{{BO}}{{CO}}=\frac{\mathrm{5}}{\mathrm{6}};\:{BO}=\mathrm{5}{x};{CO}=\mathrm{6}{x}…

Given-is-a-square-with-side-length-15-We-need-to-find-exactly-17-smaller-squares-to-fill-the-big-one-How-many-solutions-are-possible-Note-it-s-not-enough-to-find-squares-with-the-sum-of-their-are

Question Number 206806 by Frix last updated on 26/Apr/24 $$\mathrm{Given}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{side}\:\mathrm{length}\:\mathrm{15}. \\ $$$$\mathrm{We}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{exactly}\:\mathrm{17}\:\mathrm{smaller}\:\mathrm{squares} \\ $$$$\mathrm{to}\:\mathrm{fill}\:\mathrm{the}\:\mathrm{big}\:\mathrm{one}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{solutions}\:\mathrm{are} \\ $$$$\mathrm{possible}? \\ $$$$\left(\mathrm{Note}:\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{find}\:\mathrm{squares}\:\mathrm{with}\right. \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{areas}\:\mathrm{being}\:\mathrm{225},\:\mathrm{they}\:\mathrm{must} \\ $$$$\mathrm{fit}\:\mathrm{into}\:\mathrm{the}\:\mathrm{15}×\mathrm{15}\:\mathrm{square}.\:\mathrm{Example}\:\mathrm{with} \\ $$$$\mathrm{3}\:\mathrm{squares}:\:\mathrm{2}×\mathrm{2}+\mathrm{5}×\mathrm{5}+\mathrm{14}×\mathrm{14}=\mathrm{225}\:\mathrm{but} \\…

Question-206750

Question Number 206750 by mr W last updated on 23/Apr/24 Answered by HeferH24 last updated on 23/Apr/24 $$\:{By}\:{similar}\:{triangles}: \\ $$$$\:\frac{{d}}{{c}}\:=\:\frac{{x}}{{b}}\:;\:\:\frac{{d}}{{c}}\:=\:\frac{{a}}{{x}}\: \\ $$$$\:\frac{{a}}{{x}}\:=\:\frac{{x}}{{b}}\:\Leftrightarrow\:{x}^{\mathrm{2}} \:=\:{ab}\:\Leftrightarrow\:{x}=\sqrt{{ab}} \\ $$…

Question-206729

Question Number 206729 by mr W last updated on 23/Apr/24 Answered by A5T last updated on 23/Apr/24 $$\frac{{sin}\mathrm{84}}{{AD}}=\frac{{sin}\mathrm{54}}{{AC}}\Rightarrow{AD}=\frac{{ACsin}\mathrm{84}}{{sin}\mathrm{54}}…\left({i}\right) \\ $$$$\frac{{sin}\left(?\right)}{{AD}}=\frac{{sin}\left(\mathrm{54}−?\right)}{{BD}={AC}}\Rightarrow{AD}=\frac{{ACsin}\left(?\right)}{{sin}\left(\mathrm{54}−?\right)}…\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow\frac{{sin}\mathrm{84}}{{sin}\mathrm{54}}=\frac{{sin}\left(?\right)}{{sin}\left(\mathrm{54}−?\right)}\Rightarrow?=\mathrm{30}° \\ $$ Commented…