Category: Geometry

Question Number 219255 by Spillover last updated on 21/Apr/25 Answered by A5T last updated on 21/Apr/25 $$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{the}\:\mathrm{side},\mathrm{s},\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{be}\:\mathrm{1} \\ $$$$\Rightarrow\mathrm{b}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{a}×\mathrm{b}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\overset{/\mathrm{b}^{\mathrm{2}} }…

Question Number 219232 by Spillover last updated on 20/Apr/25 Answered by A5T last updated on 21/Apr/25 Commented by A5T last updated on 21/Apr/25 $$\mathrm{AD}=\mathrm{bsin}\theta\:;\:\mathrm{AE}=\mathrm{bcos}\theta\:;\:\mathrm{AC}=\mathrm{asin}\theta\:;\:\mathrm{AB}=\mathrm{acos}\theta \\…

Question Number 219068 by Spillover last updated on 19/Apr/25 Answered by mr W last updated on 20/Apr/25 Commented by mr W last updated on 20/Apr/25…

Question Number 219071 by Spillover last updated on 19/Apr/25 Answered by mr W last updated on 19/Apr/25 Commented by mr W last updated on 19/Apr/25…

Question Number 219066 by Spillover last updated on 19/Apr/25 Answered by Hamada1969 last updated on 19/Apr/25 Commented by Spillover last updated on 19/Apr/25 $${thank}\:{you} \\…