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Category: Geometry

Question-189715

Question Number 189715 by Rupesh123 last updated on 20/Mar/23 Answered by som(math1967) last updated on 21/Mar/23 $$\bigtriangleup{ABC}\cong\bigtriangleup{DBE} \\ $$$$\Rightarrow{AB}={DB} \\ $$$${BC}={BE}\:\Rightarrow\angle{BEC}=\angle{BCE} \\ $$$$\angle{ACB}=\mathrm{34}\:\left[\angle{ACB}=\angle{CBE}\right]\:\: \\ $$$$\angle{DEB}=\angle{ACB}=\mathrm{34}…

A-regular-polygon-of-2k-1-sides-has-140-as-the-size-of-each-interior-angle-Find-k-

Question Number 58567 by olalekan2 last updated on 25/Apr/19 $$\:{A}\:{regular}\:{polygon}\:{of}\:\left(\mathrm{2}{k}+\mathrm{1}\right)\:{sides} \\ $$$${has}\:\mathrm{140}\:{as}\:{the}\:{size}\:{of}\:{each}\:{interior} \\ $$$${angle}.{Find}\:{k} \\ $$ Commented by mr W last updated on 25/Apr/19 $$\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{180}−\mathrm{140}\right)=\mathrm{360}…

Question-189615

Question Number 189615 by Rupesh123 last updated on 19/Mar/23 Answered by HeferH last updated on 19/Mar/23 $$\mathrm{16}−\mathrm{x}\sqrt{\mathrm{5}}\:=\:\frac{\mathrm{8x}}{\mathrm{3}\sqrt{\mathrm{5}}}−\frac{\mathrm{x}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{16}−\mathrm{x}\sqrt{\mathrm{5}}\:=\:\frac{\mathrm{5x}}{\mathrm{3}\sqrt{\mathrm{5}}} \\ $$$$\:\mathrm{48}\sqrt{\mathrm{5}}\:−\:\mathrm{15x}\:=\:\mathrm{5x} \\ $$$$\:\mathrm{x}\:=\:\frac{\mathrm{48}\sqrt{\mathrm{5}}}{\mathrm{20}} \\ $$$$\mathrm{x}\sqrt{\mathrm{5}}\:=\:\frac{\mathrm{48}\centerdot\mathrm{5}}{\mathrm{20}}\:=\frac{\mathrm{48}}{\mathrm{4}}\:=\:\mathrm{12}\:…