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Category: Mensuration

Question-222800

Question Number 222800 by fantastic last updated on 07/Jul/25 Answered by mr W last updated on 07/Jul/25 $$\mathrm{2}×\frac{\pi{a}^{\mathrm{2}} }{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}=\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right){a}^{\mathrm{2}} \approx\mathrm{0}.\mathrm{614}{a}^{\mathrm{2}} \\ $$ Commented by…

Question-222798

Question Number 222798 by fantastic last updated on 07/Jul/25 Answered by mr W last updated on 07/Jul/25 $${r}=\frac{{a}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{4}} \\ $$$${area}\:{of}\:{semi}\:{circle}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}\pi{a}^{\mathrm{2}} }{\mathrm{32}}\:\checkmark \\ $$ Terms…

Question-222639

Question Number 222639 by Mingma last updated on 03/Jul/25 Answered by gabthemathguy25 last updated on 03/Jul/25 $$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{8}\:{cm}=\mathrm{4}\:{cm} \\ $$$$\mathrm{Slant}\:\mathrm{height}\:=\:\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\mathrm{100}+\mathrm{16}}=\sqrt{\mathrm{116}}\approx\mathrm{10}.\mathrm{77}\:\mathrm{cm} \\ $$$$\mathrm{cos}\left(\theta\right)=\frac{\mathrm{10}.\mathrm{77}^{\mathrm{2}} +\mathrm{10}.\mathrm{77}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}}…

Question-221870

Question Number 221870 by fantastic last updated on 11/Jun/25 Answered by mehdee7396 last updated on 12/Jun/25 $${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{AB}×{OM}\:\:\:\:\&\:\:\:{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{CD}×{ON} \\ $$$${AB}={CD}\:\:\Rightarrow{S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{AB}×\left({OM}+{ON}\right)=\frac{{AB}×{MN}}{\mathrm{2}}=\mathrm{16} \\ $$$$\Rightarrow{S}={AB}×{MN}=\mathrm{32}…