Category: Mensuration

Question Number 43128 by Cheyboy last updated on 07/Sep/18 Commented by Cheyboy last updated on 07/Sep/18 $${Find}\:{the}\:{area} \\ $$ Commented by ajfour last updated on…

Question Number 43023 by ajfour last updated on 06/Sep/18 Commented by ajfour last updated on 06/Sep/18 $${Find}\:{location}\:{of}\:{centre}\left(\mathrm{0},{h}\right)\:{and} \\ $$$${radius}\:{R}\:{of}\:{circle}\:{such}\:{that} \\ $$$${all}\:{four}\:{bounded}\:{areas}\: \\ $$$${O}\overset{\frown} {{AD}}\:,\:\overset{\frown} {{APB}A}\:,\:{A}\overset{\frown}…

Question Number 42866 by Cheyboy last updated on 03/Sep/18 Answered by MJS last updated on 04/Sep/18 $${x}\rightarrow{r}\:\mathrm{to}\:\mathrm{not}\:\mathrm{get}\:\mathrm{confused} \\ $$$$\mathrm{small}\:\mathrm{circle}\:\mathrm{touching}\:\:\mathrm{half}\:\mathrm{circle}\:\mathrm{from}\:\mathrm{inside} \\ $$$${r}^{\mathrm{2}} +{p}^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{2}}−{r}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{p}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}}…

Question Number 173124 by peter frank last updated on 06/Jul/22 Commented by aleks041103 last updated on 06/Jul/22 $${For}\:{which}…{what}? \\ $$ Commented by peter frank last…

Question Number 172269 by Mikenice last updated on 25/Jun/22 $${A}\:{rectangular}\:{picture}\:\mathrm{6}{cm}\:{by}\:\mathrm{8}{cm}\:{is} \\ $$$${enclosed}\:{by}\:{a}\:{frame}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:{wide}.\: \\ $$$${calculate}\:{the}\left[{area}\:{of}\:{the}\:{frame}.\right. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com