Question Number 220375 by Tawa11 last updated on 12/May/25
Commented by Tawa11 last updated on 12/May/25
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{and}\:\mathrm{perimeter}. \\ $$$$\mathrm{Please},\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}\:\mathrm{correct}? \\ $$
Commented by AntonCWX8 last updated on 12/May/25
$${The}\:{triangle}\:{looks}\:{equilateral}… \\ $$
Answered by AntonCWX8 last updated on 12/May/25
$${I}\:{assume}\:{the}\:{triangle}\:{is}\:{equilateral} \\ $$$${Area}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{12}^{\mathrm{2}} \right)\left({sin}\left(\mathrm{60}°\right)\right)+\mathrm{30}\left(\mathrm{12}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\pi\right)\left(\frac{\mathrm{12}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${Area}=\mathrm{36}\sqrt{\mathrm{3}}+\mathrm{360}+\mathrm{18}\pi\approx\mathrm{478}.\mathrm{9025}{cm}^{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 12/May/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$