Question Number 166715 by alephzero last updated on 26/Feb/22
$$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{24}.\:\mathrm{Find}\:\mathrm{it}'\mathrm{s}\:\mathrm{area}. \\ $$
Answered by Rasheed.Sindhi last updated on 26/Feb/22
$${Let}\:{each}\:{of}\:{sides}:\: \\ $$$${a}={b}={c}={l} \\ $$$${perimeter}=\mathrm{3}{l}=\mathrm{24} \\ $$$${l}=\mathrm{8} \\ $$$$\begin{array}{|c|}{{s}=\frac{{a}+{b}+{c}}{\mathrm{2}}\:;\:\blacktriangle=\:\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\:}\\\hline\end{array} \\ $$$${s}=\frac{\mathrm{3}{l}}{\mathrm{2}} \\ $$$$\blacktriangle=\sqrt{\frac{\mathrm{3}{l}}{\mathrm{2}}\left(\frac{\mathrm{3}{l}}{\mathrm{2}}−{l}\right)^{\mathrm{3}} }\:=\sqrt{\frac{\mathrm{3}{l}}{\mathrm{2}}\centerdot\frac{{l}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }}\:=\frac{{l}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }\sqrt{\mathrm{3}} \\ $$$$\:\:\:=\frac{\mathrm{8}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}}\:=\mathrm{16}\sqrt{\mathrm{3}} \\ $$
Commented by alephzero last updated on 26/Feb/22
$$\left(\frac{\mathrm{3}{l}}{\mathrm{2}}−{l}\right)^{\mathrm{3}} =\:\frac{{l}}{\mathrm{2}}? \\ $$
Commented by Rasheed.Sindhi last updated on 26/Feb/22
$${Sorry}\:{for}\:{mistake}!\:{have}\:{corrected} \\ $$$${now}.\:\mathcal{T}{hank}\:{you}! \\ $$