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Question Number 166715 by alephzero last updated on 26/Feb/22

The perimeter of an equilateral triangle is 24. Find it′s area.

$$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{24}.\:\mathrm{Find}\:\mathrm{it}'\mathrm{s}\:\mathrm{area}. \\ $$

Answered by Rasheed.Sindhi last updated on 26/Feb/22

Let each of sides: a=b=c=l perimeter=3l=24 l=8 determinant (((s=((a+b+c)/2) ; ▲= (√(s(s−a)(s−b)(s−c))) ))) s=((3l)/2) ▲=(√(((3l)/2)(((3l)/2)−l)^3 )) =(√(((3l)/2)∙(l^3 /2^3 ))) =(l^2 /2^2 )(√3) =(8^2 /4)(√3) =16(√3)

$${Let}\:{each}\:{of}\:{sides}:\: \\ $$$${a}={b}={c}={l} \\ $$$${perimeter}=\mathrm{3}{l}=\mathrm{24} \\ $$$${l}=\mathrm{8} \\ $$$$\begin{array}{|c|}{{s}=\frac{{a}+{b}+{c}}{\mathrm{2}}\:;\:\blacktriangle=\:\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\:}\\\hline\end{array} \\ $$$${s}=\frac{\mathrm{3}{l}}{\mathrm{2}} \\ $$$$\blacktriangle=\sqrt{\frac{\mathrm{3}{l}}{\mathrm{2}}\left(\frac{\mathrm{3}{l}}{\mathrm{2}}−{l}\right)^{\mathrm{3}} }\:=\sqrt{\frac{\mathrm{3}{l}}{\mathrm{2}}\centerdot\frac{{l}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }}\:=\frac{{l}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }\sqrt{\mathrm{3}} \\ $$$$\:\:\:=\frac{\mathrm{8}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}}\:=\mathrm{16}\sqrt{\mathrm{3}} \\ $$

Commented by alephzero last updated on 26/Feb/22

(((3l)/2)−l)^3 = (l/2)?

$$\left(\frac{\mathrm{3}{l}}{\mathrm{2}}−{l}\right)^{\mathrm{3}} =\:\frac{{l}}{\mathrm{2}}? \\ $$

Commented by Rasheed.Sindhi last updated on 26/Feb/22

Sorry for mistake! have corrected now. Thank you!

$${Sorry}\:{for}\:{mistake}!\:{have}\:{corrected} \\ $$$${now}.\:\mathcal{T}{hank}\:{you}! \\ $$

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