Respuesta :

[tex]\begin{gathered} \text{Triangle A is }an\text{ equilaterial triangle hence} \\ F^2=1.5^2+(\frac{F}{2})^2 \\ \\ F^2=1.5^2+\frac{F}{4}^2 \\ F^2-\frac{F}{4}^2=1.5^2 \\ \frac{3F}{4}^2=1.5^2 \\ F^2=\frac{4(1.5^2)}{3} \\ F=\sqrt[]{\frac{4(1.5^2)}{3}} \\ F=1.73cm \\ \text{From triangle B} \\ D=F=1.73cm \\ \tan (55\text{\degree)=}\frac{G}{D} \\ \text{Solving G} \\ G=D\tan (55\text{)} \\ G=(1.73cm)\tan (55\text{)} \\ G=2.47cm \\ \cos (55)=\frac{D}{E} \\ Solv\text{ing E} \\ E=\frac{D}{\cos(55)} \\ E=\frac{1.73cm}{\cos(55)} \\ E=3.01cm \\ \text{From triangle L} \\ \sin (55)=\frac{J}{G} \\ \text{Solving J} \\ J=G\sin (55) \\ J=(2.47cm)\sin (55) \\ J=2.02cm \\ \cos (55)=\frac{K}{G} \\ \text{Solving K} \\ K=G\cos (55) \\ K=(2.47cm)\cos (55) \\ K=1.42cm \\ \text{Triangle M=L} \\ Hence \\ I=N=J \\ I=2.02cm \\ N=2.02cm \\ H=O=K \\ H=1.42cm \\ O=1.42cm \\ Perimeter=2(1.73cm)+3.01cm+3(2.02cm)+1.42cm \\ Perimeter=13.95cm\approx14cm \\ \text{The perimeter is }14cm \end{gathered}[/tex]

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