Question 1

Find the perimeter of the figure shown. Round your answer to the nearest centimeter. This is for problem number 9

Find the perimeter of the figure shown. Round your answer to the nearest centimeter-example-1

Question 2

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

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