Question

I need help with question number 9. Find the perimeter of the figure shown. Round your answer to the nearest CENTIMETER.

Answer 1

Each letter represents one side of a right triangle.

We can find the length of side c as follows:

`\begin{gathered} \sin (60\degree)=(1.5)/(c) \\ c=(1.5)/(\sin(60\degree)) \\ c=\frac{1.5}{\frac{\sqrt[]{3}}{2}} \\ c\approx1.732\text{ }cm \end{gathered}`

Sides c and d are congrent, then:

`d\approx1.732\text{ }cm`

Applying the Pythagorean theorem to the triangle formed by sides c = 1.732, h, and the side of 1.5 cm, we get:

`\begin{gathered} 1.732^2=1.5^2+h^2 \\ 3=2.25+h^2 \\ 3-2.25=h^2 \\ \sqrt[]{0.75}=h \\ 0.866\text{ }cm\approx h \end{gathered}`

Applying the Pythagorean theorem to the triangle formed by sides 1.5 cm, d, and i, the same result is got, then:

`i\approx0.866\text{ }cm`

We can find the length of side e as follows:

`\begin{gathered} \cos (55\degree)=(h+i)/(e) \\ \cos (55\degree)=(0.866+0.866)/(e) \\ e=(0.866+0.866)/(\cos (55\degree)) \\ e\approx3.02\text{ }cm \end{gathered}`

Applying the Pythagorean theorem, we can find the length of side j as follows:

`\begin{gathered} e^2=j^2+(h+i)^2 \\ 3.02^2=j^2+(0.866+0.866)^2 \\ 9.1204=j^2+3 \\ 9.1204-3=j^2 \\ \sqrt[]{6.1204}=j \\ j\approx2.474\text{ }cm \end{gathered}`

We can find the length of side b as follows:

`\begin{gathered} \sin (55\degree)=(b)/(j) \\ b=\sin (55\degree)\cdot2.474 \\ b\approx2.026\text{ }cm \end{gathered}`

We can find the length of side l as follows:

`\begin{gathered} \cos (55\degree)=(l)/(2.474) \\ l=\cos (55\degree)\cdot2.474 \\ l\approx1.42\text{ }cm \end{gathered}`

Considering that sides k and j are congruent, if we would apply the Pythagorean theorem to triangles with sides b, l, j, and a, l, k, we will conclude that sides a and b are congruent, then:

`a\approx2.026\text{ }cm`

Sides a, f, g, l form a rectangle, then f and a are congruent, and l and g are also congruent, that is,

`\begin{gathered} f\approx2.026\text{ }cm \\ g\approx1.42\text{ }cm \end{gathered}`

Finally, the perimeter of the figure shown is:

`\begin{gathered} P=a+b+c+d+e+f+g \\ P\approx2.026+2.026+1.732+1.732+3.02+2.026+1.42 \\ P\approx14\text{ }cm \end{gathered}`