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juin [17]
3 years ago
14

This figure consists of a rectangle and semicircle. What is the perimeter of this figure? Use 3.14 for pi. 60.84 m 72.84 m 79.68

m 98.52 m A shape made up of a rectangle of the left and a semicircle on the right. The rectangle has a width of 15 meters and a height of 12 meters.

Mathematics
2 answers:
viva [34]3 years ago
8 0
60.84 (A)
Hope this helps..   :)
Have a nice day!
nignag [31]3 years ago
6 0

Answer:

The perimeter of the figure is 60.84\ m

Step-by-step explanation:

we know that

The perimeter of the figure is equal to the perimeter of a rectangle plus the circumference of a semicircle minus the diameter of the circle

so

<u>Find the perimeter of the rectangle</u>

The perimeter of the rectangle is equal to

P=2(L+W)

substitute the values

P=2(15+12)=54\ m

<u>Find the circumference of a semicircle</u>

The circumference of a semicircle is equal to

C=\pi r

substitute the value of r

r=12/2=6\ m -----> the radius is half the diameter

C=(3.14)(6)=18.84\ m

<u>Find the perimeter of the figure</u>

54\ m+18.84\ m-12\ m=60.84\ m


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The area of the triangle is A=\sqrt{\frac{4027}{2}}

Step-by-step explanation:

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{\bf AB} \times {\bf AC}=\begin{pmatrix}2&0&-9\end{pmatrix}\times \begin{pmatrix}-7&-5&-7\end{pmatrix}

This is the definition of cross product of two vectors in space:

Let {\bf u} = u_1{\bf i}+u_2{\bf j}+u_3{\bf k} and {\bf v} = v_1{\bf i}+v_2{\bf j}+v_3{\bf k} be vectors in space. The cross product of {\bf u} and {\bf v} is the vector

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{\bf AB} \times {\bf AC}=\begin{pmatrix}2&0&-9\end{pmatrix}\times \begin{pmatrix}-7&-5&-7\end{pmatrix}

\begin{pmatrix}0\cdot \left(-7\right)-\left(-9\left(-5\right)\right)&-9\left(-7\right)-2\left(-7\right)&2\left(-5\right)-0\cdot \left(-7\right)\end{pmatrix}\\\\\begin{pmatrix}-45&77&-10\end{pmatrix}

||{\bf AB} \times {\bf AC}||=\sqrt{(-45)^2+(77)^2+(-10)^2} \\\\||{\bf AB} \times {\bf AC}||=\sqrt{2025+5929+100}\\\\||{\bf AB} \times {\bf AC}||=\sqrt{8054}

The area of the triangle is

A=\frac{1}{2}||{\bf AB} \times {\bf AC} ||=\frac{1}{2}\sqrt{8054}=\sqrt{\frac{4027}{2}}

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