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fomenos
3 years ago
6

What is the sector with a central angle of 185 degrees and a diameter of 6.4 m? Round to the nearest tenth.

Mathematics
2 answers:
mr_godi [17]3 years ago
6 0
The area of a sector is 
     A=\frac{1}{2}r^2\theta =\frac{1}{2}\left(3.2\:m\right)^2\left(185\cdot \frac{\pi }{180}\right)=16.5\:m^2

The answer is 16.5 square meters. 
Greeley [361]3 years ago
5 0

Geometry B  Unit 5: Area - Lesson 10: Area Unit Test

1. What is the area of the trapezoid? The diagram is not drawn to scale.

72 cm^2

     

2.   Given the regular polygon, what is the measure of each numbered angle?

m∡1 = 36°; m∡2 = 72°

   

3.  What are a) the ratio of the perimeters and b) the ratio of the areas of the larger figure to the smaller figure? The figures are not drawn to scale.  

5/2 and 25/4

   

4.   What is the area of a regular pentagon with a side of 12 in.? Round the answer to the nearest tenth.

247.7 in.2

 

5.   Name the minor arc and find its measure.

AB; 162°

 

6.   What is the circumference of the given circle in terms of pi_symbol?

28pi in.

7.   What is the area of the given circle in terms of pi?

10.89pi m^2

8.   What is the area of a sector with a central angle of 185° and a diameter of 6.4 m? Round the answer to the nearest tenth.

16.5 m^2

9.   What is the area of the shaded region in the given circle in terms of pi_symbol and in simplest form?

(270 pi + 81 Root 3) m^2

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tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

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3 years ago
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We have that

<span>The -4 will lower the y-value by 4.
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zavuch27 [327]
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