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alekssr [168]
2 years ago
15

Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit. (Hint: Assume that

the central point of each arc is its corresponding vertex.) 

Mathematics
2 answers:
Vadim26 [7]2 years ago
8 0
So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
that simply means, 3 circles with a radius of 3, overlapping the triangle

now, the area in the middle, the shaded one, will be, the whole area of the triangle MINUS those 3 circle sectors

hmmmm each sector has 60°, that means, all three of them will then be 60+60+60 or 180°, so the area of those three sectors, can be combined into a 180° sector, well, hell, 180° is really half a circle

so.... the area of those three sectors of 60° each, all three combined, is the same area of half a circle with a radius of 3

so    \bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
\textit{area of half a circle}\\\\
A=\cfrac{\pi r^2}{2}\\\\
-----------------------------\\\\

\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
\end{array}\impliedby \textit{area of shaded area}\\\\
-----------------------------\\\\
\boxed{\cfrac{36\sqrt{3}}{4}-\cfrac{9\pi }{2}}

you can, add the fractions if you want, or leave them like that, or get their difference by using their decimal format

Mamont248 [21]2 years ago
7 0

Answer:

9 sqrt 3 - 4.5 pi

Step-by-step explanation:

So, first off, you need to find the area of the whole triangle. Since this is an equilateral triangle, the formula for it would be:  

A = 1/4 * s^2 * sqrt 3 ----> A = 1/4 * 6^2 * sqrt 3 ----> A = 9 sqrt 3. So the area of the whole triangle is 9 sqrt 3.

Next, imagine that there are three circles that are part of the sectors (I am using the same picture as jdoe0001 attached). So to find the area of each sector, you would do 60o (o = degree sign) divided by 360o --> 60/360. You should get 1/6. Then you would multiply this by the area of the circle, which you find by doing the following: A = pi * r^2 ----> A = 9 pi. So when you would multiply 1/6 by the area of the circle, you would get 1.5 pi. Since this is just one sector, you need to multiply 1.5 pi by 3, because there are 3 sectors. You should get 4.5 pi.

Finally, to get the area of the shaded portion, you would simply do this: the area of the whole triangle minus the area of the sectors ----> <u>9 sqrt 3 - 4.5 pi</u>. This is the exact answer. The approximate answer would be 1.45.

<em>Hope this helps! :)</em>

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In 4 hours, Naoya must have read

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There were 350 pages remaining at the time, so the total number of pages is 220+350=570.

For each hour of reading, Naoya reads 55 pages, so the number of pages he's finished reading by the t'th hour is 55t. So the number of pages left to read is given by

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6 0
2 years ago
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Group of baseball fans can see home plate from a 40 meter tall building outside the stadium. The angle of vision has a tangent o
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Answer:

17.78 meters

Step-by-step explanation:

Let

x ----> the horizontal distance, in meters, to home plate

\theta ----> the angle of vision

we know that

tan(\theta)=\frac{40}{x} ----> by TOA (opposite side divided by the adjacent side)

we have

tan(\theta)=\frac{9}{4}

substitute

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2 years ago
b. Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box.
Ainat [17]

First box = 14 pounds

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<u>Step-by-step explanation:</u>

Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:

We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.

⇒ 2x

The third box contained two more pounds than the first box i.e.

⇒ x+2

The last box contained one-fourth the amount in the second box i.e.

⇒ (\frac{1}{4})2x = \frac{x}{4}

Therefore , Sum of pounds of candy are :

⇒ \frac{x}{2} +x+2+2x+x=65

⇒ \frac{x}{2} +4x=63

⇒ \frac{9x}{2}=63

⇒ x=63(\frac{2}{9} )

⇒ x=14

Therefore , Candy in each box is :

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

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Answer:

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3. A=π(81)

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