The key to solving this problem is drawing a rectangle outside the figure and subtracting the reas of the triangles formed from the are of the rectangle. You can look at the attched image if you want a visual.So first let's find the area of the rectangle, the length is 11 and the width is 5 so the area is 55. Now let's find the area of the trinagles. Triangle 1 has sides of 4 and 3 so the area is 6. Triangle 2 has sides of 5 and 5 so the area is 25/2. Triangle 3 has sides of 2 and 5 so the area is 5. Now let's add the area of the triangles up to get 47/2. Now let's subtract 47/2 from 55 to get the area of the figure which is *drumroll*...63/2. that's all.
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First, let's get the area of the entire triangle since we'll need it later. The area of a triangle is A=1/2*b*h
We can find the height with the Pythagorean theorem by splitting the triangle in half.
3^2+b^2=6^2
9+b^2=36
b^2=27
b=√27
Then we can find the area:
A=1/2*6*√27
A=3√27 or =9√3 or 15.59
Now we can find the area of each region in the triangle other than the shaded region because they are all portions of a circle.
Each region has an angle of 60 because this is an equilateral triangle. Therefore the area of each region other than the shaded region will be 1/6 the area of a circle with a radius of 3 because a full circle is 360 degrees.
A=pi*r^2/6
A=pi*9/6
A=4.71
So three of these regions would have an area of 14.14
We do the area of the triangle minus the area of these regions to get the area of the shaded region
15.59-14.14 = 1.45
Hope this helps!
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Answer:
it defines an important concept of standard deviation used in probability theory and statistics. It has a major use in the formula for roots of a quadratic equation; quadratic fields and rings of quadratic integers, which are based on square roots, are important in algebra and have uses in geometry.