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Sunny_sXe [5.5K]
3 years ago
13

Area addition and subtraction

Mathematics
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

3.8 in^{2}

Step-by-step explanation:

We are given a square of size 6x6 in. So, area of this square is equal to 6 x 6 = 36 in^{2}. Now, the shaded region is \frac{1}{2} x (Area of square - area of semicircles)

The diameter of both semicircles = side of the square = 6in

So, radius (r) = \frac{1}{2} x diameter = \frac{1}{2} x 6 = 3in

And hence, area of semicircle is  = \frac{1}{2} x πr^{2}

= \frac{1}{2} x π3^{2}

Since, there are two semicircles we multiply above by 2, so area of both semicircles = 2 x \frac{1}{2} π3^{2} = 9π

Area of shaded region =  \frac{1}{2} (36 - 9π) = 3.8628 = 3.8 in^{2} to the nearest tenth.

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C would be able to verify the equation.

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Two vertical posts stand side by side. One post is 8 feet tall and the other is 17 feet tall. If a 24 foot wire is stretched bet
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The distance between the posts is 3\sqrt{55} feet.

Step-by-step explanation:

In the attached figure, let DB and AC be the posts and let AD be the wire attached to the top of the posts.

BC is the distance between the posts and we need to find BC.

Note that BC = DE

Also, EC = DB = 8m and AE = AC - EC = 17 - 8 = 9 m.

Now, in the right triangle ADE,

AD^{2} =AE^{2} +DE^{2}

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DE^{2} =24^{2} -9^{2}

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Hence, the distance between the posts is 3\sqrt{55} feet.



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