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antiseptic1488 [7]
4 years ago
12

Find the area of a circle with a circumference of 6.28 units. _______units^2

Mathematics
1 answer:
hichkok12 [17]4 years ago
6 0

Answer:

Area of the circle= 3.14units^2

Step-by-step explanation:

Circumference of the circle=6.28 units

Finding the radius:

Circumference of a circle =2*\pi *r

6.28=2*\pi *r\\

As, \pi =\frac{22}{7}=3.14

So,    

           6.28=2*3.14*r

          6.28=6.28*r

Dividing by '6.28' both side:

        r=6.28/6.28

           r=1unit

Area of the circle= \pi *r*r

                      =3.14*1*1\\\\=3.14*1\\\\=3.14units^2

Area of the circle= 3.14units^2

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An acute triangle has sides measuring 10 cm and 16 cm. The length of the third side is unknown.
saveliy_v [14]

Answer: Choice B

12.5 < x < 18.9

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Explanation:

We have a triangle with these side lengths:

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Let's assume that b = 16 is the largest side of this triangle.

By the converse of the pythagorean theorem, we need b^2 < a^2+c^2 to be true in order for an acute triangle to happen.

So,

b^2 < a^2 + c^2\\\\c^2 > b^2 - a^2\\\\c > \sqrt{b^2-a^2}\\\\x > \sqrt{16^2-10^2}\\\\x > \sqrt{156}\\\\x > 12.4899959967968 \ \text{(approximate)}\\\\x > 12.5

Now let's consider the possibility that the missing side x is actually the longest side.

Using the same theorem as before, we would say,

c^2 < a^2 + b^2\\\\c < \sqrt{a^2 + b^2}\\\\x < \sqrt{10^2 + 16^2}\\\\x < \sqrt{356}\\\\x < 18.8679622641132 \ \text{(approximate)}\\\\x < 18.9\\\\

We found that x > 12.5 and x < 18.9

This is the same as saying 12.5 < x and x < 18.9

Put together, they form the approximate answer of 12.5 < x < 18.9

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To solve this problem, we need to first find the dimensions of the side of the blue and purple squares.

We're given that the purple (smaller) square has a side length of x inches.
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Since the blue band surrounds the purple square on both sides, the length of the blue square is x+2(5)=x+10 inches.

The net area of the band is therefore the difference of the area of the blue square and the purple square, namely take out the area of the purple square from the blue.

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