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Vladimir79 [104]
3 years ago
6

A square is inscribed in a circle as shown. If the radius of the circle is 9, what is the area of the shaded region, rounded to

the nearest hundredth?

Mathematics
1 answer:
djyliett [7]3 years ago
3 0

Answer:

Area of shaded region is:

A = 92.41

Step-by-step explanation:

(As the diagram is not shown, consider the diagram attached below.)

r = 9

So

Diagonal of the square = 2r = 18

As all angles of square are of 90°

The diagonal is dividing the angles into 2 halves of 45° at point where diagonal is joining the corner.

Taking sine:

sin\theta=\frac{perpendicular}{hypotenuse}\\perpendicular = s\\hypotenuse = 2r = 18\\\theta = 45^\circ\\Substitute:\\sin45=\frac{s}{18}\\sin45\cdot18=s\\s=12.73

<h3>Area of Circle:</h3>

A=\pi{r}^2\\A=\pi{9}^2\\A=254.34

<h3>Area of Square:</h3><h3>Area = s\cdot{s}\\Area=12.73\cdot12.73\\Area = 161.93</h3><h3>Area of Shaded Region:</h3>

Area of circle - Area of square

A = 254.34-161.93\\A=92.41

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Philip made a total of 9 bracelets and necklaces from 120 inches of cord. He used 8 inches of cord for each bracelet and 20 inch
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Answer:

So Philip made 5 bracelets and 4 necklaces.

Step-by-step explanation:

Let x = number of bracelets and y = number of necklaces.

Since we have a total of 9 bracelets and necklaces,

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Writing equations (1) and (3) in matrix form, we have

\left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}9\\30\end{array}\right]

Using Cramer's rule to solve for x and y,

x = det \left[\begin{array}{ccc}9&1\\30&5\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

x = (9 × 5 - 30 × 1) ÷ (1 × 5 - 1 × 2)

x = (45 - 30) ÷ (5 - 2)

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y = det \left[\begin{array}{ccc}1&9\\2&30\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

y = (30 × 1 - 9 × 2) ÷ (1 × 5 - 1 × 2)

y = (30 - 18) ÷ (5 - 2)

y = 12 ÷ 3

y = 4

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