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Alekssandra [29.7K]
3 years ago
11

Mary is inscribing a square in the circle shown, but she is having difficulty remembering the process. Which summary describes t

he basic process for constructing a square inscribed in the circle? A. Create the perpendicular bisector of FO. Then use the points of intersection of the perpendicular bisector with the circle, along with points F and G, to draw the square. B. Set the compass width to the length of FO, place the point of the compass on the circle, and draw an arc. Place the compass on the intersection of the arc and the circle, and repeat the process until the circle is intersected by four arcs. Connect the intersection points to form the square. C. Create the perpendicular bisector of FG. Then use the points of intersection of the perpendicular bisector with the circle, along with points F and G, to draw the square. D. Set the compass width to the length of FG, place the point of the compass on the circle, and draw an arc. Place the compass on the intersection of the arc and the circle, and repeat the process until the circle is intersected by four arcs. Connect the intersection points to form the square.
Mathematics
1 answer:
Ivenika [448]3 years ago
4 0

Answer:

C. Create the perpendicular bisector of FG. Then use the points of intersection of the perpendicular bisector with the circle, along with points F and G, to draw the square.

Step-by-step explanation:

The steps are below:

1. A diameter of the circle is drawn.

2. A perpendicular bisector of the diameter is drawn using the method described in the perpendicular bisector of a segment. This is also a diameter of the circle.

3. The resulting four points on the circle are the vertices of the inscribed square.

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Answer:

2\ cm

Step-by-step explanation:

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As\ we\ are\ not\ said\ that\ the\ cylindrical\ vessel\ was\ empty,\ let\ its\ initial\\ height\ be\ h_1\\Radius\ of\ the\ Cylindrical\ vessel=10\ cm\\Initial\ height\ of\ the\ cylindrical\ vessel=h_1\\Initial\ volume\ of\ the\ cylindrical\ vessel=\pi *10^2*h_1=100h_1\pi \\Initial\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi \\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=Final\ amount\ of\ water\\ in\ the\ cylindrical\ vessel +Amount\ of\ water\ in\ conical\ vessel

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8 0
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Answer:

x=5/2

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We note that 30=6*5 and 12=6*2, so 30/12=5/2.

So, \boxed{x=5/2} and we're done!

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