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Geometry Session 4: Solutions
 
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A B C D 
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Solutions for Session 4, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5


Problem C1

The angle has measure 90°. As the vertex moves, the inscribed angle does not change in size. You could make the conjecture that an angle inscribed in a semicircle has measure of 90°.

<< back to Problem C1

<< back to Problem C1 non-interactive activity


 

Problem C2

The angle has measure 135°. As the vertex moves, the inscribed angle does not change in size. You could make the conjecture that an angle inscribed in a quarter-circle has measure of 135°.

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Problem C3

a. 

Answers will vary.

b. 

Answers will vary. This was the construction from Problem A2.

c. 

Answers will vary. You should form a triangle by drawing these three segments.

d. 

The two circles have the same radius. The radius's length equals the distance between the two centers. Since the distance between a circle's center and any point on it, including the point where the two circles intersect, is the same as the length of the radius, the base of the triangle (the segment connecting the centers) is the same length as its two other sides. Therefore, the triangle is equilateral.

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Problem C4

Answers will vary. One way to construct an isosceles triangle is to start with two circles as in Problem C3. Then connect the two intersection points to each other. Choose the center of one of the circles, and connect it to each of the intersection points. The two radii of the circle are congruent. The angle between them is more than 60° (measure it!), so the third side must be longer than the other two. The triangle is isosceles but not equilateral.

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Problem C5

Start with the construction of two circles that pass through each other's centers. Then connect the following points (with segments in between them) in the following order: first intersection point, center of first circle, second intersection point, center of second circle, and back to first intersection point.

All segments are radii of one circle or the other. Since the circles are the same size, all the radii are the same, so the sides of the quadrilateral are all congruent. It must be a rhombus.

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