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Geometry Session 3: Solutions
Session 3 Part A Part B Part C Homework
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Session 3 Materials:



Solutions for Session 3, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7

Problem C1

Answers will vary. One possible way of describing convex figures is that they are figures that are not dented, or that a rubber band stretched around the figure will touch the figure entirely.

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Problem C2

Definitions (a), (d), (f), (g), and (h) are all equivalent descriptions of convex polygons. Statement (b) is true for all polygons, so it would apply to concave ones as well (triangle inequality!). Statement (c) is not true for all convex polygons. For example, try drawing a convex quadrilateral with one very long side; this side will probably be longer than the shortest diagonal of the quadrilateral. Statement (e) certainly does not describe convex polygons. (Consider an obtuse triangle which is convex but whose longest side is opposite to its largest angle.)

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Problem C3

Answers will vary. You can find some more examples in Problem C4.

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Problem C4

The answers may vary for the two figures on the right.

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Problem C5

If an n-sided polygon is convex, we can pick a vertex and connect it to all other vertices, thereby creating n - 2 triangles. If the polygon is concave, we pick a vertex corresponding to an interior angle greater than 180° and connect that vertex to all the vertices so that all the resulting diagonals are in the interior of the polygon. Repeat the process if necessary. This has the effect of subdividing the original polygon into some number of convex polygons. We then divide each of the convex polygons into triangles. In the end, we will end up with n - 2 triangles.

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Problem C6

Number of Sides of the Polygon

Number of Triangles Formed

Sum of the Angles in the Polygon

















n - 2

(n - 2) • 180°

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Problem C7

Assuming that the interior of an n-sided polygon can be divided into n - 2 triangles, observe that all of the angles of the triangles actually make up the interior angles of the polygon. This is true since the vertices of all the triangles coincide with the vertices of the polygon. Therefore, since each triangle contributes 180° to the overall sum of the angles, the sum is (n - 2) • 180°.

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