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Part A: In the figure below, points E,F,G, and H are the midpoints of the sides

ID: 2980970 • Letter: P

Question

Part A: In the figure below, points E,F,G, and H are the midpoints of the sides of ABCD. Prove that EFGH is a parallelogram. (Hint: If a line segment has its endpoint the midpoints of two sides of a triangle, then the segment is contained in a line that is parallel to the third side and the segment is one-half the length of the third side)

Part B: Suppose that ABCD is a square and that line m contains the point where the diagonals of the square intersect. Prove that m partitions the square into regions of equal area. (Hint: SASAS congruence condition for quadrilaterals)

-m is a line that contians the intersection point but that is not the diagonal of the square.

Part C: A regular n-gon is a polygon with n sides in which all sides are the same length and all interior angles have the same measure. Prove that there is a point in the interior of every regular n-gon that is equidistant from each vertex of the n-gon. (This point serves as the center of the circle which "circumscribes" the n-gon)

Part D: Prove that the point that serves as the center of the circle that circumscribes a regular n-gon also serves as the center of a circle that contains the midpoint of each side of the n-gon. (This is called the inscribed circle)

Part E: The length of the radius of the inscribed circle of a regular n-gon is called the apothem of the n-gon. Prove that the area, A, of a regular n-gon can be computed using the formula A=(aP)/2, where a is the apothem and P is the perimeter of a regular n-gon.


Explanation / Answer

a) Join the BD and then we have two triangles ABD and CBD in ABD, G,H are midpoints of Sides AB and AD . so GH || BD similarly in the case of Triangle CBD EF || BD thus EF|| DB and DB||GH => GH || EF... now join AC we have two triangles ABC and CAD... in ABC, E,F are midpoints of Sides CD and BC . so GF || AC similarly in the case of Triangle CAD EH || CA thus EH|| AC and AC||GF => GF || EH........ Thus EFGH is a parallelogram (coz. the opposite sides are parallel... ..