Given: ΔABC Prove: The three medians of ΔABC intersect at a common point. When written in the correct order, the two-column proo
f below describes the statements and justifications for proving the three medians of a triangle all intersect in one point: Statements Justifications Point F is a midpoint of Line segment AB Point E is a midpoint of Line segment AC Draw Line segment BE Draw Line segment FC by Construction Point G is the point of intersection between Line segment BE and Line segment FC Intersecting Lines Postulate Draw Line segment AG by Construction Point D is the point of intersection between Line segment AG and Line segment BC Intersecting Lines Postulate Point H lies on Line segment AG such that Line segment AG ≅ Line segment GH by Construction I Line segment BD ≅ Line segment DC Properties of a Parallelogram (diagonals bisect each other) II Line segment FG is parallel to line segment BH and Line segment GE is parallel to line segment HC Midsegment Theorem III Line segment GC is parallel to line segment BH and Line segment BG is parallel to line segment HC Substitution IV BGCH is a parallelogram Properties of a Parallelogram (opposite sides are parallel) Line segment AD is a median Definition of a Median Which is the most logical order of statements and justifications I, II, III, and IV to complete the proof? II, III, I, IV III, II, I, IV II, III, IV, I III, II, IV, I
We are given an area and three different widths and we need to determine the corresponding length and perimeter.
The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards. This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.
The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards. This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.
The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10. This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.
Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.