True or False If AM≅MB, then M must be the midpoint of AB
1 answer:
You might be interested in
First, we circumscribe a rectangle about the parallelogram as shown in the figure. The size of this rectangle is 10 by 6 clearly.
Between the rectangle and the parallelogram 4 triangles are formed, 2 by 2 congruent. We are going to remove the area of these 4 triangles from 60 (the area of the rectangle) and thus get the area of the parallelogram.
Note that one of the triangles has side lengths 4; 4. Thus, the area of this triangle is (4*4)/2=16/2=8 (square units.)
Each of the other two non-congruent triangles to the above have sides of length 6;2, so the area of each is (6*2)/2 =6 (square units.)
Thus, the area of the parallelogram is 60-8-8-6-6=60-16-12=60-28=32 (square units.)
Answer: 32 (square units.)
Between the rectangle and the parallelogram 4 triangles are formed, 2 by 2 congruent. We are going to remove the area of these 4 triangles from 60 (the area of the rectangle) and thus get the area of the parallelogram.
Note that one of the triangles has side lengths 4; 4. Thus, the area of this triangle is (4*4)/2=16/2=8 (square units.)
Each of the other two non-congruent triangles to the above have sides of length 6;2, so the area of each is (6*2)/2 =6 (square units.)
Thus, the area of the parallelogram is 60-8-8-6-6=60-16-12=60-28=32 (square units.)
Answer: 32 (square units.)