Find the area of a parallelogram with the given vertices. P(1, 3), Q(3, 3), R(7, 8), S(9, 8) A. 10 units2 B. 5 units2 C. 20 unit
2 answers:
You might be interested in
There is a little-known theorem to solve this problem.
The theorem says that
In a triangle, the angle bisector cuts the opposite side into two segments in the ratio of the respective sides lengths.
See the attached triangles for cases 1 and 2. Let x be the length of the third side.
Case 1:
Segment 5cm is adjacent to the 7.6cm side, then
x/7.6=3/5 => x=7.6*3/5=4.56 cm
Case 2:
Segment 3cm is adjacent to the 7.6 cm side, then
x/7.6=5/3 => x=7.6*5/3=12.67 cm
The theorem can be proved by considering the sine rule on the adjacent triangles ADC and BDC with the common side CD and equal angles ACD and DCB.
The theorem says that
In a triangle, the angle bisector cuts the opposite side into two segments in the ratio of the respective sides lengths.
See the attached triangles for cases 1 and 2. Let x be the length of the third side.
Case 1:
Segment 5cm is adjacent to the 7.6cm side, then
x/7.6=3/5 => x=7.6*3/5=4.56 cm
Case 2:
Segment 3cm is adjacent to the 7.6 cm side, then
x/7.6=5/3 => x=7.6*5/3=12.67 cm
The theorem can be proved by considering the sine rule on the adjacent triangles ADC and BDC with the common side CD and equal angles ACD and DCB.