+90 Here is the problem, please enter the answer in this forum
https://docs.google.com/presentation/d/1YIYm4ABMI-GvdhtrZYU6RehvvfRTRa8e3CP9ssSmpI4/edit?usp=sharing
+90 copy and paste the link in a new tab in your browser to see the problem.
[APCB'QD] = 213 in2
[CQP] = 75 in2
[PRQ] = [B'CQ]
[ADRP] = [B'CP]
[B'CQ] = 30 in2
The area of triangle B'CQ is 30 square inches.
Since the paper was folded along CP, the area of triangle B'CQ is equal to the area of triangle BCP. Since P is located one-third of the way from A to B, the length of CP is 32 the length of AB. Therefore, the area of triangle BCP is 1/2*1/3⋅AB⋅CP=1/6*6AB⋅CP. Since the total area covered by the folded paper is 213 square inches, the area of triangle BCP is 213−75=138 square inches. Therefore, the area of triangle B'CQ is 1/6⋅138=23 square inches.
