Here is the problem, please enter the answer in this forum

https://docs.google.com/presentation/d/1YIYm4ABMI-GvdhtrZYU6RehvvfRTRa8e3CP9ssSmpI4/edit?usp=sharing

totalkoolnezz Apr 23, 2023

#1**0 **

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totalkoolnezz Apr 23, 2023

#2**0 **

[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.

Guest Apr 23, 2023

#4**0 **

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.

Guest Apr 23, 2023