Let APQRS be a pyramid, where the base PQRS is a square of side length 20. The total surface area of the pyramid (including the base) is 1200. Let W, X, Y, and Z be the midpoints of AP, AQ, AR, and AS, respectively. Find the total surface area of frustum PQRSWXYZ (including the bases).
I remember that someone else posted a problem like this but it turns out the guest said the answer was incorrect, and I can't find it again.
The total area of the frustum is (1200 - 400)/2 + 400 + 200 = 1000.
The two pyramids are similar to one another.
the lengths of the sides are in the ration 1:2
so the surface areas are in the ratio 1:4 (I squared them)
The surface area of the whole thing is 1200 if I subtract the area of the base it is
so the surface area of the top bit is 200
and the surface area of the base of the top bit is 100
So i think the surface area of the frustum is (800-200)+400+100 = 1100
I have not put any huge amount of thought into this. So maybe it is not right.
I assume you can plug the answer in and tell me if it is correct?
YOU ARE CORRECT!
Thank you so much. 5 stars, 5 thumbs up
I do hope you understand it and could repeat it under exam conditions.
Melody is just... great. She's a pro mathematician, and I cannot express the gratitude. You are so helpful! :)