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.

Guest Apr 1, 2020

#6**+2 **

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

1200-400=800

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?

Melody Apr 1, 2020

#7**+1 **

drumroll please.

....

...

..

YOU ARE CORRECT!

Thank you so much. 5 stars, 5 thumbs up

epic

thanks

Guest Apr 1, 2020

#8

#9**+1 **

Melody is just... great. She's a pro mathematician, and I cannot express the gratitude. You are so helpful! :)

MathProblemSolver101
Apr 2, 2020