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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.

 Apr 1, 2020
 #1
 #3
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Except what makes absolutely no sense about the guest's answer is that base PQRS is 400 and the total is 170? BRUH

Guest Apr 1, 2020
 #5
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Even so, I appreciate the willingness to help.

Guest Apr 1, 2020
 #2
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The total area of the frustum is (1200 - 400)/2 + 400 + 200 = 1000.

 Apr 1, 2020
 #4
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1000 is incorrect.

Guest Apr 1, 2020
 #6
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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?

 Apr 1, 2020
 #7
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drumroll please.

....

...

..

 

YOU ARE CORRECT!
Thank you so much. 5 stars, 5 thumbs up

epic

thanks

Guest Apr 1, 2020
 #8
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Thanks,

I do hope you understand it and could repeat it under exam conditions.

Melody  Apr 1, 2020
 #9
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Melody is just... great. She's a pro mathematician, and I cannot express the gratitude. You are so helpful! :)

 #10
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Thanks for the accolade MPSolver   cool

Melody  Apr 2, 2020

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