Answered: Square base pyramid is given. Then side…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Dear Sirs,

Square base pyramid is given. Then side of base is equal to pyramid height. Pyramid is filled up with water up to 3/8 of it's height.

Find: ratio of wetted surfaces when it sits on the base to when it sits on it's apex.

The best I could do is attached. I found that the upper base of trapezoid is equal of 5/8a. Then further, numbers became weird when I proceeded to find height of the prapezoid in order to calculate surface area.

Please help to find right approach/idea.

Thanks Kosta.

vertical
Cut view
h
D
B
h
h = a
base is square
a 1. Side of frustum
SABC-SDBE + SADEC
2
¾/₂h 1 a·a=DE⋅ 1. §a+a+DE. 22 a
2
2
2
1.6
16
a/² = DE• 2a + ² a ² + 39 DE
59,
за
2² - ² ² ² = D E (²2² +²46)
DE +39)
a
2
16
16
m/w
89
{a² = DE (²6) ; {√ √ a ² = 1aDE (DE=a

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Follow-up Question

I just would like to clarify some details. As per question pyramid is filled up with water up to 3/8 of it's height. Then wetted surfaces area when pyramid sits on it's base will be the sum: base area + area of 4 trapezoids. Base area is a^2. But trapezoid surface area is my first question.

Second question is that find wetted surface when it sits on it's apex. Apex is not triangle. Apex is just a one point/peak of the pyramid. It means pyramid is turned upside down and then total wetted surface needs to be found. Volume of water is the same as when it sits on it's base. But height will be different. Then total surface will consist of 4 triangles (wetted lateral surfaces).

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