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ENGAA Paper 2019

can someone give me hand with question 19 on the 2019 ENGAA paper for Section 1? completely baffled as to how the answer is 8 for the height.
Original post by VendibleWolf
can someone give me hand with question 19 on the 2019 ENGAA paper for Section 1? completely baffled as to how the answer is 8 for the height.

Can you post the question?
Original post by RogerOxon
Can you post the question?

https://www.undergraduate.study.cam.ac.uk/files/publications/engaa_2019.pdf
the question is on that link. i have no clue how to put it on here otherwise
Original post by VendibleWolf
https://www.undergraduate.study.cam.ac.uk/files/publications/engaa_2019.pdf
the question is on that link. i have no clue how to put it on here otherwise

Pyramid.jpg
What expression did you get for the area of each triangular side?
Original post by RogerOxon
What expression did you get for the area of each triangular side?

I'd start with the perpendicular distance from the midpoint of an edge of the base to the apex. Draw a triangle with vertices at the midpoint of the base, the midpoint of one of its edges, and the apex.
Original post by RogerOxon
I'd start with the perpendicular distance from the midpoint of an edge of the base to the apex. Draw a triangle with vertices at the midpoint of the base, the midpoint of one of its edges, and the apex.

for the area of each triangle, i got 0.5xsquare root(h^2+36)x 12 using Pythagoras theorem to find the height of each triangle and then using 0.5xBxH
(edited 3 years ago)
Original post by VendibleWolf
for the area of each triangle, i got 0.5xsquare root(h^2x36)x 12 using Pythagoras theorem to find the height of each triangle and then using 0.5xBxH

The timesed it by 4 and then add 144 (area of the base) and then equate it to 48h which I believe is the volume of the pyramid. for some reason, i am only able to get 2root(6) as an answer which is apparently wrong as it's not even an option. am i being special?
Original post by VendibleWolf
The timesed it by 4 and then add 144 (area of the base) and then equate it to 48h which I believe is the volume of the pyramid. for some reason, i am only able to get 2root(6) as an answer which is apparently wrong as it's not even an option. am i being special?

Agreed, the volume of the pyramid is 48h.

Can you post your working?
Let the perpendicular face height be x:

x2=h2+62x^2=h^2+6^2

122h3=122+2∗12∗x\frac{12^2h}{3}=12^2+2*12*x

∴x=2(h−3)\therefore x=2(h-3)

∴x2=h2+62=22(h−3)2\therefore x^2=h^2+6^2=2^2(h-3)^2

This simplifies to give two solutions.
Original post by RogerOxon
Agreed, the volume of the pyramid is 48h.

Can you post your working?

IMG_2109[1484].jpgexcuse the awful handwriting. IDK why its sideways.
Original post by VendibleWolf
IMG_2109[1484].jpgexcuse the awful handwriting. IDK why its sideways.

You went from:
a+b=ca+b=c
to
a2+b2=c2a^2+b^2=c^2
to get rid of the square root. You can't.

See my post above for a simpler approach. Using x to denote the perpendicular height of the triangular faces reduces the amount of writing. You've got to keep these things as simple as possible for as long as possible.
(edited 3 years ago)
Original post by RogerOxon
You went from:
a+b=ca+b=c
to
a2+b2=c2a^2+b^2=c^2
to get rid of the square root. You can't.

See my post above for a simpler approach. Using x to denote the perpendicular height of the triangular faces reduces the amount of writing. You've got to keep these things as simple as possible for as long as possible.

I don't understand why I can't write the perpendicular height as the square root of h^2 plus 6^2. I understand what you have done but I don't see the correlation between what you have said I can't do and what I have done. if that makes sense...
Original post by VendibleWolf
I don't understand why I can't write the perpendicular height as the square root of h^2 plus 6^2. I understand what you have done but I don't see the correlation between what you have said I can't do and what I have done. if that makes sense...

You went from:

6+h2+36=2h6+\sqrt{h^2+36}=2h

To:

36+h2+36=4h236+h^2+36=4h^2

That's your mistake.
Original post by RogerOxon
You went from:

6+h2+36=2h6+\sqrt{h^2+36}=2h

To:

36+h2+36=4h236+h^2+36=4h^2

That's your mistake.

why cant I do that?
Original post by VendibleWolf
why cant I do that?

Because it would be:
(6+h2+36)2=36+h2+36+12h2+36(6+\sqrt{h^2+36})^2=36+h^2+36+12\sqrt{h^2+36}

Try it with h=8, and see what you get before and after.
Original post by VendibleWolf
why cant I do that?

would I have to do (6+Squareroot(h^2+36))^2 in order to do it correctly?
Original post by RogerOxon
Because it would be:
(6+h2+36)2=36+h2+36+12h2+36(6+\sqrt{h^2+36})^2=36+h^2+36+12\sqrt{h^2+36}

Try it with h=8, and see what you get before and after.

Ok yeah, I'm with you. thanks for the help. Not sure how I forgot how that's how it works. much appreciated
Original post by VendibleWolf
Ok yeah, I'm with you. thanks for the help. Not sure how I forgot how that's how it works. much appreciated

No problem. We all get brain failure from time to time.

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