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April 18, 2015

April 18, 2015

Posted by **sheila** on Sunday, January 23, 2011 at 5:42am.

- pyramid -
**MathMate**, Sunday, January 23, 2011 at 8:36amIt's similar to the previous question.

Start with drawing a diagram if the book did not supply one.

O is the vertex of the pyramid.

Let D be the centre of the rectangle PQRS, and A=centre of side PQ.

OAD is a right triangle where OD is the height=h, and OA is the height of the slant face OPQ.

By Pythagoras theorem, we find

OA²=sqrt(OD²+AD²)

=sqrt(h²+3.5²)

Area of one slant face

=AQ*OA

=3.5sqrt(h²+3.5²)

Area of 4 slant faces

=14sqrt(h²+3.5²)

Area of rectangular base

=7*7

=49

Equate the sum of areas to total surface area

14sqrt(h²+3.5²)+49=161

Solve for h. I get 7.2 approx.

- MATH! PLEASE REPLY ASAP! -
**no**, Monday, April 6, 2015 at 10:50amSe the methos

H^2= B^2+P^2

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