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