OPQRS is a regular square-pyramid whose base has sides of 7 cm. Given that the total surface area of the pyramid is 161 cm2, find its slant height.

Answer 1

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.

Answer 2

Se the methos
H^2= B^2+P^2

Answer 3

answer will be 8

Answer 4

You people are just desyroying the math. be ashame ofvyourself you cannot give correct answer of even 1 question shame on yourself

Answer 5

Ddfc

Answer 6

EXERCISE 7.2