Asked by sheila
                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.
            
            
        Answers
                    Answered by
            MathMate
            
    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.
    
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.
                    Answered by
            no
            
    Se the methos
H^2= B^2+P^2
    
H^2= B^2+P^2
                    Answered by
            Abdullah
            
    answer will be 8
    
                    Answered by
            Abdullah
            
    You people are just desyroying the math. be ashame ofvyourself you cannot give correct answer of even 1 question shame on yourself
    
                    Answered by
            Ddf
            
    Ddfc
    
                    Answered by
            ABEBE DEGU
            
    EXERCISE 7.2
    
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