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Pyramid Problems
Surface area and volume of pyramid problems along with with detailed solutions are presented.
Volume of a Pyramid
We consider here a pyramid whose base is a rectangle.
Volume = (1 / 3) L W h
where L and W are the length and width of the base and h is the height of the pyramid.
Problem 1
Find a formula for the total area of the surface of the pyramid shown above
Solution to Problem 1:
The surface of the pyramid is made up of four triangles congruent in pairs and a rectangular base. We need to find the area of each all these figures in order to find the area of the surface of the pyramid. Let S be the middle of the diagonal of the base ABCD of the pyramid. Let O' be the middle of CD and O" be the middle of AD. OS is orthogonal to the base ABCD of the pyramid and is therefore perpendicular to SO' and SO". OSO' is a right triangle and the length of the altitude H of triangle DOC is given by: (using Pythagora's theorem)
H = sqrt [ h
^{ 2}
+ (L/2)
^{ 2}
]
We now use the length W of the base and the length H of the altitude of triangle DOC to find its area. (1/2 * base * height)
A(DOC) = (1 / 2) * W * sqrt [ h
^{ 2}
+ (L/2)
^{ 2}
]
Similarly, using triangle OSO" we can find a formula for the altitude H' of triangle AOD as follows:
H' = sqrt [ h
^{ 2}
+ (W/2)
^{ 2}
]
In a similar way as above, the area of triangle AOD is given by:
A(AOD) = (1 / 2) * L * sqrt [ h
^{ 2}
+ (W/2)
^{ 2}
]
Triangle AOB is congruent to triangle DOC and have equal areas. Triangle BOC is congruent to triangle AOD and have equal areas. The total lateral area of the surface made up of the four triangles is given by:
A(lateral surface) = W * sqrt [ h
^{ 2}
+ (L/2)
^{ 2}
] + L * sqrt [ h
^{ 2}
+ (W/2)
^{ 2}
]
Now the total area is obtained by adding the area of the base W * L to the area of the lateral surface.
Total Area = W * sqrt [ h
^{ 2}
+ (L/2)
^{ 2}
] + L * sqrt [ h
^{ 2}
+ (W/2)
^{ 2}
] + W * L
Problem 2
Below is shown a pyramid with square base, side x, and height h. Find the value of x so that the volume of the pyramid is 1000 cm
^{ 3}
the surface area is minimum.
Solution to Problem 2:
We first use the formula of the volume given above to write the equation:
(1 / 3) h x
^{ 2}
= 1000
We now use the formula for the surface area found in problem 1 above to write a formula for the surface area S of the given pyramid. In this problem we have L = W = x, hence:
S = x * sqrt [ h
^{ 2}
+ (x/2)
^{ 2}
] + x * sqrt [ h
^{ 2}
+ (x/2)
^{ 2}
] + x * x
= 2 x sqrt [ h
^{ 2}
+ (x/2)
^{ 2}
] + x
^{ 2}
Solve the equation (1 / 3) h x
^{ 2}
= 1000 for h to obtain:
h = 3000 / x
^{ 2}
Substitute h in the surface area formula by 3000 / x
^{ 2}
to obtain a formula in terms of x only:
S = 2 x sqrt [ (3000 / x
^{ 2}
)
^{ 2}
+ (x/2)
^{ 2}
] + x
^{ 2}
To find the x value that will minimize the surface area, we graph S as a function of x and locate the minimum point using a graphic calculator.
The approximate value of x was found to be:
x = 12.9 cm. (approximated to 1 decimal place).
Note that locating the
locating the minimum point
on the graph of S above can be done rigorously using calculus methods.
More References and Links to Geometry Problems
Geometry Tutorials, Problems and Interactive Applets.
Surface Area and Volume of Pyramid - Geometry Calculator.
Online calculator to calculate the surface area, the volume and many other parameters of a pyramid given the dimensions of its rectangular base and its height.
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