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 Question from Paul: trying to determine volume of an attic base is rectangular 29 ft by 37 ft peak runs parallel to longest base side and is 7 ft long roof lines run from each end of the peak to the respective corners the peak is 3.5 ft above the base (so seen from above there are 5 visible lines: the peak and 1 line to each of the 4 corners)

Hi Paul,

I see you attic as composed of two pieces, a rectangular pyramid where the apex of the pyramid is not above the center of the rectangle

and a dormer $PQRS.$

The volume of the attic is the sum of the volumes of the two pieces.The volume of each of these pieces can be calculated from a theorem I have always found surprising.

Suppose $A$ s a region in the plane with an area and $B$ is a point which is $h$ units above the plane. Construct a three dimensional region by drawing a line segment fro $B$ to each point in $A.$ The volume of this region is

$\frac13 \times \mbox{ the area of the base A } \times h$

The rectangular pyramid part of your attic is such three dimensional region. The region $A$ in the plane is a 29 feet by 37 feet rectangle and $h$ is 3.5 feet. Hence the volume of this part of your attic is

$\frac13 \times (29 \times 27) \times 3.5 \mbox{ cubic feet.}$

The dormer is also a region whose volume can be calculated in a similar fashion. Think of the triangle $SQR$ as the base and $S$ as the point not in the plane of the base. The height of $S$ above the base plane is the length of $PS.$ The base of the triangle is $QR$ which has length 29 feet and its height, the perpendicular distance from $QR$ to $S$ is 3.5 feet. Hence the area of the triangle $SQR$ is $\large \frac12 \normalsize \times 27 \times 3.5$ square feet. Finally the volume of the dormer is

$\frac13 \times \left(\frac12 \times 27 \times 3.5 \right) \times 7 \mbox{ cubic feet.}$

I hope this helps,
Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.