Editing 2509: Useful Geometry Formulas

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This illustration is commonly used to depict a rectangular prism, with 'b' denoting the 'breadth', 'd' the 'depth' and 'h' the 'height'.  However, the labeled angle θ, which is necessary for the area calculation of the 2D shape, would not normally be used in a diagram of a rectangular prism, as all angles are assumed to be right angles.  A rhomboidal prism could be accurately described by this diagram with the assumption that the 'base' parallelogram is perpendicular to the 'front' and that the only non-right angle is θ.  In that case 'd' would not accurately describe the depth of the solid, which would be d sin θ.
 
This illustration is commonly used to depict a rectangular prism, with 'b' denoting the 'breadth', 'd' the 'depth' and 'h' the 'height'.  However, the labeled angle θ, which is necessary for the area calculation of the 2D shape, would not normally be used in a diagram of a rectangular prism, as all angles are assumed to be right angles.  A rhomboidal prism could be accurately described by this diagram with the assumption that the 'base' parallelogram is perpendicular to the 'front' and that the only non-right angle is θ.  In that case 'd' would not accurately describe the depth of the solid, which would be d sin θ.
  
The area of the 2D shape comprises the rectangle at the lower left, the parallelogram above it, and the parallelogram on the right.  The area of the rectangle representing the front face of the prism is bh. The area of the upper parallelogram is db sin θ. The area of the right parallelogram is dh cos θ.  The equation for this area is  A = bh + d(b sinθ + h cosθ) as is given below the figure.  
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The area of the 2D shape is comprised of the rectangle at the lower left, the parallelogram above it, and the parallelogram on the right.  The area of the rectangle representing the front face of the prism is bh. The area of the upper parallelogram is db sin θ. The area of the right parallelogram is dh cos θ.  The equation for this area is  A = bh + d(b sinθ + h cosθ) as is given below the figure.  
  
 
The surface area of the prism would be 2bh + 2db sin θ + 2dh. The volume is bdh sin θ. Assuming a 3D shape, θ can be artificially altered by the projection; the assumption could be made that θ is 90 degrees, and sin θ is 1 (and therefore can be eliminated from the formulas), but since θ is marked, such an assumption might not be valid.
 
The surface area of the prism would be 2bh + 2db sin θ + 2dh. The volume is bdh sin θ. Assuming a 3D shape, θ can be artificially altered by the projection; the assumption could be made that θ is 90 degrees, and sin θ is 1 (and therefore can be eliminated from the formulas), but since θ is marked, such an assumption might not be valid.

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