In the basis of many geometrical figures rectangles and squares lie. The parallelepiped is most widespread among them. Also the cube, a pyramid and the truncated pyramid belong to them. All these four figures have the parameter called height.
Instruction
1. Draw the simplest isometric figure called a rectangular parallelepiped. It received the name for the reason that its sides are rectangles. The basis of this parallelepiped also is the rectangle having width of an and length of b.
2. The volume of a rectangular parallelepiped is equal to the work of the area of the basis on height: V = S*h. As in the basis of a parallelepiped the rectangle lies, the area of this basis is equal to S=a*b where an is length, b is width. From here, volume is equal to V=a*b*h where h is height (and, h = c where with - a parallelepiped edge). If it is required to find parallelepiped height in a task, transform the last formula as follows: h=V/a*b.
3. There are rectangular parallelepipeds in which bases squares lie. All its sides represent rectangles from which squares are two. It means that its volume is equal to V=h*a^2 where h is height of a parallelepiped, an is the square length equal to width. Respectively, find height of this figure as follows: h=V/a^2.
4. At a cube squares with identical parameters are all six sides. The formula for calculation of its volume looks so: V=a^3. To calculate any of its parties if another is known, it is not required as all of them are equal among themselves.
5. All above-mentioned ways assume calculation of height through parallelepiped volume. However there is also other way allowing to calculate height with the set width and length. Use it in case the area is given in a statement of the problem instead of volume. The area of a parallelepiped is equal to S=2*a^2*b^2*c^2. From here, c (parallelepiped height) is equal with =sqrt (s / (2*a^2*b^2)).
6. There are also other tasks of calculation of height with the set length and width. Pyramids appear in some of them. If in a task the corner at the plane of the basis of a pyramid is given and also its length and width, find height, using Pythagorean theorem and properties of corners.
7. To find pyramid height, at first determine basis diagonal. From the drawing it is possible to draw a conclusion that diagonal is equal to d= √ to a^2+b^2. As height falls in the center of the basis, find a half of diagonal as follows: d/2= √ a^2+b^2/2. Find height, using properties of a tangent: tgα=h / √ a^2+b^2/2. From this it follows that height is equal to h= √ to a^2+b^2/2*tgα.