# Thread: Volume & total surface area

1. ## Volume & total surface area

how do i found the volume and total surface area of the sqaure pyramid if the perpendicular high is 14.0cm

2. ok, the formula to working out the VOLUME of any pyramid is:
1/3 x area of base x height (perpendicular height)
so if we substitue the numbers in...

we will have ((5 x 5) / 3) x 14
= 116.66666 (recurring) cm^3

now to find the total surface area...
that means, adding up the areas of the square base and the 4 other triangles around it

the area of the base = 5x5 = 25
the area of each triangle = (14 x 5)/2 = 35
so 4 triangles = 4 x 35 = 140

140 + 25 = 165
so total surface area = 165cm^2

3. Originally Posted by turbod15b
how do i found the volume and total surface area of the sqaure pyramid if the perpendicular high is 14.0cm

well, the manual way to do it is to just find the area of the base, which is a sqaure and add it to the area of all the sides which are triangles. you can find the height of the triangle by using Pythagoras' theorem to find the hypotenuse of triangle ABC in the diagram (i assume the vertex is labeled A).

but there are nice compact formulas for the surface area and volume of square-based pyramids. see the link below.

see Surface Area and Volume

4. Hello, turbod15b!

I assume that the top vertex is A.

For the surface area:

The base is a square. .Its area is: 5² = 25 cm².

In the base, triangle BCE is 45° right triangle.
Since CE = 2.5, then BC = 2.5 cm

The sides of the pyramid are four congruent triangles.
Triangle ADE has base DE = 5 and height AC.
In right triangle ABC: AC² = .AB² + BC² .= .14² + 2.5² .= .809/4
. . Hence: .AC = ½√809
. . . . . . . . . . . . . . . . . . . . . .___ . . . . . . . ___
The area of ∆ADE .= .½(5)(½√809) .= .¼·5√809
. . . . . . . . . . . . . . . . . . . . . . . . . .___
The area of the four triangles is: .5√809

. . - . . . - . . . . . . . . . . . . . ___
The total surface area is: .5√809 + 25 . .167.2 cm²