Sides Of A Square Pyramid

The area of the $4$ triangles is the lateral area of the pyramid. The lateral area of a regular pyramid is given by the formula: $A_L=\frac{1}{2}PL$ where: $P$ = perimeter of the base and $L$ = slant height, the slant height is the height of one face

From the figure, $L=\sqrt{3^2+4^2}=5$ . It follows that $P=4\times 6 =24$ .

Finally,

$A_L=\frac{1}{2}PL=\frac{1}{2}(24)(5)=$ $\boxed{\color{#D61F06}\large 60}$

Comment: The formula I used is a derived formula.

The total area of the sides is clearly 4 times the area of one of the sides, since all the triangles are identical.

The area of a triangle is $\dfrac12 \times \text{ base } \times\text{ height}$ . The base is given as 6 m, so this simplifies to $3\times \text{ height}$ .

To find the height: Picture a line drawn from the center of the square base directly out to the center of the base of the triangle. The length of this line is 1/2 the length of the side of the square, or 3 m. The vertical line from the point of the pyramid straight down to the base is perpendicular to this, and is given as 4 m. The height of the face is the hypotenuse of the right triangle with those as legs. We see that I rigged the problem to use the 3-4-5 right triangle. So the height is 5 m. $3 \times 5 = 15$ square meters per side, times 4 sides gives 60 square meters.

The total area of the four triangles is the lateral area of the pyramid and the formula is $\dfrac{1}{2}pL$ where $p$ is the perimeter of the base and $L$ is the slant height. Note that the slant height is the height of one face. If we see the figure the slant height is $5$ since it is a $3-4-5$ right triangle. Substitute:

$A=\dfrac{1}{2}(4)(6)(5)=\boxed{60}~\text{m}$

It is asking for the lateral area of the pyramid. The formula is

$A_{L}$ $=$ $\frac{1}{2}$ $p$ $L$ ; where: p = perimeter of the base, L = slant height

$L$ $=$ $\sqrt{4^2 + 3^2}$ $=$ $5$

$A_{L}$ $=$ $\frac{1}{2}$ $24$ $*$ $5$ $=$ $\boxed{60}$