2. find the lateral area of the pyramid to the nearest whole number.
a) 200 m^2
b) 204 m^2
c) 208 m^2
d) 212 m^2

area of lateral surface

Area=0.5bh×4(because there are four lateral faces)

b=12m

Area=0.5×12m×h

Area=6m×h

using Pythagoras theorem to find height of the triangle of lateral surface.

h²=6²+H²(height of the pyramid)

h²=6²+6²

h²=36+36

h²=72

h=root of 72

h=8.5cm(rounded to nearest tenth)

SA=6m×8.5m×4

Sa=51m²×4

Sa=201m²

answer.. 204m²

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.