Geometry #1

For this question, we can use the formula {1/2 ab sin c} to calculate the area of a triangle, where c is an arbitrary angle. So we know that [ABC] =3. Using the above formula, we can find the arbitrary angle c to be roughly 14.478 deg. As the arbitrary angle for triangles CDE and ABC are same, then employ the same formula so the answer would be 1/2 x (4+8) x (3+2) x sin 14.478 which is 12.0004... Which is rounded to 12. ! Do note that this method may have some rounding off errors and adjustments so please use this method when you have to employ this equation for only a few times for accuracy and precision.

The area of $\triangle ABC$ is 3, and the base is 6. So, we can get the equation for area $\frac{6h}{2}=3$ . Solving for $h$ , we get $h=1$ . Now, $CD$ is just segment $CA$ , but 3 times. It has the exact same slope the whole time. That means that if we were to extend segment $CE$ to 8 units long, while retaining its slope and position of point $C$ , then the height from the new point $e$ to the base would be 2. Since it is $3*CA$ , the height of $\triangle CDE$ is 3. The base, obviously, is 8, and we get $a=\frac{8*3}{2}=\frac{24}{2}=12$ . So 12 is the answer!