Htykuut

Having a hell of a lot of issues with these problems, supposed to be on the topic of triangle area ratio theorem (ratio area of triangles = ratio of triangles' heights x ratio of triangles' bases.)

Here are that are causing me trouble:

Any help would be greatly appreciated.

I'll get you started:

For (2), we use $text{Area} = frac{1}{2}absin(C)$, where $C$ is the angle between the sides with length $a$ and $b$. So,
$$
[CDE] = frac{1}{2} times 3 times 5 times sin(angle ACB) = 7,
$$
Use this to find $sin(angle ACB)$, from which you can use the same formula to find $[ABC]$.

For (3), the same formula can get you $angle ACB$, from which you can determine everything about $triangle ABC$ - in particular, the information you need to find $x$.

Without using any trigonometry, here is a solution to (2):

To start off, I created a new point called F that lies on segment AC so that AF = 12.

Since △CDE and △EFE share an altitude, we can assign [DFE] = 7/3 because CD = 3 and EF = 1, so [DFE] must be 1/3 of [CDE].

△CDE and △FAE also share an altitude, and the ratio of CD to FA is 3:12, or 1:4, so [FAE] must be 28.

7+(7/3)+28 = 35 + 7/3 = [CAE]. Now, we have [CAE]. Using the ratio of CE to EB, 5:12, we can find the area of EAB. 12(35 + 7/3)/5 = (420+28)/5 = 89 + 3/5 = [AEB].

Now, we add up all the parts. [ABC] =7 + 89 + (3/5) + 28 + (7/3) = 126.93̅

So, [ABC] = 126.93̅.