Round 3308.89 to the nearest Australian dollar - $4336 AUD .
Short leg = 18 Long leg= 24 Hypotenuse= 30
My calculations were simple trial and error. I started with a trial of 10 for the short leg to begin with, averaging and estimating the appropriate proportions based off 6 additional increments. Sorry I don't have an algebraic method for you. Hope this has helped somewhat :)
Answer:

Find the midsegment of the triangle which is parallel to CA.

Tip
- A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle.
- This segment has two special properties. It is always parallel to the third side, and the length of the midsegment is half the length of the third side.
- If two segments are congruent, then they have the same length or measure.In other words, congruent sides of a triangle have the same length.

We have to find the segment which is parallel to CA.
From the given data,
The segment EG is the midsegment of the triangle
ABC.
So we have,
A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side.

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In this problem, you apply principles in trigonometry. Since it is not mentioned, you will not assume that the triangle is a special triangle such as the right triangle. Hence, you cannot use Pythagorean formulas. The only equations you can use is the Law of Sines and Law of Cosines.
For finding side a, you can answer this easily by the Law of Cosines. The equation is
a2=b2 +c2 -2bccosA
a2 = 11^2 + 8^2 -2(11)(8)(cos54)
a2 = 81.55
a = √81.55
a = 9
Then, we use the Law of Sines to find angles B and C. The formula would be
a/sinA = b/sinB = c/sinC
9/sin54° = 11/sinB
B = 80.4°
9/sin54° = 8/sinC
C = 45.6°
The answer would be: a ≈ 9, C ≈ 45.6, B ≈ 80.4
It is c because 8x2=16 and 8x4=32 and so on