Calculating Triangle Area: S1, S2 and ABF

[Semo727]

Triangle Area

Hello!
I came across this: On the picture there is triangle ABC with the unknown area S. All we know is, that area of triangle ABF is S1 and area of FGC is S2. And line AB is parallel with line FG. What is the area of triangle ABC? Thanks for your help!

Im sorry for title of this thread: Triangle volume, I wanted to say "area"(a^2).

And now, I will write what I managed to get:
$$S=\frac{S_2}{2}+S_1+\sqrt{\frac{S_2^2}{4}+S_1 S_2 }$$
but, I'm not sure if its right and I don't like the way I got it, could you please tell me what was your way?

 

[topsquark]

Hello!
I came across this: On the picture there is triangle ABC with the unknown area S. All we know is, that area of triangle ABF is S1 and area of FGC is S2. And line AB is parallel with line FG. What is the area of triangle ABC? Thanks for your help!

Im sorry for title of this thread: Triangle volume, I wanted to say "area"(a^2).

And now, I will write what I managed to get:
$$S=\frac{S_2}{2}+S_1+\sqrt{\frac{S_2^2}{4}+S_1 S_2 }$$
but, I'm not sure if its right and I don't like the way I got it, could you please tell me what was your way?

I don't know how you got it, but my answer is the same, so at least you know it's correct.

-Dan

 

[Architeuthis Dux]

Hi there,

Thank you for sharing your question on calculating the area of triangle ABC. I understand the importance of accuracy in calculations and I am happy to help you with this problem.

To calculate the area of triangle ABC, we can use the formula A = (1/2)bh, where A is the area, b is the base, and h is the height of the triangle. In this case, we do not know the base or the height of triangle ABC, but we can use the fact that line AB is parallel to line FG to our advantage.

First, we can find the ratio of the areas of triangles ABF and FGC. Since line AB is parallel to line FG, we can use the fact that corresponding sides of parallel lines are proportional. Therefore, we can write the ratio of the areas as S1/S2 = (AB/FG)^2. We can rearrange this equation to solve for AB: AB = FG * sqrt(S1/S2).

Next, we can use the Pythagorean theorem to find the height of triangle ABC. Since line AB is parallel to line FG, we can draw a perpendicular line from point A to line FG, creating a right triangle with sides AB, h (the height of triangle ABC), and FG. By the Pythagorean theorem, we can write the equation AB^2 + h^2 = FG^2. We can substitute the value of AB that we found in the previous step and solve for h:

(FG * sqrt(S1/S2))^2 + h^2 = FG^2
(S1/S2)FG^2 + h^2 = FG^2
h^2 = FG^2 - (S1/S2)FG^2
h = FG * sqrt(1 - S1/S2)

Now that we have found the base and height of triangle ABC, we can plug these values into the formula for the area to get the final answer:

A = (1/2)(FG * sqrt(S1/S2))(FG * sqrt(1 - S1/S2))
= (FG^2/2) * sqrt(S1/S2) * sqrt(1 - S1/S2)
= (FG^2/2) * sqrt(S1(1 - S1/S2))

Therefore, the area of triangle ABC is (FG^2/2) * sqrt(S1(1 - S1/S2)).