what is the area of triangle ABC?

Question

http://img.photobucket.com/albums/v680/soulball/triangle1.jpg

triangle ABC is divied into 6 smaller triangles by lines drawn from the verticies through a common interior point. the areas of four of these traiangles are as indicated. find the area of triangle ABC.

sorry for the bad pic, but the triangle is not equilateral

a hint that was given to me: The segment AC is cut into two portions; find the ratio of their lengths. Do the same thing for the line from A going through the interior point. Use the information thus obtained to write down a system of two equations for the areas of the two unknown subtriangles.

Answer 1

There is a line from B to a point along AC. Let's call the point D where it intersects with AC. Since the base AC has two triangles on it with a common height, the ratio of the areas is equal to the ratio of the length of the two bases, namely 4/3. Similarly, the ratio of the area of the three triangles to the left of the line BD, to the area of the three triangles on the right is also 4/3 (in total they have a common height also).

However, since the common point in the interior (let's call it P) is an arbitrary one along the line BD, the individual triangles do not all have the ratio of areas of 4/3 for left to right. Let's call the areas of the two subtriangles area x and y. The one on the left is x and the one on the right is y.

Then

(40 + x + 84) / (y + 35 + 30) = 4/3
(124 + x) / (y + 65) = 4/3
3(124 + x) = 4(y + 65)
372 + 3x = 4y + 260
3x + 112 = 4y
y = (3x + 112)/4 = 3x/4 + 28 = .75x + 28

The area of triangle ABC is

Area = 40 + x + 84 + y + 35 + 30 = 189 + x + y
Area = 189 + x + .75x + 28
Area = 217 + 1.75x

So far we have not fixed the point of P. So let's divide the triangle into two more groups of three triangles each. Let's take the line from C to a point on AB. Then we have:

(40 + 30 + x)/(84 + y + 35) = x/84
84(70 + x) = x(119 + y) = x(119 + .75x + 28) = x(147 + .75x)
5880 + 84x = 147x + .75x²

0 = .75x² + 63x - 5880
.75x² + 63x - 5880 = 0
3x² + 252x - 23,520 = 0
x² + 84x - 7840 = 0
(x - 56)(x + 140) = 0
x = 56, -140
The negative solution for x is rejected because area must be positive.
x = 56

The area of triangle ABC is

Area = 217 + 1.75x = 217 + 1.75(56) = 315

Answer 2

draw a line down the middle providing you with 2 proper triangles with 15 degrees the size of the bottom is two x 20 x sin(15) the size of the line by the middle is 20 x cos(15) the realm is a million/2 base x genuine so A = 20 x sin(15) x 20 x cos(15) = one hundred ya what he said

Answer 3

Its actually quite simple if you look at it properly. The line for A ends up half way from C to B, i.e, ( 40 + 30 + 35 ) = Area of half triange.
Therefore, the area of triangle ABC is 2( 40 + 30 + 35) = 210cm square.

Answer 4

maybe a proportion. maybe?

(x/40)=(35/30)??

try it. idk?