geometrytriangles
How to find the area of green region in terms of yellow, blue and red region in the following figure?
The triangle is any random triangle and an arbitrary point $P$ is taken where all the colored regions coincide.
The answer given is $$G = \frac{BY(Y + B + 2R)}{R^2 – BY}$$
But I'm not able to find out. Please help.
$$ \begin{align} \text{red area} &= \text{area of }\Delta ABC + \text{area of semi-circle} - \text{area of quadrant} \\ &= \tfrac12 \times 21 \times 28 + \tfrac12 \pi \left(\tfrac{35}{2}\right)^2 - \tfrac14 \pi (21)^2 \\ &= 294 + 481.0563 - 346.3606 \\ &= 428.6957 \end{align} $$ So it looks like you are right and your book is wrong.
See the image with blue parts shifted:
The single blue figure is the same part of its small square as the red plus both blue of the big square, hence areas $$\frac{2\cdot blue + red}{blue}=\frac{big\ square}{small\ square}=4$$ so $$2\cdot blue + red = 4\cdot blue$$ hence $$red = 2\cdot blue$$ Q.E.D.
Without the formula for the area of a triangle, and without the formula for the area of a circle...
Best Answer
The following solution is not very elegant, but it gets the job done. Then again, the desired answer is not itself very pretty, so this will do.
Let the vertices of the triangle adjacent to the green, yellow, and blue areas be $A$, $B$, and $C$ respectively. Suppose the barycentric coordinates of the point $P$ are $\alpha$, $\beta$, and $\gamma$, so that $$\begin{align} \alpha A + \beta B + \gamma C &= P, \\ \alpha + \beta + \gamma &= 1. \end{align}$$ Let the area of the whole triangle be $\triangle$. Then the area of the red region is $\alpha\triangle$, because the height of $P$ from $BC$ is $\alpha$ times that of $A$. The yellow and red together form a triangle whose third vertex divides $BC$ in the ratio $\alpha:\beta$, so its area is $\triangle\alpha/(\alpha+\beta)$. Similarly, the area of the blue and red together is $\triangle\alpha/(\alpha+\gamma)$. (All this should become clear from looking at the diagrams in the MathWorld article on barycentric coordinates.) Solving, you get $$\begin{align} \beta &= \frac B{B+R}, \\ \gamma &= \frac Y{Y+R}, \\ \alpha &= 1 - \beta - \gamma = \frac{R^2-BY}{(B+R)(Y+R)}, \\ \triangle &= \frac R\alpha = \frac{R(B+R)(Y+R)}{R^2-BY}. \end{align}$$ The remaining green area is of course $\triangle - R - B - Y$, which simplifies to the desired result.