Find area of the fourth triangle given the area of three triangles.
area
This is the question that I got in TCS Ninja under the Quantitative section.
How shall I do this? Help !!
Best Answer
Denote by $[...]$ the area of the polygon $...$
Lema 1
Let $ABCD$ be a rectangle a $P$ a point inside as shown. It follows that $$[APB]+[DPC]=[APD]+[CPB]$$
Proof
Draw the parallels to $AD$ and $AB$ through $P$. Then
$$[APB]+[DPC]=\frac{AB·FP}{2}+\frac{DC·EP}{2}=\frac{AB·FP+AB·DC}{2}=\frac{AB·EF}{2}=\frac{[ABCD]}{2}$$
Now back to your problem $$A1+A3=2081=A2+A4=1016+A4\iff A4=2081-1016=\color{red}{1065}$$
Note that, $AB=AD=6$ units (tangents from a point to a circle are equal in length), and $BC=DC=2$(radius) units. Join $AC$. $\triangle ACD$ is congruent to triangle $\triangle ABC$ (tangents make $90^0$ with the radius, AB=AD, BC=DC, hence they're congruent by the RHS criteria). Hence,
Best Answer
Denote by $[...]$ the area of the polygon $...$
Proof
Draw the parallels to $AD$ and $AB$ through $P$. Then $$[APB]+[DPC]=\frac{AB·FP}{2}+\frac{DC·EP}{2}=\frac{AB·FP+AB·DC}{2}=\frac{AB·EF}{2}=\frac{[ABCD]}{2}$$
Now back to your problem $$A1+A3=2081=A2+A4=1016+A4\iff A4=2081-1016=\color{red}{1065}$$
Thanks for the correction @Jean Marie.