To note: There is a question on site with same diagram but the questions I ask regarding this diagram , are different.
Find $ S1$ in the parallelogram below
The Question:
Given:
$S_1 = 10\ \mathrm{ m^2}$ , $S_2=3\ \mathrm{ m^2}$. Find $S_1$.
$S$ in the diagram is the shaded region. Answer is $7\ \mathrm{ m^2}.$
I read an answer & it wrote the given relations to be true:
$S_1+S_2=S_3$
$S_{BOC}+ S_{AOD} =\frac12\cdot\text{Area of parallelogram}$
Q1: I am not able to figure out as to how the 1st relation is true.
For the 2nd point. I have Been trying to figure it out for long , I noticed something.
Point is $O$ is not a center of the parallelogram. From this point , four lines are drawn which touch the vertices of the parallelogram . Therefore, 4 types of different areas can be formed.
The relation states that :
$S_{AOD} + S_{BOC} = S_{AOB} + S_{DOC} =\frac12\cdot\text{Area of parallelogram}$
Q2: If I take any point in a parallelogram like the point $O$, 4 types of areas will be formed. Can we say that out of these 4 areas, sum of 2 = sum of other 2 areas always ? I’m not sure if this is true.
You need to get why $\text{Eq.2}$ and $\text{Eq.3}$ are true, after which $\text{Eq.1}$ is a piece of cake.
Consider a vertical line segment $\overline{EF}$ passing through $O$, perpendicular to $\overline{AD}$ and $\overline{BC}$, with $E$ lying on $\overline{BC}$ and $F$ on $\overline{BC}.$
$\begin{align} \frac12\cdot [ABCD] &= \frac12\cdot EF\cdot AD\\ &= \frac12\cdot (EO + OF) \cdot AD \\ &=\frac12 \cdot EO\cdot AD + \frac12\cdot OF \cdot BC \quad (AD = BC)\\ &= [\triangle AOD] + [\triangle BOC] \end{align}$
(where $AB$ represents length of line segment $\overline{AB}$ and $[\quad]$ represents area of polygon)
$\text{Eq3}$ is very easy now, with either similar logic or subtraction from total area of the parallelogram.
Finally, for $\text{Eq.1}$, let $S_4$ = area of triangle formed when $S_2$ is removed from $\triangle AOD$:
$\begin{align} S_1 + S_2 &= S_1 + (S_2 + S_4) - S_4 \\ &= [\triangle AOD] + [\triangle BOC] - S_4 \\ &= \frac12\cdot [ABCD] - S_4 \\ &= [\triangle ACD] - S_4 \quad (\because\triangle ACD\cong\triangle CAB)\\ &= S_3 \end{align}$