Question

State true or false:

The given figure shows a parallelogram $ABED$. Points $P$ and $Q$ trisect the side $AB$, then
$area(△)=area\left(BQC\right)$
$=\frac{1}{6}\times area\left(/{/}_{gm}ABCD\right)$

• A. True
• B. False

Answer 1

The correct option is A True

Given,a parallelogram ABED. Points P and Q trisect the side AB .
So, AP$=$PQ$=$QB,
So, AB$=$3AP$=$3PQ$=$3QB
Join A and C to make diagonal AC.
As diagonal divides the parallelogram into two triangles of equal area.
$\therefore$ Area of $△ABC=\frac{1}{2}\times$Area of parallelogram (1)
Now, In $△ACQ$,
AP=PQ (given)
P is mid point of AQ
Therefore, PC is median of $△ACQ$. As, median divides triangle into two triangle of equal area.
Area of $△APC=$ Area of $△PCQ$ (2)
Again in $△PCB$,
PQ$=$ QB (given)
Q is the mid point of PB.
Therefore, QC is median of $△PCB$. As, median divides triangle into two triangle of equal area.
Area of $△PCQ=$ Area of $△BQC$ (3)
From (2) & (3),
Area of $△APC$$=$Area of $△PCQ$$=$Area of $△QCB$
So, Area of $△APC$$=$Area of $△PCQ$$=$Area of $△BQC$$=\frac{1}{3}\times$Area of $△ABC$
$\because$ Area of $△ABC=\frac{1}{2}\times$Area of parallelogram
Area of $△BQC$$=\frac{1}{3}\times$Area of $△ABC$
=$\frac{1}{3}\times \frac{1}{2}$Area of parallelogram
=$\frac{1}{6}\times$Area of parallelogram