91Ó°ÊÓ

$PS=20,SR=30.$

Height of parallelogram$h=16$ .

Area of parallelogram can be found using the formula

$A=bh$

Where, b= base of parallelogram and h= height of parallelogram.

Area of parallelogram$PQRS$will be

$\begin{array}{l}b=30\\ h=16\\ A=bh\\ A=\left(30\right)\left(16\right)\\ A=480\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Therefore, the area of$â–\pm PQRS$ is$480sq.unit$ .

$PS=20,SR=30.$

Height of triangle$h=16$ .

Area of triangle can be found using the formula

$A=\frac{1}{2}bh$

Where,b= base of triangle and h= height of triangle.

Area of triangle$PSR$will be

$\begin{array}{l}b=30\\ h=16\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(30\right)\left(16\right)\\ A=240\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Therefore, the area of $â–³PSR$is$240sq.unit$ .

$PS=20,SR=30.$

Height of triangle$h=16$ .

Area of triangle can be found using the formula

$A=\frac{1}{2}bh$

Where, b= base of triangle and h= height of triangle.

Area of triangle$PSR$will be

$\begin{array}{l}b=30\\ h=16\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(30\right)\left(16\right)\\ A\left(PSR\right)=240\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Since 0 is the midpoint of PR .Thus,$OP=OR$. So, the area of$OSR=\frac{1}{2}$of area$PSR$

$\begin{array}{l}A\left(OSR\right)=\frac{1}{2}A\left(PSR\right)\\ A\left(OSR\right)=\frac{1}{2}\left(240\right)\\ A\left(OSR\right)=120\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Therefore, the area of$â–³OSR$ is$120sq.unit$ .

$PS=20,SR=30.$

Height of triangle$h=16$ .

Area of triangle can be found using the formula

$A=\frac{1}{2}bh$

Where, b= base of triangle and h= height of triangle.

Area of triangle$PSR$will be

$\begin{array}{l}b=30\\ h=16\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(30\right)\left(16\right)\\ A\left(PSR\right)=240\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Since 0 is the midpoint of$PR$.Thus,$OP=OR$. So, the area of$PSO=\frac{1}{2}$of area$PSR$

$\begin{array}{l}A\left(PSO\right)=\frac{1}{2}A\left(PSR\right)\\ A\left(PSO\right)=\frac{1}{2}\left(240\right)\\ A\left(PSO\right)=120\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Therefore, the area of $â–³PSO$is $120sq.unit$.

$PS=20,SR=30.$

Height of triangle$h=16$ .

Area of triangle can be found using the formula

$A=\frac{1}{2}bh$

Where,b= base of triangle and h= height of triangle.

Area of triangle$PSR$will be

$\begin{array}{l}b=30\\ h=16\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(30\right)\left(16\right)\\ A\left(PSR\right)=240\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Since 0 is the midpoint of$PR$.Thus,$OP=OR$.

So, the area of$OSR=\frac{1}{2}$of area$PSR$

$\begin{array}{l}A\left(OSR\right)=\frac{1}{2}A\left(PSR\right)\\ A\left(OSR\right)=\frac{1}{2}\left(240\right)\\ A\left(OSR\right)=120\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

So, the area of$PSO=\frac{1}{2}$of area$PSR$

$\begin{array}{l}A\left(PSO\right)=\frac{1}{2}A\left(PSR\right)\\ A\left(PSO\right)=\frac{1}{2}\left(240\right)\\ A\left(PSO\right)=120\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit\end{array}$

Since$\mathrm{Δ}POQâ‰\dots \mathrm{Δ}ORS$, thus$A\left(POQ\right)=A\left(ORS\right)=120\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit$

Similarly ,

$\mathrm{Δ}OQRâ‰\dots \mathrm{Δ}OPS$, thus$A\left(OQR\right)=A\left(OPS\right)=120\text{ â¶Ä‰}sq.\text{ â¶Ä‰}unit$

Therefore, the area of $â–³POQ$and $â–³OQR$are$120sq.unit$.

The diagonals of$â–\pm PQRS$can divide the$â–\pm PQRS$into four triangle.

We know that the opposite sides are equal in length, which means$PQ=SR$and $PS=QR$, and 0 is the midpoint ofPR .Thus, we can conclude that the diagonals divide the parallelogram into four equal triangle as shown below:

Therefore, the diagonals of a parallelogram divide the parallelogram into four equal triangle.