Q32. - Geometry

Question

a. Use the diagram at the right. Find the area of$▱PQRS$ .

b. Use the diagram at the right. Find the area of $△PSR$.

c. Use the diagram at the right. Find the area of$△OSR$ . (Hint: Refer to $△PSR$ and use Exercise28 ).

d. Use the diagram at the right. What is the area of$△PSO$ ?

e. Use the diagram at the right. What must the area of  $△POQ$be? Why? What must the area of$△OQR$ be?

f. Use the diagram at the right. State what you have shown in parts (a)-(e) about how the diagonals of a parallelogram divide the parallelogram.

Step-by-Step Solution

Answer

a. The area of $▱PQRS$ is$480sq.unit$ .

b. The area of $△PSR$ is$240sq.unit$ .

c. The area of $△OSR$ is$120sq.unit$ .

d. The area of $△PSO$ is$120sq.unit$ .

e. The area of $△POQ$ and  are $120sq.unit$.

The diagonals of a parallelogram divide the parallelogram into four equal triangle.

Step 1. Given information.

$PS=20,SR=30.$

Height of parallelogram$h=16$ .

Step 2. Concept Used.

Area of parallelogram can be found using the formula

$A=bh$

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

Step 3. Find the area of parallelogram.

Area of parallelogram$PQRS$  will be

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

Therefore, the area of$▱PQRS$  is$480sq.unit$ .

Step 1. Given information.

$PS=20,SR=30.$

Height of triangle$h=16$ .

Step 2. Concept Used.

Area of triangle can be found using the formula

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

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

Step 3. Find the area of triangle.

Area of triangle$PSR$  will be

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

Therefore, the area of $△PSR$ is$240sq.unit$ .

Step 1. Given information.

$PS=20,SR=30.$

Height of triangle$h=16$ .

Step 2. Concept Used.

Area of triangle can be found using the formula

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

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

Step 3. Find the area of triangle.

Area of triangle $PSR$ will be

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

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

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

Therefore, the area of$△OSR$  is$120sq.unit$ .

Step 1. Given information.

$PS=20,SR=30.$

Height of triangle$h=16$ .

Step 2. Concept Used.

Area of triangle can be found using the formula

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

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

Step 3. Find the area of triangle.

Area of triangle$PSR$  will be

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

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

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

Therefore, the area of $△PSO$ is $120sq.unit$.

Step 1. Given information.

$PS=20,SR=30.$

Height of triangle$h=16$ .

Step 2. Concept Used.

Area of triangle can be found using the formula

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

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

Step 3. Find the area of triangle.

Area of triangle $PSR$ will be

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

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

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

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

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

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

Step 4. Use property of congruent triangles.

Since$\Delta POQ\cong \Delta ORS$ , thus $A\left(POQ\right)=A\left(ORS\right)=120sq.unit$

Similarly ,

$\Delta OQR\cong \Delta OPS$, thus$A\left(OQR\right)=A\left(OPS\right)=120sq.unit$

Therefore, the area of $△POQ$ and $△OQR$ are$120sq.unit$ .

The diagonals of $▱PQRS$ can divide the $▱PQRS$ 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.