Question

Name the quadrilateral formed by joining the points $(1,2)$, $(5,4)$, $(3,8)$ and $(-1,6)$ in order. Also, find the area of the region formed by joining the midpoints of the sides of this quadrilateral.

• A. Square
• B. $10$ sq. units
• C. Rhombus
• D. $20$ sq. units

Answer 1

Answer: (A), (B)

The correct options are
A Square
B $10$ sq. units

Solution-

We shall use the distance formula to calculate the lengths of $AB,BC,CD$ & $DA$.

$AB=\sqrt{{(5-1)}^2+{(4-2)}^2}=\sqrt{20}$ units.

$BC=\sqrt{{(5-3)}^2+{(4-8)}^2}=\sqrt{20}$ units.

$CD=\sqrt{{(5-1)}^2+{(4-2)}^2}=\sqrt{20}$ units.

$DA=\sqrt{{(3+1)}^2+{(8-6)}^2}=\sqrt{20}$ units.

So, $AB=BC=CD=DA.$

$\therefore ABCD$ is either a rhombus or a square.

Now, we calculate the slopes of each line.

$m_{AB}=\frac{4-2}{5-1}=\frac{1}{2},$

$m_{BC}=\frac{8-4}{3-5}=-2$,

$m_{CD}=\frac{8-6}{3+1}=\frac{1}{2}$,

$m_{DA}=\frac{2-6}{1+1}=-2$,

$\therefore m_{AB}\times m_{BC}=\frac{1}{2}\times (-2)=-1$

$\Rightarrow \angle ABC={90}^o.$

$m_{BC}C\times m_{CD}=(-2)\times \frac{1}{2}=-1$

$\Rightarrow \angle BCD={90}^o.$

$m_{CD}C\times m_{DA}=\frac{1}{2}\times (-2)=-1$

$\Rightarrow \angle CDA={90}^o.$

So, all the angles are ${90}^o$ and the sides are equal.

$\therefore ABCD$ is a square (also it is a rhombus).

$\therefore ar.ABCD={\left(AB\right)}^2={\left(\sqrt{20}\right)}^{2}sq.units=20$ sq. units.

We know that the figure, made by joining the mid points of a square, has an area $=$ half of the square.

$\therefore A\left(\square PQRS\right)=\frac{1}{2}\times A\left(\square ABCD\right)=\frac{1}{2}\times 20=10$ sq. units.

So, $ABCD$ is a square (also it is a rhombus) and $A\left(\square PQRS\right)=10$ units.