- Ais $2$
- ✓is $4$
- Cis $8$
- Dcannot be determined from the data
We have,
$A B C D$ is a trapezium.
$A B$ is parallel to $C D$.
Area of trapezium $=12$
$\frac{1}{2} \times h(A B+C D)=12$
$A B+C D=\frac{24}{h}$
Sides and height of trapezium are integer.
$\therefore h$ is a factor of $24$
$h=1,2,3,4,6,8,12,24$
$A B+C D=24,12,8,6,4,3,2,1$
But $A B+C D > h$
$A B+C D=24,12,8,6$
In $\triangle B E C$,
$B E C$ is a right angled triangle.
$\therefore h$ must be $3$ and $4$
When $h=3, B E=4, C E=5$
$A B+C D=8$
$A E+B E+A E=8$
$2 A E=8-B E=8-4$
$A E=2$
$\therefore A B=4+2=6, C D=2$
$\therefore|A B-C D|=|6-2|=4$
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$S_1=\{z \in C:|z|<4\}, S_2=\left\{z \in C: \operatorname{Im}\left[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right]>0\right\} \text { and } $
$S_3:\{z \in C: \operatorname{Re} z>0\} .$
$1.$ Area of $S=$
$(A)$ $\frac{10 \pi}{3}$ $(B)$ $\frac{20 \pi}{3}$ $(C)$ $\frac{16 \pi}{3}$ $(D)$ $\frac{32 \pi}{3}$
$2.$ $\min _{z \in S}|1-3 i-z|=$
$(A)$ $\frac{2-\sqrt{3}}{2}$ $(B)$ $\frac{2+\sqrt{3}}{2}$ $(C)$ $\frac{3-\sqrt{3}}{2}$$(D)$ $\frac{3+\sqrt{3}}{2}$
Give the answer question $1$ and $2$
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