Question

In $A\left(\overline{a}\right),B\left(\overline{b}\right),C\left(\overline{c}\right),D\left(\overline{d}\right)$ are two vertices of trapezium .$E\left(\overline{e}\right),F\left(\overline{f}\right)$ are midpoints of sides AD and BC, then

• A. $\overline{EF}=K\left(\overline{AB}\right)$
• B. $\overline{e}=\frac{\overline{a}+\overline{d}}{2}$
• C. $\overline{f}=\frac{\overline{b}+\overline{c}}{2}$
• D. $\overline{EF}=K\left(\overline{CD}\right)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\overline{EF}=K\left(\overline{AB}\right)$
B $\overline{e}=\frac{\overline{a}+\overline{d}}{2}$
C $\overline{f}=\frac{\overline{b}+\overline{c}}{2}$
D $\overline{EF}=K\left(\overline{CD}\right)$

$ABCD$ is trapezium

$AB$ is parallel to $CD$

$E$ is mid point of $AD$

$\overline{e}=\frac{\overline{a}+\overline{d}}{2}$

Hence option $B$ is correct

$F$ is mid point of $BC$

$\overline{f}=\frac{\overline{b}+\overline{c}}{2}$

Hence option $C$ is correct

line $EF$ is the joining of both mid point of both sides

So $EF$ is parallel to both sides $AB$ and $CD$

$\overline{EF}=K\left(\overline{AB}\right)$

hence option $A$ is correct

and

$\overline{EF}=K\left(\overline{CD}\right)$

hence option $D$ is correct