Question

$ABCD$ is a trapezium with parallel sides $AB=a$ cm and $DC=b$ cm. $E$ and $F$ are the mid-points of the non-parallel sides. The ratio of $\text{ar}\left(ABFE\right)$ and $\text{ar}\left(EFCD\right)$ is

• A. $a:b$
• B. $(3a+b):(a+3b)$
• C. $(a+3b):(3a+b)$
• D. $(2a+b):(3a+b)$

Answer 1

The correct option is B $(3a+b):(a+3b)$

Given- $ABCD$ is a trapezium. $EF$ is the line joining the mid points\\ of $AD$ and $BC$ at $E$ and $F$ respectively. $AB=a,CD=b$

To find out- The ratio of $\text{ar}ABFE$ and $\text{ar}EFCD)$

Construction- A perpendicular $mn$ is dropped from $m$ on $DC$ to $AB$.

$mn$ meets $AB$ at $n$ and intersects $EF$ at $O$.

Solution-

Let the height of the trapezium $ABCD$ be $h$.

$\therefore$ its area $=\frac{1}{2}\times (AB+CD)\times h=\frac{1}{2}\times (a+b)\times h$ ..........(i)

Since $E$ and $F$ are the mid points of $AD$ and $BC$, the line $EF$ divides the height of the trapezium into two equals.

$\therefore mO=On=\frac{h}{2}$ .........(ii)

Let $EF=x$

$\therefore \text{ar}ABFE=\frac{1}{2}\times (a+x)\times \frac{h}{2}=\frac{h}{4}\times (a+x)$ ......(from (i) )

and $\text{ar}DCFE=\frac{1}{2}\times (b+x)\times \frac{h}{2}=\frac{h}{4}\times (b+x)$ ,,,,,,,,,,,,(iii)

Now $\text{ar}ABFE+\text{ar}DCFE=\text{ar}ABCD$

$\Rightarrow \frac{h}{4}\times (a+x)+\frac{h}{4}\times (b+x)=\frac{h}{2}\times (a+b)$ .....[from (i), (ii) and (iii)]

$\Rightarrow \frac{a+b}{2}+x=a+b$

$\Rightarrow x=\frac{a+b}{2}$

Substituting this value of $x$ (i) and (ii), we have

$\text{ar}ABFE=\frac{1}{4}h(a+\frac{a+b}{2})=\frac{1}{8}h(3a+b)$

And $\text{ar}DCFE=\frac{1}{4}h(b+\frac{a+b}{2})=\frac{1}{8}h(a+3b)$

$\therefore \text{ar}ABFE:\text{ar}DCFE=\frac{1}{8}h(3a+b):\frac{1}{8}h(a+3b)=(3a+b):(a+3b)$