Question

The parallel sides of a trapezoid are $3$ $cm$ and $9$ $cm$. The non-parallel sides are $4$ $cm$ and $6$ $cm$. A line parallel to the base divides the trapezoid into two trapezoids of equal perimeters. What is the ratio in which each of the non-parallel sides is divided?

• A. $4:3$
• B. $3:2$
• C. $4:1$
• D. $3:1$

Answer 1

The correct option is C $4:1$

ABCD is a trapezium in which, $AB=3cm,CD=9cm,AD=4cm,BC=6cm$ & $EF$ $\parallel$ $CD$.

Now, Let $AE=x$, then

$DE=4-x$ & $BF=y,CF=6-y$

we know that line parallel to the base divides the two non - parallel lines in the same ratio. Therefore,

$AE:DE::BF:CF$ $\Rightarrow \frac{x}{4-x}=\frac{y}{6-y}$

$\Rightarrow 6x-xy=4x-xy\Rightarrow 6x-4y=0⟶\left(1\right)$

Perimeter of both trapezoids are same, then

$EF+FB+BA+AE=EF+FC+CD+DE$

$FB+BA+AE=FC+CD+DE$

$y+3+x=6-y+9+4-x$ $\Rightarrow$ $2x+2y=16$

$\Rightarrow x+y=8⟶\left(2\right)$

Solving (1) and (2), we get $x=3.2y=4.8$

Thus required ratio $=\frac{x}{4-x}=\frac{3.2}{4-3.2}=\frac{3.2}{0.8}=\frac{4}{1}or4:1$