Question

Consider a trapezium as shown in the figure then

• A. $\mathrm{CD}=12 \mathrm{~cm}$
• B. $\mathrm{CD}=13 \mathrm{~cm}$
• C. $\frac{\text { Area of trapezium }}{\text { Area of } \triangle \mathrm{ADE}}=6$
• D. Area of triangle $B F C=6 \mathrm{~cm}^2$

Answer 1

Answer: (B), (D)

In given trapezium, $\mathrm{AB}|| \mathrm{CD}$ and $\mathrm{AE} \perp \mathrm{CD}$ and $\mathrm{BF} \perp \mathrm{CD}$
In triangle $\mathrm{ADE}$, by Pythagoras theorem
$\mathrm{AD}^2=\mathrm{AE}^2+\mathrm{DE}^2 $
$ \Rightarrow(5)^2=(4)^2+\mathrm{DE}^2 $
$ \Rightarrow \mathrm{DE}^2=9 $
$\Rightarrow \mathrm{DE}=3 \mathrm{~cm}$
Similarly $\mathrm{FC}=3 \mathrm{~cm}$
$[\mathrm{AS}, \mathrm{BF}=4 \mathrm{~cm}]$
$\therefore \mathrm{CD} =\mathrm{DE}+\mathrm{EF}+\mathrm{FC} $
$ =3+\mathrm{AB}+3 $
$=3+7+3$
$=13 \mathrm{~cm}$
Area of triangle $\mathrm{BFC}=\frac{1}{2} \times B F \times C F$
$ {\left[A s, \operatorname{ar}(\Delta)=\frac{1}{2} \times b \times h\right] }$
$= \frac{1}{2} \times 4 \times 3 $
$= 6 \mathrm{~cm}^2$
Area of triangle $\mathrm{BFC}=$ area of triangle $\mathrm{ADE}$
$=6 \mathrm{~cm}^2$
$\text { Area of trapezium }\mathrm{ABCD} =\frac{1}{2}(A B+C D) \times h $
$ =\frac{1}{2} \times(7+13) \times 4 $
$ =40 \mathrm{~cm}^2$
$\frac{\text { Area of Trapezium } \mathrm{ABCD}}{\text { Area of } \triangle \mathrm{ADE}}=\frac{40}{6}=\frac{20}{3}$