Question

Consider a trapezium as shown in the figure then

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

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 $\mathrm{BFC}=\frac{1}{2}\times BF\times CF$
$\left[As,\mathrm{ar}(\Delta )=\frac{1}{2}\times b\times h\right]$
$=\frac{1}{2}\times 4\times 3$
$=6{\mathrm{cm}}^2$
Area of triangle $\mathrm{BFC}=$ area of triangle $\mathrm{ADE}$
$=6{\mathrm{cm}}^2$
$\text{ Area of trapezium }\mathrm{ABCD}=\frac{1}{2}(AB+CD)\times h$
$=\frac{1}{2}\times (7+13)\times 4$
$=40{\mathrm{cm}}^2$
$\frac{\text{ Area of Trapezium }\mathrm{ABCD}}{\text{ Area of }△\mathrm{ADE}}=\frac{40}{6}=\frac{20}{3}$