Question

$ABCD$ is a trapezium is which $AB=7$cm, $AD=BC=5$cm and distance between $AB$ and $CD$ is $4$cm. then the length of $CD$ is

• A. $10$cm
• B. $11$cm
• C. $12$cm
• D. $13$cm

Answer 1

The correct option is D $13$cm

Draw AL$\perp$DC,

BM$\perp$DC.

Then AL$=$BM$=4$cm and LM$=7$cm

In $\Delta$ADL, we have

$A{D}^2=A{L}^2+D{L}^2$

$\Rightarrow 25=16+D{L}^2$

$\Rightarrow D{L}^2=25-16$

$DL=\sqrt{9}cm$

$DL=3$cm

$MC=\sqrt{B{C}^2-B{M}^2}=\sqrt{25-16}=3cm$

$\therefore x=DC=CD=CM+ML+LD$

$DL+LM+MC$

$=3+7+3=13cm$.