Question

In a quadrilateral ABCD, if AB || CD, $\angle$D $=2\angle$B, AD = b and CD = a, then the side AB is of length:

• A. $\frac{a}{2}+2b$
• B. $a+2b$
• C. $2a-b$
• D. $a+b$

Answer 1

Answer: (D)

$a+b$

Given, $AB\parallel CD$, $\angle D=2\angle B$, $AD=b$, $CD=a$

Let $\angle B=x$ and draw a line segment, $CM$ equal and parallel to $AD$, as shown in the above figure.

Hence, $AMCD$ is a parallelogram. [Quadrilateral having one pair of opposite sides equal and parallel]

Now, $AD=b$

$\therefore CM=AD=b$
and $CD=a$

$\therefore AM=CD=a$

$\therefore \angle D=\angle AMC$ [Opposite angles of a parallelogram]
Hence, $\angle AMC=2x[\angle D=2\angle B=2x]$

Now, in $△MCB$
$2x=\angle MCB+\angle x(\because$ Exterior angle is equal to sum of opposite interior angle$)$
$\therefore \angle MCB=x$
Hence, $MB=MC[$As sides opposite to equal angles are equal$]$
$\therefore MB=b$

Now, $AB=AM+MB$

$\therefore AB=a+b$

Hence, option D is correct.