Question

Construct a quadrilateral ABCD where AB = 8 cm, BC = 6 cm, DC = 4 cm. DC is parallel to AB and $\angle DCB$ is twice the angle ABC.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Our aim is to construct the quadrilateral ABCD.

Given, DC || AB and $\angle DCB$ is twice the $\angle ABC$ ...(i)

Suppose, the following figure is the rough sketch.


Let, $\angle ABC=x$
From (i), $\angle DCB=2x$

Since DC || AB,
$\angle DCB+\angle ABC=180°$ (Co-interior angles)
$\Rightarrow 2x+x=180°$
$\Rightarrow 3x=180°$
$\Rightarrow x=\frac{180°}{3}=60°$

From (i), $\angle ABC=x=60°$, and $\angle DCB=2x=120°$.

Steps of construction:

Step 1: Draw a line segment AB of length 8 cm.

Step 2: At B, construct an angle of 60° and extend the ray.

Step 3: Keeping B as centre, draw an arc of radius 6 cm on the extended line. Mark the intersecting point to be C.

Step 4: Construct 120° at the point C and extend the line.

Step 5: Keeping C as centre, draw an arc of radius 4 cm on the extended line. Mark the intersecting point to be D.

Step 6: Join D and A.

After construction, we get the following quadrilateral.

Hence, ABCD is the required quadrilateral.