Question

Find the angle x from the figure if ABCD is a parallelogram in which $\angle A:\angle B=1:4$. Also, $\Delta$DEF is an isosceles triangle with DE = DF.

• A. 36°
• B. 18°
• C. 54°
• D. 44°

Answer 1

The correct option is B

18°

As per the question,
$\angle A$ : $\angle B$ = 1 : 4.
$\Rightarrow$$\frac{\angle A}{\angle B}$ = $\frac{1}{4}$
$\Rightarrow$ $\angle B$ = 4$\angle A$ ...(1)

We know that, BC || AD and AB is the transversal.
$\angle A$ + $\angle B$ = 180° ( Co-interior angles)

From (1), $\angle A$ + 4$\angle A$ = 180°
$\Rightarrow$ 5$\angle A$ = 180°
$\Rightarrow$ $\angle A$ = 36°
$\angle B$ = 4 $\times$ 36° = 144° ( from(i) )

In a parallelogram, opposite angles are equal.
So, $\angle B$ = $\angle D$
$\angle D$ = $\angle EDF$ = 144° (Vertically opposite angles)

$△DEF$ is an isosceles triangle and DE = DF (Given)
So, $\angle E$ = $\angle F$ (Base angles of an isoceles triangle are equal)

$\angle EDF$ + $\angle E$ + $\angle F$ = 180° (Angle sum property of a triangle).
$\Rightarrow$ 144° + x + x = 180°
$\Rightarrow$ 2x = 180° - 144° = 36°
$\Rightarrow$ x = $\frac{36°}{2}$ = 18°.

Hence, x = 18°.