In the figure, ABCDE and AEPQ are a regular pentagon and a square respectively. What is the measure of ∠AQB? (ABQ is isosceles)

9°

We know, measure of each interior angle of a regular pentagon = (n - 2)/n × 180°

Where n is the number of sides.

Here, n = 5

So, each interior angle = (5 - 2)/5 × 180°

= 3/5 × 180°

= 3 × 36

= 108°

So, ∠BAE = 108°

Each angle of square = 90°

We know, ∠QAB = 360° - (∠EAQ + ∠BAE)

∠QAB = 360° - (90° + 108°)

= 360° - 198°

= 162°

Considering triangle ABQ,

By angle sum property of a triangle,

∠QAB + ∠ABQ + ∠AQB = 180°

162° + ∠ABQ + ∠AQB = 180°

∠ABQ + ∠AQB = 180° - 162°

∠ABQ + ∠AQB = 18°

ABQ is an isosceles triangle, so AQ = AB

We know that the angles opposite to the equal sides are equal.

So, ∠ABQ = ∠AQB

Now, ∠ABQ + ∠AQB = 18°

∠ABQ + ∠AQB = 18°

2∠ABQ = 18°

Therefore, ∠ABQ = 9°

The correct answer is Option (1) → 9°