Question

o is the center of the regular nonagon below. find its area. round to the nearest tenth if necessary.
12
answer attempt 1 out of 2
$a = \square$ square units submit answer

Explanation:

Step1: Find central angle per triangle

A regular nonagon has 9 equal triangular slices from the center. The total central angle is $360^\circ$, so each central angle is:
$\theta = \frac{360^\circ}{9} = 40^\circ$

Step2: Find base of one triangle

The apothem (distance from center to side) is 12, which is the height of the isosceles triangle. Split the triangle into two right triangles, so half the central angle is $20^\circ$. Let half the base be $x$:
$\tan(20^\circ) = \frac{x}{12}$
$x = 12\tan(20^\circ)$
Full base $s = 2x = 24\tan(20^\circ)$

Step3: Area of one triangular slice

Area of one triangle is $\frac{1}{2} \times \text{base} \times \text{apothem}$:
$A_{\text{triangle}} = \frac{1}{2} \times 24\tan(20^\circ) \times 12 = 144\tan(20^\circ)$

Step4: Total area of nonagon

Multiply by 9 triangles:
$A = 9 \times 144\tan(20^\circ) = 1296\tan(20^\circ)$
Calculate $\tan(20^\circ) \approx 0.3640$:
$A \approx 1296 \times 0.3640 \approx 471.7$

Answer:

$471.7$ square units