Honors Geometry Unit 11 Review: Areas of Polygons and Circles

A regular polygon has all sides congruent and all angles congruent, so you can split it into $n$ congruent isosceles triangles that meet at the center. Each triangle has a base equal to one side $s$ and a height equal to the apothem $a$, the perpendicular distance from the center to a side. Adding all $n$ triangles gives $A = \frac{1}{2}ap$, where $p = ns$ is the perimeter.

If you only know the side length, find the apothem with trig. The central angle is $\frac{360^\circ}{n}$, and the apothem cuts it in half. That makes a right triangle with legs $a$ and $\frac{s}{2}$, so $\tan\left(\frac{180^\circ}{n}\right) = \frac{s/2}{a}$.

Worked example. Find the area of a regular hexagon with side length 6. Half the central angle is $\frac{180^\circ}{6} = 30^\circ$, so $a = \frac{3}{\tan 30^\circ} = 3\sqrt{3} \approx 5.20$. The perimeter is $6 \cdot 6 = 36$. Then $A = \frac{1}{2}(3\sqrt{3})(36) = 54\sqrt{3} \approx 93.5$ square units.

Composite figures have no single formula. Break the figure into shapes you know, add the pieces that are attached, and subtract the pieces that are cut out. A 10 by 6 rectangle with a semicircle of radius 3 cut out of one short side has area $60 - \frac{1}{2}\pi(3)^2 = 60 - 4.5\pi \approx 45.9$ square units.

Every circle has the same ratio of circumference to diameter, and that ratio is $\pi \approx 3.14159$. That gives you circumference, the distance around the circle, as $C = \pi d$ or $C = 2\pi r$. Area is $A = \pi r^2$. Only $r$ gets squared, not $\pi$.

Leave answers in terms of $\pi$ unless the problem asks you to round. A circle with radius 7 has $C = 14\pi$ and $A = 49\pi$, and those exact forms are easier to check than 43.98 and 153.94.

You also need to work backward. From circumference, $r = \frac{C}{2\pi}$. From area, $r = \sqrt{\frac{A}{\pi}}$.

Worked example. A circular track has circumference $20\pi$ meters. Find the area it encloses. First, $r = \frac{20\pi}{2\pi} = 10$ m. Then $A = \pi(10)^2 = 100\pi \approx 314.2$ square meters.

Watch how scaling works. Doubling the radius doubles the circumference but multiplies the area by 4, because area depends on $r^2$. For partial circles, perimeter includes the straight edges. A semicircle with diameter 4 has perimeter $\frac{1}{2}(4\pi) + 4 = 2\pi + 4 \approx 10.28$, not just $2\pi$.

Arc length and sector area both start with one question. What fraction of the full $360^\circ$ does the central angle cover? Multiply that fraction by the circumference to get arc length, or by the area to get sector area.

$\text{arc length} = \frac{\theta}{360^\circ} \cdot 2\pi r \qquad \text{sector area} = \frac{\theta}{360^\circ} \cdot \pi r^2$

If the angle is in radians, these simplify to $s = r\theta$ and $A = \frac{1}{2}r^2\theta$.

Arc measure and arc length are different things. Arc measure equals the measure of the matching central angle and can be expressed in degrees or radians. Arc length is a distance in units like cm. Two arcs can both measure $60^\circ$ and have different lengths if the circles have different radii.

Worked example. In a circle with radius 9 cm and a central angle of $80^\circ$, the fraction is $\frac{80}{360} = \frac{2}{9}$. Arc length is $\frac{2}{9} \cdot 18\pi = 4\pi \approx 12.57$ cm. Sector area is $\frac{2}{9} \cdot 81\pi = 18\pi \approx 56.55$ cm².

A segment is the region between a chord and its arc. For a minor segment, subtract the triangle formed by the radii and chord from the matching sector. For the major segment, subtract the minor segment from the full circle. For $r = 6$ and $\theta = 60^\circ$, the sector is $\frac{1}{6} \cdot 36\pi = 6\pi$ and the triangle is $\frac{1}{2}(6)(6)\sin 60^\circ = 9\sqrt{3}$. The segment is $6\pi - 9\sqrt{3} \approx 3.26$ square units.

Several area formulas pair a base with a perpendicular height. For a triangle, $A = \frac{1}{2}bh$, and any side can be the base as long as the height is perpendicular to it. In an obtuse triangle, the height to a short side can land outside the triangle, on an extension of the base.

A parallelogram has $A = bh$, because you can slide the slanted end over to form a rectangle. The slanted side is not the height. A trapezoid has $A = \frac{1}{2}(b_1 + b_2)h$, which is the average of the two parallel sides times the height. Rhombuses and kites have perpendicular diagonals, so $A = \frac{1}{2}d_1d_2$.

When the height is missing, honors problems give you other tools. With two sides and the included angle, $A = \frac{1}{2}ab\sin C$. With all three sides, Heron's formula gives $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is half the perimeter.

Worked example. An isosceles trapezoid has bases 8 and 12 and legs of 5. The bases differ by 4, so each end has a right triangle with horizontal leg 2 and hypotenuse 5. That gives $h = \sqrt{5^2 - 2^2} = \sqrt{21} \approx 4.58$. Then $A = \frac{1}{2}(8 + 12)\sqrt{21} = 10\sqrt{21} \approx 45.8$ square units.