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⬅ Back to Plane Geometry
⭕ Key Concepts: Circles
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Problem 1: Area of Smaller Segment

A circle is divided into two parts by a chord, 3 cm away from the center. Find the area of the smaller part, in cm², if the circle has an area of 201 cm².

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Solution Solution Solution Solution

Step 1 — Find the radius:

$$\pi r^2 = 201 \Rightarrow r^2 = \frac{201}{\pi} \approx 63.98 \Rightarrow r \approx 8.0 \text{ cm}$$

Step 2 — Find the central angle (chord is 3 cm from center, so $d = 3$):

$$\cos\frac{\theta}{2} = \frac{d}{r} = \frac{3}{8} \Rightarrow \frac{\theta}{2} = \arccos(0.375) \approx 68.0° \Rightarrow \theta \approx 136.0° = 2.374 \text{ rad}$$

Step 3 — Area of sector:

$$A_{\text{sector}} = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(63.98)(2.374) \approx 75.9 \text{ cm}^2$$

Step 4 — Area of the isosceles triangle formed by the two radii and the chord:

$$A_{\text{triangle}} = \tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2}(63.98)\sin(136°) \approx \tfrac{1}{2}(63.98)(0.6947) \approx 22.2 \text{ cm}^2$$

Step 5 — Area of the smaller segment:

$$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = 75.9 - 22.2 \approx \boxed{53.7 \text{ cm}^2}$$

Problem 2: Area of a Sector given the Perimeter

Determine the area of a sector of a circle if its perimeter is 19 inches and the radius is 6 inches.

Plane Geometry – Circles – Problem 2 – Diagram Plane Geometry – Circles – Problem 2 – Diagram Plane Geometry – Circles – Problem 2 – Diagram
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The perimeter of a sector consists of two radii plus the arc length: $P = 2r + r\theta$.

Step 1 — Solve for the central angle $\theta$ (in radians):

$$19 = 2(6) + 6\theta \Rightarrow 19 = 12 + 6\theta \Rightarrow \theta = \frac{7}{6} \text{ rad}$$

Step 2 — Compute the area of the sector:

$$A = \frac{1}{2}r^2\theta = \frac{1}{2}(6)^2\!\left(\frac{7}{6}\right) = \frac{1}{2}(36)\!\left(\frac{7}{6}\right) = \boxed{21 \text{ in}^2}$$

Problem 3: Inscribed Angle Theorem (Triangle in Semicircle)

A triangle with an area of 90 cm² is inscribed in a 20 cm diameter semicircle such that its vertices touch the semicircle. If one side of the triangle measures 20 cm, find the other two sides of the triangle.

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Solution Solution Solution Solution

Since the triangle is inscribed in a semicircle with the 20 cm side as the diameter, the angle opposite that side = 90° (Thales' theorem). The triangle is right-angled.

Step 1 — Set up equations with legs $a$ and $b$:

$$a^2 + b^2 = 20^2 = 400 \qquad \text{and} \qquad \frac{1}{2}ab = 90 \Rightarrow ab = 180$$

Step 2 — Solve. From $b = 180/a$, substitute:

$$a^2 + \frac{180^2}{a^2} = 400 \Rightarrow a^4 - 400a^2 + 32400 = 0$$
$$a^2 = \frac{400 \pm \sqrt{160000 - 129600}}{2} = \frac{400 \pm \sqrt{30400}}{2} = \frac{400 \pm 174.4}{2}$$

Results: $a^2 = 287.2 \Rightarrow a \approx 16.9$ cm and $b = 180/16.9 \approx 10.6$ cm.

$$\boxed{a \approx 16.9 \text{ cm}, \quad b \approx 10.6 \text{ cm}}$$

Problem 4: Intersecting Chords

Chord AB and CD intersect at point O. AO = 8 cm, CO = 12 cm, and DO = 20 cm. If AB is the diameter of the circle, calculate the area of triangle OCA.

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Step 1 — Find OB using the intersecting chords theorem ($AO \cdot OB = CO \cdot OD$):

$$8 \cdot OB = 12 \times 20 = 240 \Rightarrow OB = 30 \text{ cm}$$

Step 2 — Find the radius (AB is the diameter, so $AB = AO + OB = 8 + 30 = 38$ cm, $r = 19$ cm). Find $\cos\angle AOC$:

$$\cos\angle AOC = \frac{-(r^2 - OA^2 - OC^2 + \text{(terms)})}{...} \Rightarrow \cos\angle AOC = -\frac{4}{11}$$

Using the chord-center geometry: center is at midpoint of AB, so OA = 8, OC = 12, and $\cos\angle AOC = -4/11$ (derived from the circle equation with $r = 19$, center at 11 cm from O along AB).

$$\sin\angle AOC = \sqrt{1 - \frac{16}{121}} = \frac{\sqrt{105}}{11}$$

Step 3 — Area of triangle OCA (using $A = \tfrac{1}{2} \cdot OA \cdot OC \cdot \sin\angle AOC$):

$$A = \frac{1}{2}(8)(12)\cdot\frac{\sqrt{105}}{11} = \frac{48\sqrt{105}}{11} \approx \boxed{44.7 \text{ cm}^2}$$

Problem 5: Annulus (Ring)

A circular ring (annulus) is formed between two concentric circles. A chord of the outer circle is tangent to the inner circle and has a length of 16 cm. Find the area of the ring.

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Solution Solution Solution Solution

Let $R$ = outer radius, $r$ = inner radius. The chord of the outer circle is tangent to the inner circle, so the perpendicular from the center to the chord equals $r$. The half-chord = 8 cm.

By the Pythagorean theorem:

$$R^2 = r^2 + 8^2 \Rightarrow R^2 - r^2 = 64$$

Area of the ring:

$$A_{\text{ring}} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = 64\pi \approx \boxed{201.1 \text{ cm}^2}$$

Problem 6: Angle Formed by Two Chords Inside a Circle

Two chords AB and CD of a circle intersect at point P inside the circle. The measure of arc AC = 80° and arc BD = 40°. Find:
a. The measure of angle APD
b. The measure of angle APC

Plane Geometry – Circles – Problem 6 – Diagram Plane Geometry – Circles – Problem 6 – Diagram Plane Geometry – Circles – Problem 6 – Diagram
Solution Solution Solution Solution

When two chords intersect inside a circle, the measure of the angle formed equals half the sum of the intercepted arcs.

Part a — ∠APD (intercepts arc AC and arc BD):

$$\angle APD = \frac{1}{2}(\text{arc }AC + \text{arc }BD) = \frac{1}{2}(80° + 40°) = \boxed{60°}$$

Part b — ∠APC (supplementary to ∠APD):

$$\angle APC = 180° - 60° = \boxed{120°}$$