2.5. Problem-Solving Techniques

Example 1: A circle has a radius of 7 cm. Find the area and circumference. Solution: $\text{Area} = \pi \times 7^2 = 154 \text{ cm}^2 \\ \text{Circumference} = 2\pi \times 7 = 44 \text{ cm}$

In this example, we learn to calculate the area and circumference of a circle given its radius. First, the area is calculated using the formula A = πr². Substituting the value of the radius (7 cm), we calculate the area: A = π × 7² = π × 49 = 154 cm² (using π ≈ 3.14). Next, we find the circumference using the formula C = 2πr. For a radius of 7 cm, we calculate C = 2 × π × 7 = 44 cm (again using π ≈ 3.14). These formulas provide useful measurements for the circle.

Think of a hula hoop. If you know how far it reaches (the radius), you can easily find out how big the area of the circle it makes is—like the space covered on the ground when you lay it down. You can also determine how far you’d need to walk just to go once around the hula hoop! That’s the circumference.

Example 2: An arc subtends an angle of 60° at the center of a circle with radius 10 cm. Find the arc length. Solution: 60∘ \frac{1}{6} 20𝜋 Arc length = \times2\pi\times10 = \times20\pi = ≈ 10.47 cm 360∘ 6 6

To find the length of an arc, we can use the formula for arc length, which relies on the circle's radius and the angle subtended by the arc. Here, the radius is 10 cm and the arc subtends an angle of 60°. We start by stating that the arc length is the fraction of the circumference defined by the angle. The full circle is 360°, so the fraction becomes 60°/360°, which simplifies to 1/6. The circumference of the circle is calculated first as C = 2πr = 20π cm. Then, we find the arc length using the fraction: Arc Length = (1/6) × 20π ≈ 10.47 cm.

Imagine slicing a pizza! If the entire pizza is a circle and you take one slice out, that slice's crust can be likened to the arc. If that slice covers a smaller angle, like in this case with 60°, you can think of it as taking a smaller piece out from the whole pizza.

Example 3: Prove that tangents drawn from an external point to a circle are equal. Solution: Let the external point be 𝑃, and tangents touch the circle at points 𝐴 and 𝐵. Join 𝑂𝑃, 𝑂𝐴, 𝑂𝐵, 𝑃𝐴, and 𝑃𝐵. Using congruent triangles: • 𝑂𝐴 = 𝑂𝐵 = radius • 𝑂𝑃 is common • ∠𝑂𝐴𝑃 = ∠𝑂𝐵𝑃 = 90∘ (radius ⊥ tangent) Thus, △OAP ≅ △OBP ⇒ 𝑃𝐴 = 𝑃𝐵

In this example, we are proving that the lengths of the tangents from an external point to a circle are equal. We label the external point P and the points where the tangents touch the circle as A and B. By joining certain lines, we form two triangles, OAP and OBP. Since both triangles share side OP, the lengths OA and OB (both being radii of the circle) are equal. Also, the angles at points A and B formed with the tangents are right angles. This means we can conclude that triangle OAP is congruent to triangle OBP, indicating PA = PB. Thus, the lengths of tangents from point P are equal.

Consider two ropes being pulled tight from a point outside a circular running track to where they touch the track. If you have two runners at the points where the ropes touch the ground, the distance they run away from the point will be the same, even though they head in different directions at that moment.

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