Math Problem Statement

Figure shows the sector, AOB of a circle with centre O and radius 8 cm. A circle of radius 2cm touches the lines OA and OB and the arc AB. Angle AOB is 2θ radians, 0 < θ < π 4 . (a) Find, to 4 significant figures, the value of θ

[3]

(b) Find, to 3 significant figures, the area of the region Shaded in Figure. [3]

Solution

Let's break down the problem and solve it step by step.

Part (a) Finding the value of θ\theta

Given:

  • Radius of the large circle R=8cmR = 8 \, \text{cm}
  • Radius of the smaller circle r=2cmr = 2 \, \text{cm}
  • Angle AOB=2θ\text{AOB} = 2\theta radians

The smaller circle is inscribed in the sector AOBAOB, touching the lines OAOA and OBOB, and the arc ABAB. The distance from the center OO to the center of the smaller circle is Rr=82=6cmR - r = 8 - 2 = 6 \, \text{cm}.

For the inscribed circle, the distance from OO to the center of the smaller circle (let's call it CC) can also be represented as the hypotenuse of a right triangle where:

  • The distance from OO to the tangent point on OAOA is OCcos(θ)OC\cos(\theta),
  • The distance from CC to the tangent point on ABAB is OCsin(θ)OC\sin(\theta).

Given OC=RrOC = R - r, we have: OC=6cmOC = 6 \, \text{cm}

The radius rr can also be expressed in terms of θ\theta using trigonometric relationships: r=R(1cos(θ))r = R(1 - \cos(\theta))

However, by focusing on the geometry: rOC=sin(θ)\frac{r}{OC} = \sin(\theta) sin(θ)=rOC=26=13\sin(\theta) = \frac{r}{OC} = \frac{2}{6} = \frac{1}{3} θ=arcsin(13)\theta = \arcsin\left(\frac{1}{3}\right)

Thus, 2θ2\theta is approximately: θarcsin(13)0.3398radians\theta \approx \arcsin\left(\frac{1}{3}\right) \approx 0.3398 \, \text{radians}

Part (b) Finding the area of the shaded region

The shaded region is the area of the sector AOBAOB minus the area of the inscribed circle.

  • Area of sector AOBAOB: Area of sector=12R2×2θ=R2θ=82×0.339821.7472cm2\text{Area of sector} = \frac{1}{2} R^2 \times 2\theta = R^2 \theta = 8^2 \times 0.3398 \approx 21.7472 \, \text{cm}^2

  • Area of the inscribed circle: Area of smaller circle=πr2=π×22=4π12.5664cm2\text{Area of smaller circle} = \pi r^2 = \pi \times 2^2 = 4\pi \approx 12.5664 \, \text{cm}^2

  • Area of the shaded region: Shaded area=Area of sectorArea of inscribed circle21.747212.5664=9.1808cm2\text{Shaded area} = \text{Area of sector} - \text{Area of inscribed circle} \approx 21.7472 - 12.5664 = 9.1808 \, \text{cm}^2

To 3 significant figures, the area of the shaded region is: Shaded area9.18cm2\text{Shaded area} \approx 9.18 \, \text{cm}^2

Summary:

  • (a) The value of θ\theta is approximately 0.33980.3398 radians (to 4 significant figures).
  • (b) The area of the shaded region is approximately 9.18cm29.18 \, \text{cm}^2 (to 3 significant figures).

Would you like to see more details on any part of this solution?


Here are some additional related questions for you:

  1. How would the area of the shaded region change if the radius of the smaller circle were increased?
  2. What would the area of the shaded region be if the angle 2θ2\theta were doubled?
  3. How would the value of θ\theta change if the radius of the smaller circle were halved?
  4. Can you derive a general formula for the area of the shaded region in terms of RR, rr, and θ\theta?
  5. What is the relationship between the angle θ\theta and the radius of the smaller circle?
  6. How does the area of the sector change if θ\theta is increased by 10%10\%?
  7. What is the perimeter of the shaded region?
  8. Can you find the length of the arc ABAB for the given value of θ\theta?

Tip: When dealing with problems involving sectors and circles, understanding trigonometric relationships is key to finding distances and areas effectively.

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Math Problem Analysis

Mathematical Concepts

Circle Geometry
Trigonometry
Sector Area

Formulas

Area of a sector
Area of a circle

Theorems

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Suitable Grade Level

High School