Math Problem Statement

OPAQ is a sector of a circle with centre 0 and radius cm. The angle at the centre is 0. The sector is then formed into a cone by joining the two radii, OP and OQ, together. The radius of the base of the cone is

r/3 cm. The cone has a height of 6 cm and a volume of 50π cm'. Find the value of 0.

Solution

We are given that OPAQ is a sector of a circle with radius rr cm, and the sector is formed into a cone by joining the two radii. The radius of the base of the cone is r3\frac{r}{3} cm, the height of the cone is 6 cm, and the volume of the cone is 50π50\pi cm3^3.

We are asked to find the angle θ\theta at the center of the sector.

Step 1: Volume of the Cone

The formula for the volume VV of a cone is:

V=13πrb2hV = \frac{1}{3} \pi r_b^2 h

Where:

  • rbr_b is the radius of the base of the cone,
  • hh is the height of the cone.

Substitute the given values:

  • V=50πV = 50\pi cm3^3,
  • rb=r3r_b = \frac{r}{3},
  • h=6h = 6 cm.

50π=13π(r3)2(6)50\pi = \frac{1}{3} \pi \left( \frac{r}{3} \right)^2 (6)

Simplify this equation: 50π=13πr29650\pi = \frac{1}{3} \pi \cdot \frac{r^2}{9} \cdot 6

50π=2πr2950\pi = \frac{2\pi r^2}{9}

Cancel π\pi from both sides: 50=2r2950 = \frac{2r^2}{9}

Multiply both sides by 9: 450=2r2450 = 2r^2

Divide by 2: r2=225r^2 = 225

So: r=15cmr = 15 \, \text{cm}

Step 2: Relation Between the Sector and the Cone

The arc length of the sector forms the circumference of the base of the cone when the sector is rolled into a cone. The formula for the arc length LL of a sector is:

L=rθL = r \theta

And the circumference CC of the base of the cone is:

C=2πrbC = 2\pi r_b

Substitute rb=r3r_b = \frac{r}{3}: C=2π×r3=2πr3C = 2\pi \times \frac{r}{3} = \frac{2\pi r}{3}

Since the arc length LL of the sector equals the circumference of the base of the cone, we have:

rθ=2πr3r \theta = \frac{2\pi r}{3}

Cancel rr from both sides: θ=2π3\theta = \frac{2\pi}{3}

Thus, the angle at the center of the sector is:

2π3radians.\boxed{\frac{2\pi}{3}} \, \text{radians}.

Final Answer:

The angle θ\theta is 2π3\frac{2\pi}{3} radians.


Would you like more details or have any questions?

Here are 5 related questions for practice:

  1. What is the surface area of the cone in this problem?
  2. How would the volume change if the radius of the cone's base doubled?
  3. How does the height of the cone affect its volume?
  4. What would happen to the angle θ\theta if the radius rbr_b of the base was halved?
  5. How can we derive the formula for the lateral surface area of the cone from the sector?

Tip: Always double-check units and simplify calculations step by step for accuracy in geometry problems.

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Math Problem Analysis

Mathematical Concepts

Geometry
Volume of a Cone
Arc Length
Sector of a Circle

Formulas

Volume of a Cone: V = (1/3)πr²h
Arc Length: L = rθ
Circumference of a Circle: C = 2πr

Theorems

Relation between sector and cone's base (arc length equals circumference)

Suitable Grade Level

Grades 10-12