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Surface Areas and Volumes - Volume of a Right Circular Cone

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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A right circular cone is a three-dimensional solid figure generated by rotating a right-angled triangle around one of its legs. Visually, it consists of a flat circular base and a curved surface that narrows smoothly from the base to a point called the apex or vertex.

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The volume of a cone is defined as the total space occupied by the cone in three dimensions. Visually, if you compare a cone and a cylinder with the same base radius (rr) and vertical height (hh), the cone occupies exactly one-third of the space of the cylinder.

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The core dimensions of a cone are the radius (rr) of its circular base, the vertical height (hh) which is the perpendicular distance from the center of the base to the apex, and the slant height (ll) which is the distance from the apex to any point on the edge of the base.

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The vertical height (hh), radius (rr), and slant height (ll) form a right-angled triangle inside the cone. According to the Pythagoras Theorem, the relationship is l2=r2+h2l^2 = r^2 + h^2. This allows us to calculate the height if only the slant height and radius are provided.

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The volume is calculated using the area of the base. Since the base is a circle with area Ο€r2\pi r^2, the volume is expressed as 13Γ—(AreaΒ ofΒ Base)Γ—Height\frac{1}{3} \times (\text{Area of Base}) \times \text{Height}.

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In practical problems, 'capacity' refers to the volume of a substance (like liquid or grain) that a conical container can hold. Capacity is measured in cubic units like cm3cm^3 or m3m^3, often converted to liters using the relation 1000cm3=11000 cm^3 = 1 liter.

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When solving for volume, always ensure that the radius (rr) and height (hh) are in the same units. If the slant height is given instead of the vertical height, you must first solve for hh using h=l2βˆ’r2h = \sqrt{l^2 - r^2}.

πŸ“Formulae

Volume of a Right Circular Cone: V=13Ο€r2hV = \frac{1}{3} \pi r^2 h

Slant Height formula: l=r2+h2l = \sqrt{r^2 + h^2}

Vertical Height in terms of Slant Height: h=l2βˆ’r2h = \sqrt{l^2 - r^2}

Radius in terms of Volume and Height: r=3VΟ€hr = \sqrt{\frac{3V}{\pi h}}

πŸ’‘Examples

Problem 1:

Find the volume of a right circular cone whose base radius is 66 cm and vertical height is 77 cm. (Take Ο€=227\pi = \frac{22}{7})

Solution:

  1. Identify the given values: Radius r=6r = 6 cm and Height h=7h = 7 cm.
  2. Apply the volume formula: V=13Ο€r2hV = \frac{1}{3} \pi r^2 h
  3. Substitute the values: V=13Γ—227Γ—(6)2imes7V = \frac{1}{3} \times \frac{22}{7} \times (6)^2 imes 7
  4. Simplify the expression: V=13Γ—227Γ—36Γ—7V = \frac{1}{3} \times \frac{22}{7} \times 36 \times 7
  5. Cancel 77 from numerator and denominator and divide 3636 by 33: V=22Γ—12V = 22 \times 12
  6. Final calculation: V=264cm3V = 264 cm^3

Explanation:

This is a direct application of the volume formula where the base radius and vertical height are provided. We substitute the values into the formula V=13Ο€r2hV = \frac{1}{3} \pi r^2 h and simplify to find the space occupied by the cone.

Problem 2:

A conical pit has a radius of 77 m and a slant height of 2525 m. Calculate its capacity in kiloliters.

Solution:

  1. Identify given values: r=7r = 7 m, l=25l = 25 m.
  2. We need vertical height hh to find the volume. Use h=l2βˆ’r2h = \sqrt{l^2 - r^2}.
  3. h=252βˆ’72=625βˆ’49=576=24h = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 m.
  4. Calculate Volume: V=13Γ—227Γ—72Γ—24V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24
  5. V=13Γ—227Γ—49Γ—24V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24
  6. V=22Γ—7Γ—8=1232m3V = 22 \times 7 \times 8 = 1232 m^3.
  7. Since 1m3=11 m^3 = 1 kiloliter, Capacity = 12321232 kl.

Explanation:

In this problem, the slant height (ll) is given instead of the vertical height (hh). We first use the Pythagorean relationship to find hh. Once hh is found, we calculate the volume in cubic meters and convert it to kiloliters.