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Mensuration: Surface Area and Volume - Compute curved and total surface area and volume of right circular cones

Grade 9CBSE

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

🔑Concepts

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A right circular cone is a solid generated by the rotation of a right-angled triangle about one of its sides (other than the hypotenuse) which remains fixed. It consists of a circular base and a curved surface tapering to a point called the vertex.

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The Slant Height (ll) of a cone is the distance from the vertex to any point on the edge of the circular base. It forms the hypotenuse of a right-angled triangle where the legs are the vertical height (hh) and the radius (rr). The relationship is given by l2=r2+h2l^2 = r^2 + h^2.

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The Curved Surface Area (CSA) of a cone refers to the area of the lateral surface excluding the base. It is calculated using the formula CSA=πrlCSA = \pi r l. If the slant height is not given, it must be calculated first using the radius and height.

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The Total Surface Area (TSA) of a cone is the sum of its curved surface area and the area of its circular base (πr2\pi r^2). Thus, TSA=πrl+πr2=πr(l+r)TSA = \pi r l + \pi r^2 = \pi r(l + r).

Net of a cone showing the sector for curved surface and the circle for base.
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The Volume of a cone represents its capacity. It is exactly one-third of the volume of a cylinder with the same radius and height: V=13πr2hV = \frac{1}{3} \pi r^2 h. Note that volume depends on the vertical height (hh), not the slant height (ll).

📐Formulae

Volume of a Right Circular Cone: V=13πr2hV = \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=l2−r2h = \sqrt{l^2 - r^2}

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

💡Examples

Problem 1:

Find the volume of a right circular cone whose base radius is 66 cm and vertical height is 77 cm. (Take π=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=13πr2hV = \frac{1}{3} \pi r^2 h
  3. Substitute the values: V=13×227×(6)2×7V = \frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 7
  4. Simplify the expression: V=13×227×36×7V = \frac{1}{3} \times \frac{22}{7} \times 36 \times 7
  5. Cancel 77 from numerator and denominator and divide 3636 by 33: V=22×12V = 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=13πr2hV = \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=l2−r2h = \sqrt{l^2 - r^2}.
  3. h=252−72=625−49=576=24h = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 m.
  4. Calculate Volume: V=13×227×72×24V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24
  5. V=13×227×49×24V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24
  6. V=22×7×8=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.

Problem 3:

A joker's cap is in the form of a right circular cone of base radius 77 cm and height 2424 cm. Find the area of the sheet required to make 1010 such caps.

Conical cap with height 24cm and radius 7cm.

Solution:

Given: Radius r=7r = 7 cm Height h=24h = 24 cm

Step 1: Find the slant height (ll) l=r2+h2l = \sqrt{r^2 + h^2} l=72+242l = \sqrt{7^2 + 24^2} l=49+576l = \sqrt{49 + 576} $$l = \sqrt{625} = 25$ cm

Step 2: Find the Curved Surface Area (CSA) of one cap CSA=πrlCSA = \pi r l CSA=227×7×25CSA = \frac{22}{7} \times 7 \times 25 CSA=22×25=550 cm2CSA = 22 \times 25 = 550 \text{ cm}^2

Step 3: Find the area for 1010 caps Total Area=10×550=5500 cm2\text{Total Area} = 10 \times 550 = 5500 \text{ cm}^2

The total area of the sheet required is 5500 cm25500 \text{ cm}^2.

Explanation:

To make a cap, only the lateral (curved) surface is needed as the base remains open. We use the Pythagorean theorem to find the slant height because the formula for CSA requires ll, not hh.

Problem 4:

The volume of a right circular cone is 9856 cm39856 \text{ cm}^3. If the diameter of the base is 2828 cm, find the height and the slant height of the cone.

Cone with diameter 28 and unknown height h.

Solution:

Given: Volume V=9856 cm3V = 9856 \text{ cm}^3 Diameter d=28d = 28 cm Radius r=282=14r = \frac{28}{2} = 14 cm

Step 1: Find the height (hh) V=13πr2hV = \frac{1}{3} \pi r^2 h 9856=13×227×14×14×h9856 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h 9856=13×22×2×14×h9856 = \frac{1}{3} \times 22 \times 2 \times 14 \times h 9856=6163×h9856 = \frac{616}{3} \times h h=9856×3616h = \frac{9856 \times 3}{616} h=16×3=48 cmh = 16 \times 3 = 48 \text{ cm}

Step 2: Find the slant height (ll) l=r2+h2l = \sqrt{r^2 + h^2} l=142+482l = \sqrt{14^2 + 48^2} l=196+2304l = \sqrt{196 + 2304} l=2500=50 cml = \sqrt{2500} = 50 \text{ cm}

The height is 4848 cm and the slant height is 5050 cm.

Explanation:

First, the radius is derived from the diameter. Then, the volume formula is rearranged to solve for the unknown height hh. Finally, hh and rr are used to find ll using the relation between the dimensions of a cone.