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Mensuration - Combination of solids

Grade 10ICSE

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

🔑Concepts

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Combination of Solids: This involves shapes created by joining two or more basic three-dimensional solids like cylinders, cones, spheres, and hemispheres. When calculating properties, visualize the new object as a single entity where the common faces between the shapes are hidden inside the solid and do not contribute to the external surface area.

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Total Surface Area (TSA) of Combined Solids: Unlike volume, the surface area of a combined solid is the sum of only the visible surfaces. For example, if a cone is placed on a hemisphere of the same radius, the surface area of the resulting toy is the Curved Surface Area (CSA) of the cone plus the CSA of the hemisphere, because the circular bases of both shapes are pressed together and hidden.

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Volume of Combined Solids: The volume of a solid formed by joining two or more solids is always the sum of the volumes of the individual components. For instance, a capsule-shaped solid consists of a central cylinder with two hemispherical ends; its total volume is the volume of the cylinder plus the volume of two hemispheres (or one full sphere).

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Melting and Recasting: When a solid is melted and recast into another shape (or multiple smaller shapes), the total volume of the material remains constant. This principle is expressed as Voriginal=VnewV_{original} = V_{new}. If nn small identical objects are formed from a larger one, then Vlarge=n×VsmallV_{large} = n \times V_{small}.

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Concept of Slant Height (ll): In combined solids involving cones, the slant height is crucial for surface area calculations. It represents the distance from the apex to any point on the base circumference. In a visual cross-section, the height (hh), radius (rr), and slant height (ll) form a right-angled triangle where l=r2+h2l = \sqrt{r^2 + h^2}.

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Hollow Solids and Cavities: If a solid has a hole or a cavity (like a cylinder with a conical hole drilled out), the total surface area includes the exterior surface plus the surface area of the internal cavity. The volume, however, is calculated by subtracting the volume of the cavity from the volume of the original solid.

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Internal and External Dimensions: For hollow objects like pipes or shells, the thickness is the difference between the external radius (RR) and internal radius (rr). Visually, this looks like two concentric circles in cross-section. The volume of the material used is given by the difference in volumes of the outer and inner shapes.

📐Formulae

Volume of Cylinder: V=πr2hV = \pi r^2 h

Curved Surface Area of Cylinder: CSA=2πrhCSA = 2\pi rh

Total Surface Area of Cylinder: TSA=2πr(h+r)TSA = 2\pi r(h + r)

Volume of Cone: V=13πr2hV = \frac{1}{3}\pi r^2 h

Curved Surface Area of Cone: CSA=πrlCSA = \pi rl where l=r2+h2l = \sqrt{r^2 + h^2}

Total Surface Area of Cone: TSA=πr(l+r)TSA = \pi r(l + r)

Volume of Sphere: V=43πr3V = \frac{4}{3}\pi r^3

Surface Area of Sphere: SA=4πr2SA = 4\pi r^2

Volume of Hemisphere: V=23πr3V = \frac{2}{3}\pi r^3

Curved Surface Area of Hemisphere: CSA=2πr2CSA = 2\pi r^2

Total Surface Area of Hemisphere: TSA=3πr2TSA = 3\pi r^2

Volume of material in a hollow cylinder: V=π(R2−r2)hV = \pi(R^2 - r^2)h

💡Examples

Problem 1:

A toy is in the form of a cone of radius 3.5 cm3.5\text{ cm} mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm15.5\text{ cm}. Find the total surface area of the toy. (Use π=227\pi = \frac{22}{7})

Solution:

  1. Radius of cone and hemisphere, r=3.5=72 cmr = 3.5 = \frac{7}{2}\text{ cm}.
  2. Total height of toy = 15.5 cm15.5\text{ cm}. Height of cone, h=15.5−3.5=12 cmh = 15.5 - 3.5 = 12\text{ cm}.
  3. Calculate slant height ll of the cone: l=r2+h2=(3.5)2+122=12.25+144=156.25=12.5 cml = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\text{ cm}.
  4. Surface Area of toy = CSA of cone + CSA of hemisphere.
  5. Surface Area = πrl+2πr2=πr(l+2r)\pi rl + 2\pi r^2 = \pi r(l + 2r).
  6. Substituting values: S=227×72×(12.5+2×3.5)=11×(12.5+7)=11×19.5=214.5 cm2S = \frac{22}{7} \times \frac{7}{2} \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5\text{ cm}^2.

Explanation:

To find the surface area of a combined solid, we only sum the areas of the visible surfaces. The base of the cone and the base of the hemisphere are joined, so they are not part of the external surface. We use the total height to find the vertical height of the cone first, then find the slant height for the CSA calculation.

Problem 2:

A solid metallic sphere of radius 10.5 cm10.5\text{ cm} is melted and recast into a number of smaller cones, each of radius 3.5 cm3.5\text{ cm} and height 3 cm3\text{ cm}. Find the number of cones formed.

Solution:

  1. Volume of the metallic sphere Vs=43πR3=43×π×(10.5)3V_s = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \pi \times (10.5)^3.
  2. Volume of one small cone Vc=13πr2h=13×π×(3.5)2×3V_c = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times (3.5)^2 \times 3.
  3. Let nn be the number of cones. By conservation of volume: n×Vc=Vsn \times V_c = V_s.
  4. n×(13π×3.5×3.5×3)=43π×10.5×10.5×10.5n \times (\frac{1}{3} \pi \times 3.5 \times 3.5 \times 3) = \frac{4}{3} \pi \times 10.5 \times 10.5 \times 10.5.
  5. Canceling π\pi and 13\frac{1}{3} from both sides: n×3.5×3.5×3=4×10.5×10.5×10.5n \times 3.5 \times 3.5 \times 3 = 4 \times 10.5 \times 10.5 \times 10.5.
  6. n=4×10.5×10.5×10.53.5×3.5×3n = \frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3}.
  7. Since 10.53.5=3\frac{10.5}{3.5} = 3, we get n=4×3×3×10.53=4×3×10.5=12×10.5=126n = \frac{4 \times 3 \times 3 \times 10.5}{3} = 4 \times 3 \times 10.5 = 12 \times 10.5 = 126.

Explanation:

When melting one solid to form others, the total volume remains the same. We set up an equation where the volume of the large sphere equals nn times the volume of a single small cone. It is usually easier to keep π\pi as a symbol and cancel it out later to simplify calculations.