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Mensuration - Area and volume of Cylinder, Cone and Sphere

Grade 10ICSE

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

🔑Concepts

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A Right Circular Cylinder is a solid generated by the revolution of a rectangle about one of its sides. Visually, it consists of two congruent, parallel circular bases and a curved lateral surface. Imagine a stack of circular coins; the distance between the bases is the height (hh) and the radius of the circular face is rr. If you 'unroll' the curved surface, it forms a rectangle with length equal to the circumference 2pir2\\pi r and breadth equal to the height hh.

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A Right Circular Cone is a solid generated by rotating a right-angled triangle around one of its legs. It features a single circular base and a point called the apex. The vertical height (hh) is the perpendicular distance from the apex to the center of the base, while the slant height (ll) is the distance from the apex to any point on the edge of the circular base. Visually, it looks like a party hat or an ice cream cone.

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The Slant Height (ll) of a cone is an essential dimension for calculating surface area. Because the vertical height (hh), the radius (rr), and the slant height (ll) form a right-angled triangle within the cone, they are related by the Pythagorean theorem: l2=r2+h2l^2 = r^2 + h^2. Visually, ll is always the longest side (hypotenuse) of this internal triangle.

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A Sphere is a perfectly symmetrical three-dimensional shape where every point on the surface is equidistant (rr) from the center. Imagine rotating a circle 180circ180^{\\circ} around its diameter to create this 'ball' shape. A Hemisphere is exactly one-half of a sphere, created by cutting the sphere through its center. Visually, a hemisphere has two surfaces: a curved top and a flat circular base (the cross-section).

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The concept of Melting and Recasting involves transforming one solid shape into another (e.g., melting a sphere to make multiple small cones). In these problems, the most important rule is that the total volume remains constant during the transition. Therefore, textVolumeoforiginalsolid=textTotalvolumeofnewsolids\\text{Volume of original solid} = \\text{Total volume of new solids}.

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Composite Solids are objects made by combining two or more basic shapes, such as a cylinder with hemispherical ends or a cone mounted on a hemisphere. To find the Total Surface Area (TSA) of a composite solid, you must sum only the visible outer surfaces. Visually, the faces where the shapes are joined are 'hidden' inside the solid and are not included in the surface area calculation.

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A Hollow Cylinder (or Pipe) is the region bounded between two concentric cylinders of the same height. It has an external radius (RR) and an internal radius (rr). The thickness of the material is given by R−rR - r. Visually, it looks like a ring-shaped cross-section extended through a height hh.

📐Formulae

Cylinder Volume: V=pir2hV = \\pi r^2 h

Cylinder Curved Surface Area (CSA): CSA=2pirhCSA = 2\\pi rh

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

Cone Slant Height: l=sqrtr2+h2l = \\sqrt{r^2 + h^2}

Cone Volume: V=frac13pir2hV = \\frac{1}{3}\\pi r^2 h

Cone Curved Surface Area (CSA): CSA=pirlCSA = \\pi rl

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

Sphere Surface Area: A=4pir2A = 4\\pi r^2

Sphere Volume: V=frac43pir3V = \\frac{4}{3}\\pi r^3

Hemisphere Curved Surface Area (CSA): CSA=2pir2CSA = 2\\pi r^2

Hemisphere Total Surface Area (TSA): TSA=3pir2TSA = 3\\pi r^2

Hemisphere Volume: V=frac23pir3V = \\frac{2}{3}\\pi r^3

Hollow Cylinder Volume: V=pi(R2−r2)hV = \\pi(R^2 - r^2)h

💡Examples

Problem 1:

A metallic sphere of radius 10.5textcm10.5\\text{ cm} is melted and then recast into small cones, each of radius 3.5textcm3.5\\text{ cm} and height 3textcm3\\text{ cm}. Find the number of cones formed.

Solution:

  1. Volume of the sphere = frac43piR3=frac43timespitimes(10.5)3\\frac{4}{3} \\pi R^3 = \\frac{4}{3} \\times \\pi \\times (10.5)^3\n2. Volume of one small cone = frac13pir2h=frac13timespitimes(3.5)2times3=pitimes(3.5)2\\frac{1}{3} \\pi r^2 h = \\frac{1}{3} \\times \\pi \\times (3.5)^2 \\times 3 = \\pi \\times (3.5)^2\n3. Let the number of cones be nn. By the principle of conservation of volume: ntimestextVolumeofonecone=textVolumeofspheren \\times \\text{Volume of one cone} = \\text{Volume of sphere}\n4. ntimes(pitimes3.5times3.5)=frac43timespitimes10.5times10.5times10.5n \\times (\\pi \\times 3.5 \\times 3.5) = \\frac{4}{3} \\times \\pi \\times 10.5 \\times 10.5 \\times 10.5\n5. n=frac4times10.5times10.5times10.53times3.5times3.5n = \\frac{4 \\times 10.5 \\times 10.5 \\times 10.5}{3 \\times 3.5 \\times 3.5}\n6. n=4times3times3times3.5=126n = 4 \\times 3 \\times 3 \\times 3.5 = 126

Explanation:

When a solid is melted and recast, the total volume remains the same. We calculate the volume of the large sphere and divide it by the volume of a single small cone to find the total number of cones.

Problem 2:

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19textcm19\\text{ cm} and the diameter of the cylinder is 7textcm7\\text{ cm}. Find the total surface area and the volume of the solid.

Solution:

  1. Radius of cylinder and hemispheres r=frac72=3.5textcmr = \\frac{7}{2} = 3.5\\text{ cm}\n2. Height of the cylinder H=textTotalheight−2timesr=19−(3.5+3.5)=12textcmH = \\text{Total height} - 2 \\times r = 19 - (3.5 + 3.5) = 12\\text{ cm}\n3. Volume = Volume of cylinder + 2times2 \\times Volume of hemisphere\nV=pir2H+2timesfrac23pir3=pir2(H+frac43r)V = \\pi r^2 H + 2 \\times \\frac{2}{3}\\pi r^3 = \\pi r^2 (H + \\frac{4}{3}r)\nV=frac227times(3.5)2times(12+frac43times3.5)=38.5times(12+4.67)approx641.67textcm3V = \\frac{22}{7} \\times (3.5)^2 \\times (12 + \\frac{4}{3} \\times 3.5) = 38.5 \\times (12 + 4.67) \\approx 641.67\\text{ cm}^3\n4. Total Surface Area (TSA) = CSA of cylinder + 2times2 \\times CSA of hemisphere\nTSA=2pirH+4pir2=2pir(H+2r)TSA = 2\\pi rH + 4\\pi r^2 = 2\\pi r(H + 2r)\nTSA=2timesfrac227times3.5times(12+7)=22times19=418textcm2TSA = 2 \\times \\frac{22}{7} \\times 3.5 \\times (12 + 7) = 22 \\times 19 = 418\\text{ cm}^2

Explanation:

For composite solids, the height of the central cylinder is found by subtracting the radii of the two hemispherical ends from the total height. The surface area includes only the curved parts because the flat circular faces are joined together internally.