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Mensuration: Surface Area and Volume - Compute surface area and volume of pyramids from given dimensions

Grade 9CBSE

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

🔑Concepts

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A pyramid is a three-dimensional solid with a polygonal base and triangular lateral faces that meet at a common vertex called the apex.

Diagram showing the parts of a pyramid including the apex and polygonal base.
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In a right square pyramid, the vertical height (hh) is the distance from the apex to the center of the base, while the slant height (ll) is the distance from the apex to the midpoint of a base edge.

A right triangle formed inside a pyramid by the height, slant height, and half the base side.
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The Volume of any pyramid is exactly one-third of the volume of a prism with the same base area and height.

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The Lateral Surface Area (LSA) of a regular pyramid consists of the sum of the areas of all congruent triangular lateral faces.

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Total Surface Area (TSA) is calculated by adding the area of the base to the Lateral Surface Area.

📐Formulae

Volume of Pyramid=13×Base Area×h\text{Volume of Pyramid} = \frac{1}{3} \times \text{Base Area} \times h

Lateral Surface Area (LSA)=12×Perimeter of Base×l\text{Lateral Surface Area (LSA)} = \frac{1}{2} \times \text{Perimeter of Base} \times l

Total Surface Area (TSA)=LSA+Base Area\text{Total Surface Area (TSA)} = \text{LSA} + \text{Base Area}

Slant Height (for square base side a):l=h2+(a2)2\text{Slant Height (for square base side } a\text{)}: l = \sqrt{h^2 + \left(\frac{a}{2}\right)^2}

💡Examples

Problem 1:

A right pyramid has a square base with a side of 10 cm10 \text{ cm} and a height of 12 cm12 \text{ cm}. Find its volume and total surface area.

Solution:

  1. Find the Volume: Base Area B=a2=102=100 cm2B = a^2 = 10^2 = 100 \text{ cm}^2. V=13×100×12=400 cm3V = \frac{1}{3} \times 100 \times 12 = 400 \text{ cm}^3.

  2. Find the Slant Height (ll): l=h2+(a2)2=122+52=144+25=169=13 cml = \sqrt{h^2 + (\frac{a}{2})^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}.

  3. Find the Lateral Surface Area (LSA): Perimeter P=4×10=40 cmP = 4 \times 10 = 40 \text{ cm}. LSA=12×40×13=260 cm2LSA = \frac{1}{2} \times 40 \times 13 = 260 \text{ cm}^2.

  4. Find the Total Surface Area (TSA): TSA=Base Area+LSATSA = \text{Base Area} + LSA. 100+260360\begin{array}{r} 100 \\ + 260 \\ \hline 360 \end{array} TSA=360 cm2TSA = 360 \text{ cm}^2.

Explanation:

We first calculate the base area to find the volume using the 13Bh\frac{1}{3}Bh formula. To find the surface area, we calculate the slant height using Pythagoras' theorem on the triangle formed by the height, slant height, and half the base side. Finally, we sum the LSA and base area.

Problem 2:

Find the volume of a pyramid with a rectangular base of dimensions 8 cm×6 cm8 \text{ cm} \times 6 \text{ cm} and a height of 15 cm15 \text{ cm}.

Solution:

  1. Calculate Base Area (BB): B=length×width=8×6=48 cm2B = \text{length} \times \text{width} = 8 \times 6 = 48 \text{ cm}^2.

  2. Calculate Volume (VV): V=13×B×hV = \frac{1}{3} \times B \times h V=13×48×15V = \frac{1}{3} \times 48 \times 15 V=16×15=240 cm3V = 16 \times 15 = 240 \text{ cm}^3.

Explanation:

For any pyramid, the volume formula remains 13×Area of Base×Height\frac{1}{3} \times \text{Area of Base} \times \text{Height}, regardless of the shape of the base.

Problem 3:

A right pyramid has a square base with side 6 cm6 \text{ cm}. If the slant height of the pyramid is 5 cm5 \text{ cm}, find its vertical height and volume.

Right triangle showing height h, slant height 5, and base segment 3.

Solution:

  1. Let a=6 cma = 6 \text{ cm} and slant height l=5 cml = 5 \text{ cm}.
  2. The relationship between height hh, slant height ll, and base side aa is l2=h2+(a2)2l^2 = h^2 + (\frac{a}{2})^2.
  3. Substitute the values: 52=h2+(62)25^2 = h^2 + (\frac{6}{2})^2 25=h2+3225 = h^2 + 3^2 25=h2+925 = h^2 + 9 h2=16h^2 = 16 h=4 cmh = 4 \text{ cm}.
  4. Base Area =a2=6×6=36 cm2= a^2 = 6 \times 6 = 36 \text{ cm}^2.
  5. Volume =13×Base Area×h= \frac{1}{3} \times \text{Base Area} \times h Volume =13×36×4=12×4=48 cm3= \frac{1}{3} \times 36 \times 4 = 12 \times 4 = 48 \text{ cm}^3.

Explanation:

To find the volume, we first need the vertical height hh. We use the Pythagorean theorem on the triangle formed by hh, ll, and half of the base side (a/2a/2). Once hh is found, we apply the standard volume formula.

Problem 4:

The base of a pyramid is an equilateral triangle with side 4 cm4 \text{ cm}. If the height of the pyramid is 9 cm9 \text{ cm}, calculate its volume.

Triangular pyramid with height 9 and base side 4.

Solution:

  1. Side of equilateral triangle a=4 cma = 4 \text{ cm}.
  2. Base Area =34a2= \frac{\sqrt{3}}{4} a^2 Base Area =34×42=34×16=43 cm2= \frac{\sqrt{3}}{4} \times 4^2 = \frac{\sqrt{3}}{4} \times 16 = 4\sqrt{3} \text{ cm}^2.
  3. Height h=9 cmh = 9 \text{ cm}.
  4. Volume =13×Base Area×h= \frac{1}{3} \times \text{Base Area} \times h Volume =13×43×9= \frac{1}{3} \times 4\sqrt{3} \times 9 Volume =123 cm3= 12\sqrt{3} \text{ cm}^3.
  5. Using 3≈1.732\sqrt{3} \approx 1.732, Volume ≈12×1.732=20.784 cm3\approx 12 \times 1.732 = 20.784 \text{ cm}^3.

Explanation:

For a triangular pyramid, we first calculate the area of the triangular base. Since the base is equilateral, we use the formula 34a2\frac{\sqrt{3}}{4} a^2. Then, we use the general pyramid volume formula.