Pyramid

Subject: Compulsory Mathematics

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Overview

The pyramid is solid with polygonal base and triangular faces with common vertex. A line through the vertex to the centre of the base is called the height of the pyramid. Height is perpendicular to the base is called a right pyramid otherwise pyramid is oblique pyramid. Height is perpendicular to the base is called right pyramid otherwise pyramid is oblique pyramid. A pyramid is regular if it's all lateral faces are congruent isosceles triangle.Surface area of the pyramid is the total surface area of its al

Pyramid

Solid objects, as shown below are the pyramids.

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As we see above, the pyramid is solid with a polygonal base and triangular faces with a common vertex. A line through the vertex to the centre of the base is called the height of the pyramid. Height perpendicular to the base is called right pyramid otherwise, pyramid is an oblique pyramid. A pyramid is regular if it's all lateral faces are a congruent isosceles triangle.

A pyramid whose base is an equilateral triangle is a tetrahedron. In tetrahedron, all the faces are congruent equilateral triangles.

A perpendicular line segment drawn from the vertex to any side of its base is called the slant height for the face consisting that side.

A pyramid is a three-dimensional solid figure in which the base is a polygon of any number of sides, and other faces are triangles that meet at a common point.

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\( \therefore \text {Area of triangular face} = \frac {1} {2} base side \times slant \: height\)

The surface area of the pyramid is the total surface area of its all triangular faces together with the base.

Volume of a pyramid

Let's take a cubical container of side 'a' units. Take a pyramid of a square base with
a side of length 'a' units and height is same to that of the previous cube. Fill up water in cube by a pyramid.

Cube is filled up when the water is poured three times by the pyramid. By the
above experiment, we can say that the volume of the pyramid is one-third of the
volume of cube whose base and height are the same as that of pyramid. That is, if
V be the volume of the pyramid then, \( V = \frac {1} {3} a^3 \)

\(\boxed { \therefore V= \frac {1} {3} \times volume \: of \: the \: cube } \)

It can be written as, \( V= \frac {1} {3} a^2 \times a \). Hence, \( V= \frac {1} {3} \times base \: area \times height \)

Alternatively,

Take a cube of side '2a' units. Draw the space diagonal as shown in the figure.

There are six equal pyramids inside the cube, each has a square base of a side 2a units and height is half of the above cube. One of them is shown to the right of the diagram.

Let V be the volume of each pyramid. The total volume of such six pyramids is same as that of the cube. That is,
\begin{align*} 6V &= (2a)^3 \\ or, 6V &= (2a)^2 2a \\ or, V &= \frac {1} {6} (2a)^2 . 2a \\ \therefore V &= \frac {1} {3} (2a)^2 . a \\ \end{align*}

This means volume of each pyramid is equal to the one-third of product of its base area and height.

\(\therefore V =\frac {1} {3} \times base \: area \times height \)

In the adjoining figure,

Volume of solid = Volume of cuboid + volume of pyramid.
TSA of solid = base area + CSA of cuboid + LSA of the pyramid. 

Things to remember

A pyramid is the three-dimensional solid figure in which the base is a polygon of any number of sides and other faces are a triangle that meets at a common point.

Area of a triangle face = \(\frac{1}{2}\) base side × slant height

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Videos for Pyramid
Surface Area Of A Pyramid
Surface Area Pyramid + Finding Slant Height
Volume of a Pyramid
Questions and Answers

\begin{align*} {\text{Area of square base (A)}} &= (16 cm)^2\\&= 256 cm^2\\ \end{align*}

\begin{align*} {\text{Height of pyramid (h)}} &= \sqrt {(17 cm)^2 - (\frac {16}2 cm)^2}\\ &= \sqrt {289 cm^2 - 64 cm^2}\\ &= \sqrt {225 cm}\\ &= 15 cm\\ \end{align*}

\begin{align*} {\text{Volume of pyramid (V)}} &= \frac 13 Ah\\ &= \frac 13× 256 cm^2× 15 cm\\ &= 1280 cm^3_{Ans}\\ \end{align*}

Here,

\begin{align*} BC &= \sqrt {AC^2 - AB^2}\\ &= \sqrt {25^2 -24^2}\\ &= \sqrt {49}\\ &= 7 cm \end{align*}

\begin{align*} {\text{Diagonal of square base}} &= DC\\ &= 2× BC\\ &= 2 × 7 cm\\ &= 14 cm \end{align*}

Let a be the length of each side of the square base.

Then,

Diagonal = \(\sqrt 2\)a

or, 14 cm = \(\sqrt 2\)a

∴ a = \(\frac {14}{\sqrt 2}\) cm

A = area of square base = a2 = \(\frac {(14)^2}2\) = 98 cm2

h = height = 24 cm

Hence,

\begin{align*} {\text{Volume of the pyramid (V)}} &= \frac 13 Ah\\ &= \frac 13 × 98 × 24 cm^3\\ &= 784 cm^3_{Ans}\\ \end{align*}

\begin{align*} {\text{Area of base of the pyramid}} &= \text{area of equilateral triangle}\\ &= \frac {\sqrt 3}4 a^2, \text{where a = side of the triangle}\\ &= \frac {\sqrt 3}4 (6 cm)^2\\ &= 9\sqrt 3 cm^2_{Ans}\\ \end{align*}

h = height of pyramid = ?

We know that,

Volume of the pyramid = \(\frac 13\) Ah

or, 36 cm3 = \(\frac 13\)× 9\(\sqrt 3\) cm2× h

or, 36 cm3 = 3\(\sqrt 3\) cm2× h

or, h = \(\frac {36}{3\sqrt 3}\)cm

or, h = \(\frac {12}{\sqrt 3}\)cm

∴ h = 6.93 cmAns

Suppose,

PQ⊥ BC

Here,

a = BC = 12 cm

h = OP = 8 cm

\begin{align*} \therefore l &=\sqrt {(\frac a2)^2 + h^2}\\ &= \sqrt {(6 cm)^2 + (8 cm)^2}\\ &= \sqrt {100 cm^2}\\ &= 10 cm\\ \end{align*}

Hence,

\begin{align*} {\text{Total surface area of given prism}} &= a^2 + 2al\\ &= (12 cm)^2 + 2 × 12 × 10 cm^2\\ &= 144 cm^2 + 240 cm^2\\ &= 384 cm^2_{Ans}\\ \end{align*}

Side of the squared base (a) = 16 cm

Slant height (l) = 10 cm

\begin{align*} \therefore {\text{Total surface area of the pyramid}} &= a^2 + 2al\\ &= (16 cm)^2 + 2 × 16 cm × 10 cm\\ &= 256 cm^2 + 320 cm^2\\ &= 576 cm^2_{Ans} \end{align*}

Lateral surface area of pyramid = 144\(\sqrt 2\) cm2

OP = m. cm

WX = 2m. cm

Therefore,

\begin{align*} PQ &= \frac {WX}2\\ &= \frac {2m}2\\ &= m. cm\\ \end{align*}

\begin{align*} OQ &= \sqrt {OP^2 + PQ^2}\\ &= \sqrt {m^2 + m^2}cm\\ &= \sqrt {2m^2} cm\\ &= m\sqrt 2 cm \end{align*}

\begin{align*} {\text{Area of Δ OXY}} &= \frac 12× XY× OQ\\ &= \frac 12 × 2m × m\sqrt 2 cm^2\\ &= m^2\sqrt 2 cm^2\\ \end{align*}

\begin{align*} {\text{Lateral surface area}} &= 4 × {\text{Area of Δ OXY}}\\ &= 4m^2 \sqrt 2 cm^2 \end{align*}

\begin{align*}

Therefore,

4m2 \(\sqrt 2\) cm2 = 144\(\sqrt 2\)cm2

or, m2 = \(\frac {144\sqrt 2}{4\sqrt 2}\)

or, m2 = 36

∴m = 6

Now,

WX = 2m = 2× 6 = 12 cm

OP = m. cm = 6 cm

\begin{align*} {\text{Area of square base (A)}} &= (WX)^2\\ &= (12 cm)^2\\ &= 144 cm^2\\ \end{align*}

\begin{align*} {\text{Volume of pyramid}} &= \frac 13 Ah\\ &= \frac 13× 144× OP\\ &= \frac 13× 144× 6 cm^3\\ &= 288 cm^3\\ \end{align*}

Volume of pyramid (V) = 578 cm3

Height of pyramid (h) = 6 cm

Area of square base (A) = ?

Side of square base (a) = ?

We have,

V = \(\frac 13\) Ah

or, 578 = \(\frac 13\)× A× 6

or, 578 = 2A

or, A = \(\frac {578}2\)

∴ A = 289 cm2Ans

Now,

A = a2

or, a = \(\sqrt A\)

or, a = \(\sqrt (289 cm^2)\)

∴ a = 17 cmAns

Let 'l' be the slant height and 'a' be the length of the side of square base of the pyramis.

Then,

Total surface area = a2 + 2al

By question,

Total surface area = 96 cm2

a = 6 cm

So,

96 = 62 + 2× 6× l

or, 96 - 36 = 12l

or, 60 = 12l

or, l = \(\frac {60}{12}\)

∴ l = 5 cmAns

Side of square base (a) = 14 cm

Area of square base (A) = a2 = (14 cm)2 = 196 cm2

Let h be the height of the p[yramid.

Then,

Volume of pyramid = \(\frac 13\) Ah

or, 1568 = \(\frac 13\)× 196× h

or, h = \(\frac {1568 × 3}{}196\)

∴ h = 24 cm

\begin{align*} {\text{Height of the triangular face (l)}} &= \sqrt {(\frac a2)^2 + h^2}\\ &= \sqrt {(\frac {14}2)^2 + h^2}\\ &= \sqrt {7^2 + 24^2}\\ &= \sqrt {625}\\ &= 25 cm\\ \end{align*}

\begin{align*} {\text{Total surface area of the pyramid}} &= a^2 + 2al\\ &= (14 cm)^2 + 2× 14× 25 cm\\ &= 196 cm^2 + 2 × 14 × 25 cm^2\\ &= 896 cm^2_{Ans}\\ \end{align*}

Let 'a' be the side of the base.

Slant height (l) = 13 cm

Here,

Total surface area = a2 + 2al

or, 360 = a2 + 2a× 13

or, a2 + 26a - 360 = 0

or, a2 + 36a - 10a - 360 = 0

or, a(a + 36) - 10(a + 36) = 0

or, (a - 10) (a + 36) = 0

Either,

a - 10 = 0

∴ a = 10

Or,

a + 36 = 0

∴ a = -36

Since, the length of the side is always positive so a = -36 is impossible.

Hence,

a = 10 cm

\begin{align*} \therefore {\text{Perimeter of base}} &= 4a\\ &= 4× 10 cm\\ &= 40 cm_{Ans}\\ \end{align*}

Length of the side of base (CD) = 8 cm

Height (l) of a triangular face = ?

Volume (V) of pyramid = ?

Now,

Area of all triangular faces = 4(\(\frac 12\) × base × height)

or, 80 = 2× 8× l

or, l = \(\frac {80}{16}\)

∴ l = 5 cm

Also,

\begin{align*} {\text{Height of the pyramid (h)}} &= \sqrt {l^2 - (\frac {side}2)^2}\\ &= \sqrt {5^2 - (\frac 82)^2}\\ &= \sqrt {25 - 16}\\ &= 3 cm\end{align*}

\begin{align*} {\text{Area of base (A)}} &= (8 cm)^2\\ &= 64 cm^2\\ \end{align*}

\begin{align*} \therefore {\text{Volume of the pyramid}} = \frac 13 Ah\\ &= \frac 13× 64× 3\\ &= 64 cm^3_{Ans}\\ \end{align*}

Here,

PR = 2× OP = 2× 5\(\sqrt 2\) = 10\(\sqrt 2\) cm

Let 'a' be the length of a side of the square PQRS then:

PR = \(\sqrt 2\)a

or, 10\(\sqrt 2\) = \(\sqrt 2\)a

∴ a = 10 cm

If l be the slant height of the pyramid, then:

l2 + \((\frac a2\))2= AR2

or, l2 + 52 = 132

or, l2 = 132 - 52

or, l = \(\sqrt {169 - 25}\)

∴ l = 12 cm

\begin{align*} {\text{Total surface area}} &= a^2 + 2al\\ &= (10)^2 + 2× 10× 12\\ &= 100 + 240\\ &= 340 cm^2_{Ans}\\ \end{align*}

Suppose,

h = 4x

l = 5x

If 'a' be the length of a side of the square ABCD, then:

h2 + (\(\frac a2\))2 = l2

or, \(\frac {a^2}{4}\0 = l2 + h2

or, a2 = 4 (25x2 - 16x2)

or, a2 = 4× 9x2

or, a2 = 36x2

∴ a = 6x

Triangular surface area = 2al

or, 720 = 2× 6x× 5x

or, 720 = 60x2

or, x2 = \(\frac {720}{60}\)

or, x2 = 12

or, x = \(\sqrt 12\)

∴ x = 2\(\sqrt 3\)

\begin{align*} A &= a^2\\ &= 36x^2\\ &= 36× (2\sqrt 3)^2\\ &= 432\\ \end{align*}

\begin{align*} Volume (V) &= \frac 13 Ah\\ &= \frac 13× 432× 4× 2\sqrt 3\\ &= 1995.32 cm^3_{Ans}\\ \end{align*}

Here,

l = 12 cm

TSA = 340 cm2

Side of the square base = a (suppose)

Then,

TSA = a2 + 2al

or, 340 = a2 + 2a× 12

or, a2 + 24a - 340 = 0

or, a2 + 34a - 10a - 340 = 0

or, a(a + 34) - 10(a + 34) = 0

or, (a + 34) (a - 10) = 0

Either,

a + 34 = 0

∴ a = -34

Or,

a - 10 = 0

∴ a = 10

Length of a side cannot be negative. Therefore, a = -34 is impossible.

Hence,

Side of the square base (a) = 10 cm

∴ Perimeter of the base = 4a = 4× 10 cm = 40 cmAns

Here,

l = PQ = 12cm

OQ = \(\frac a2\) = \(\frac {10}2\) = 5 cm

∴ OP = \(\sqrt {PQ^2 - OQ^2}\) = \(\sqrt {12^2 - 5^2}cm\) = \sqrt {119}cm

\begin{align*} {\text{Volume of pyramid}} &= \frac 13 ×a^2× OP\\ &= \frac 13× 100× \sqrt{119} cm^3\\ &= 363.62 cm^3_{Ans}\\ \end{align*}

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