Subject: Compulsory Mathematics

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

Solid objects, as shown below are the pyramids.

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.

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

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

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

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. |

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

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Given figure is a square based pyramid. If the slant height of the pyramid is 17 cm and the side of the base 16 cm, determine the volume of the pyramid.

\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*}

In the given figure, the vertical height (AB) and length of the slant edge (AC) of the square based pyramid are 24 cm and 25 cm respectively. Find the volume of the pyramid.

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 = a^{2} = \(\frac {(14)^2}2\) = 98 cm^{2}

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*}

A pyramid has a volume of 36 cubic cm. If its base is an equilateral triangle of side 6 cm, calculate the height of the pyramid.

\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 cm^{3} = \(\frac 13\)× 9\(\sqrt 3\) cm^{2}× h

or, 36 cm^{3} = 3\(\sqrt 3\) cm^{2}× h

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

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

∴ h = 6.93 cm_{Ans}

Find the total surface area of the solid pyramid with squared base as given in the adjoining diagram where BC = CD = 12 cm and OP = 8 cm

.

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*}

In the adjoining figure the length of side of the base of a square based pyramid is 16 cm and its slant height is 10 cm. Find the total surface area of the pyramid.

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*}

In the given figure, the lateral surface area of the pyramid is 144\(\sqrt 2\) sq. cm, OP = m. cm and WX = 2m. cm, find the volume of the pyramid.

Lateral surface area of pyramid = 144\(\sqrt 2\) cm^{2}

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,

4m^{2} \(\sqrt 2\) cm^{2} = 144\(\sqrt 2\)cm^{2}

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

or, m^{2} = 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*}

The volume of the square pyramid is 578 cm^{3}. If its height is 6 cm, find the side of its square base.

Volume of pyramid (V) = 578 cm^{3}

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 cm^{2}_{Ans}

Now,

A = a^{2}

or, a = \(\sqrt A\)

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

∴ a = 17 cm_{Ans}

In the given figure, the total surface area of the square based pyramid is 96 cm^{2} and the side of square base is 6 cm, find the slant height of the pyramid.

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

Then,

Total surface area = a^{2} + 2al

By question,

Total surface area = 96 cm^{2}

a = 6 cm

So,

96 = 6^{2} + 2× 6× l

or, 96 - 36 = 12l

or, 60 = 12l

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

∴ l = 5 cm_{Ans}

In the given figure if the length of a side of the base of the pyramid having square base is 14 cm and the volume of the pyramid is 1568 cm^{3}, find the total surface area of the pyramid.

Side of square base (a) = 14 cm

Area of square base (A) = a^{2} = (14 cm)^{2} = 196 cm^{2}

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*}

In the adjoining figure, the slant height and total surface area of square based pyramid are 13 cm and 360 cm^{2} respectively. Find the perimeter of the base.

Let 'a' be the side of the base.

Slant height (l) = 13 cm

Here,

Total surface area = a^{2} + 2al

or, 360 = a^{2} + 2a× 13

or, a^{2} + 26a - 360 = 0

or, a^{2} + 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*}

The area of triangular surfaces of the square based pyramid given in the diagram is 80 cm^{2}. Find the volume of the pyramid. Where the length of the side of base is 8 cm.

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*}

Given figure is a square based pyramid in which OP = 5\(\sqrt 2\) cm and AR = 13 cm. Find its total surface area.

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:

l^{2} + \((\frac a2\))^{2}= AR^{2}

or, l^{2} + 5^{2} = 13^{2}

or, l^{2} = 13^{2} - 5^{2}

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*}

The triangular surface area of a pyramid given in the figure is 720 cm^{2} and the ratio of vertical height to slant height is 4 : 5, find the volume.

Suppose,

h = 4x

l = 5x

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

h^{2} + (\(\frac a2\))^{2} = l^{2}

or, \(\frac {a^2}{4}\0 = l^{2} + h^{2}

or, a^{2} = 4 (25x^{2} - 16x^{2})

or, a^{2} = 4× 9x^{2}

or, a^{2} = 36x^{2}

∴ a = 6x

Triangular surface area = 2al

or, 720 = 2× 6x× 5x

or, 720 = 60x^{2}

or, x^{2} = \(\frac {720}{60}\)

or, x^{2} = 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*}

In the adjoining figure, the slant height and total surface area of square based pyramid area 12 cm and 340 sq. cm respectively. Find the perimeter of the base and volume.

Here,

l = 12 cm

TSA = 340 cm^{2}

Side of the square base = a (suppose)

Then,

TSA = a^{2} + 2al

or, 340 = a^{2} + 2a× 12

or, a^{2} + 24a - 340 = 0

or, a^{2} + 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 cm_{Ans}

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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