Pyramid
Solid objects, as shown below are the pyramids.
As we see above, the pyramid is a 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 is perpendicular to the base is called a right pyramid otherwise pyramid is an oblique pyramid. Height is 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 threedimensional 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. 
Surface area of pyramid
The surface area of the pyramid is the total surface area of its all triangular faces together with the base.
\( \therefore \text {Area of triangular face} = \frac {1} {2} base side \times slant \: height \)
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 onethird 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 onethird 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 threedimensional 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
\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*}
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*}
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}
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}
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*}
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*}
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*}
\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}
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*}
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*}
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*}
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*}
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*}

Find the lateral surface area with base is an equilateral triangle of area 16√3 cm^{2} and the length of lateral edge is 5 cm.
36 cm^{2}
26 cm^{2}
>28 cm^{2}

If every edges of a triangular pyramid is 10 cm, find the lateral surfaces area of the pyramid.
65√3 cm^{2}
60√3 cm^{2}
75√3 cm^{2}
70√3 cm^{2}

In a square based pyramid the ratio of vertical height and slant height is 4:5 and the total surface area is 96 cm^{2}.Find the volume of the pyramid.
48 cm^{3}
40 cm^{3}
50 cm^{3}
41cm^{3}

In a square based pyramid, the ratio of slant height and length of base is 5:6 and the total surface area is 1536 cm^{2}.Find the volume of the pyramid.
3030 cm^{3}
3050 cm^{3}
3072 cm^{3}
3070 cm^{3}

A pyramid has squared base of side 24 cm and slant height is 13cm.Find the total surface area and the volume.
1240 cm^{2},930 cm^{3}
1230 cm^{2},980 cm^{3}
1200cm^{2},960 cm^{3}
1222 cm^{2},970 cm^{3}

A pyramid has a squared base of side 18 cm and height is 12 cm.Calculate the total surface area and the volume.
866 cm^{2},1294 cm^{3}
865 cm^{2},1293 cm^{3}
867 cm^{2},1295cm^{3}
864 cm^{2},1296 cm^{3}

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Sudheer Budha MagarWhat is theFormula to calculate volume of pyramid? 
Dec 31, 2016 
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