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
.
\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*}
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.
26 cm^{2}
36 cm^{2}
If every edges of a triangular pyramid is 10 cm, find the lateral surfaces area of the pyramid.
75√3 cm^{2}
65√3 cm^{2}
70√3 cm^{2}
60√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}
50 cm^{3}
41cm^{3}
40 cm^{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.
3072 cm^{3}
3070 cm^{3}
3050 cm^{3}
3030 cm^{3}
A pyramid has squared base of side 24 cm and slant height is 13cm.Find the total surface area and the volume.
1230 cm^{2},980 cm^{3}
1240 cm^{2},930 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.
867 cm^{2},1295cm^{3}
866 cm^{2},1294 cm^{3}
865 cm^{2},1293 cm^{3}
864 cm^{2},1296 cm^{3}
You must login to reply
In the given figure the length of the side of the base of the pyramid having square base is 14cm volume is1568 cm³ find the TSA
Mar 18, 2017
0 Replies
Successfully Posted ...
Please Wait...
Feb 28, 2017
0 Replies
Successfully Posted ...
Please Wait...
Hiranya
Ask any queries on this note.triangular piramyd base of piramd (a-2) vertical hight 12.124
Feb 28, 2017
0 Replies
Successfully Posted ...
Please Wait...
gagan
formula to calculate t.s.a. of combined prism made up of cube and pyramid
Feb 08, 2017
0 Replies
Successfully Posted ...
Please Wait...