## What is a Pyramid?

APyramid is defined as a three-dimensional structure encompassing a polygon as its base.

Every corner of a pyramid is linked to a single apex which makes it appear as a distinct shape. The edge of each base and its top makes a triangle.

Types of Pyramid

There are different types of pyramids that are named based on the shapes of their bases.

- Triangular Pyramid -The base of this pyramid has a shape of a triangle, therefore we call it as a triangular pyramid.

2. Square Pyramid – The base of this pyramid has a shape of a square, therefore we call it as a Square Pyramid.

3. Pentagonal Pyramid – The base of this Pyramid has a shape of a Pentagon, therefore we call it as a Pentagonal Pyramid.

4. Right pyramid – This apex of this pyramid is exactly over the middle of the base, hence named as Right Pyramid.

5. Oblique pyramid – The apex of this pyramid is not exactly over the middle of its base and named as Oblique Pyramid.

Regular vs Irregular Pyramid

To distinguish between regular and irregular Pyramid, you need to consider the shape of the base. If the base of a polygon is regular, its labeled as Regular Pyramid, else it is considered as Irregular Pyramid.

**Hero of Alexandra** was a great mathematician who derived the formula for the calculation of the area of a triangle using the length of all three sides. This derived formula is called as **Heron’s formula**. He also extended this idea to find the area of quadrilaterals and also higher order polygons.

**Derivation of Heron’s formula:**

Area of ∆ABC is given by

\(A~=~\frac{1}{2} bh\)

Draw a perpendicular BD on AC

Consider a ∆ADB

\(x^2~+~ h^2~ = ~c^2\)

\(x^2~ =~ c^2~ -~ h^2\)

\(⇒x~ =~\sqrt{c^2 -h^2}\)

Consider a ∆CDB

\((b-x)^2~ + ~h^2~ =~a^2\)

\((b-x)^2~ =~a^2~ -~h^2\)

\(b^2~ -~2bx~+~x^2~=~a^2 – h^2\)

Substituting the value of \(x\)

\(b^2~ – ~2b\sqrt{c^2 -h^2}~+~ c^2 -h^2~ = ~a^2~ -~ h^2\)

\(b^2~ + ~c^2~ -~a^2~ =~ 2b\sqrt{c^2 ~-~h^2}\)

Squaring on both sides

\((b^2 + c^2 – a^2 )^2~ =~ 4b^2 (c^2 -h^2)\)

\(\frac{(b^2 + c^2 -a^2 )^2}{4b^2} = c^2 – h^2\)

\(h^2~ = ~c^2 ~-~\frac{(b^2 +c^2 -a^2 )^2}{4b^2}\)

\(h^2~ =~\frac{4b^2 c^2 – (b^2 +c^2 -a^2 )^2 }{4b^2}\)

\(h^2~=~\frac{(2bc)^2 -(b^2+c^2- a^2 )^2}{4b^2}\)

\(h^2~=~\frac{[2bc+(b^2 +c^2 -a^2 )][2bc-(b^2+c^2-a^2)]}

{4b^2}\)

\(h^2~ =~\frac{[(b^2+2bc+c^2)-a^2][a^2-(b^2- 2bc+c^2)]}{4b^2}\)

\(h^2~=~\frac{[(b+c)^2 – a^2 ].[a^2 -(b-c)^2 ]}{4b^2}\)

\(h^2~=~\frac{[(b+c)+a][(b+c)-a].[a+(b-c)][a-(b-c)]}{4b^2}\)

\(h^2~=~\frac{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}{4b^2}\)

The perimeter of a ∆ABC is

\(P~=~ a+b+c\)

\(⇒~h^2~ =~\frac{P(P – 2a)(P – 2b)(P -2c)}{4b^2}\)

\(⇒~h~ = \sqrt{P(P – 2a)(P – 2b)(P -2c)}{2b}\)

Substituting the value of h in equation (i)

\(A~=~\frac{1}{2} b \frac{\sqrt{P(P – 2a)(P – 2b)(P -2c)}}{2b}\)

\(A~=~\frac{1}{4}\sqrt{(P(P – 2a)(P – 2b)(P -2c)}\)

\(A~=~\sqrt{\frac{1}{16} P(P – 2a)(P – 2b)(P -2c)}\)

\(A~=~\sqrt{\frac{P}{2}\left(\frac{P – 2a}{2}\right)\left(\frac{P – 2b}{2}\right)\left(\frac{P -2c}{2}\right)}\)

Semi perimeter(S) = \(\frac{perimeter}{2}~=~\frac{P}{2}\)

\(⇒~A~=~\sqrt{S(S – a)(S – b)(S – c)}\)

Note: Heron’s formula is applicable to all type of triangles and the formula can also be derived using the law of cosines and the law of Cotangents.

Let us now look into some example to have a brief insight about the topic:

Let’s Work Out-

Example: Find the area of a trapezium, length of whose parallel sides is given as 22cm and 12 cm and the length of other sides is 14 cm each.

Solution: Let PQRS be the given trapezium in which PQ= 22cm, SR= 12cm,

PS=QR=14cm.

Constructions: Draw OR||PS

Now, PORS is a parallelogram in which PS||OR and PO||SR

Therefore, PO=SR=12cm

⇒OQ = PQ-PO = 22 -12 = 10cm

In ∆OQR , we have

\(S~ = ~\frac{14+14+10}{2}~ =~\frac{38}{2}~=~19\)

Area of ∆OQR = \(\sqrt{S(S – a)(S – b)(S –C)}\)

= \( \sqrt{ (19(19 – 14)(19 – 14)(19 –10))}\)

= \( \sqrt{4275}\)

= \(15 \sqrt{19} cm^2\)

We know that Area = \(\frac{1}{2} \times b \times h\)

\(\Rightarrow 15\sqrt{19} = \frac{1}{2} \times 10 \times h\)

\(\Rightarrow h = 5\sqrt{19}\)

Area of trapezium = \(\frac{1}{2}~ (PQ+SR) × h \)

= \(\frac{1}{2}(22+12) × 3 \sqrt{19} \)

= \(51 \sqrt{19} cm^2\)

To solve more problems based on Heron’s Formula visit BYJU’S which provides detailed and step by step solutions to all questions in an NCERT Books. Also, take free tests to practice for exams.

Pyramid Formulas

To find out the surface area and some formulas can be used. The surface area of a pyramid is the total area of all the surfaces that pyramid has. Formulas to find out the surface area of a pyramid.

S A = (base area) + (1/2) * (perimeter) * (slant height)

The base area is the area of the base of a pyramid that can be calculated on the basis of figure type. To calculate the triangular pyramid volume, the formula to compute it is given by:

Volume = 1/ 3 x [Base Area] x height

You can visit a separate page for volume of the pyramid, that provides all the surface and volume formulas along with solved examples.

To refer more and solve work on Pyramid maths problems prism, you can visit www.byjus.com

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