These Class 9 Maths Notes and Chapter 6 Measuring Space Perimeter and Area Class 9 Ganita Manjari Notes are designed according to the latest CBSE syllabus.
Class 9 Maths Chapter 6 Measuring Space Perimeter and Area Notes
Class 9 Maths Ganita Manjari Chapter 6 Notes
Class 9 Measuring Space Perimeter and Area Notes
→ Introduction: In our day today lives, we come across many situations when we need to know the length of boundary of some specific landscape or amount of space occupied by it, means its area. In short many situations require us to measure the space. Suppose we have to fence a piece of land; in this case we need to know its perimeter. If we want to sow some seeds in a piece of land, then we must be knowing its area. In this chapter we are going to find out the ways to get perimeters and areas of various shapes in space, like triangles, quadrilaterals, circles etc.
→ Perimeter of a Shape:
- Perimeter of any shape is defined as the total length around its border.
- Perimeter of a triangle or,quadrilateral or,polygon = sum of lengths of all of its sides,
- Perimeter of an equilateral triangle whose side is of length ‘a’ = 3 × a,
- Perimeter of a square of side ‘a’ = 4 × a,
- Perimeter of a rectangle of length ‘l’ and breadth ‘b’ = 2 × (l + b),
- Perimeter of a circle,which is called its circumference,of radius ‘r’ = 2πr.
→ The pie π: π can be defined as the constant ratio of circumference to diameter for all circles. If we represent circumference by C and dimeter by D, then meaningfully, we can write π = \(\frac{C}{D}\). From the knowledge of the previous chapters, we are aware that a rational approximation of π ≈ \(\frac{22}{7}\) or 3.14, though it is irrational.
A much better approximation for π is \(\frac{355}{113}\), but will be a cumbersome assumption for the sake of calculation.
→ The circumference:
The circumference of a circle is given by C = 2πr, where V is the radius of the circle. In the same way we can write the circumference of a circle as C = πd, where ‘d’ is its diameter.
For example, the circumference of the circle whose radius is 42 cm isC = 2 × π × r = 2 × π ×42
Using π = \(\frac {22}{7}\), we get circumference of the given circle as C = 2 × \(\frac {22}{7}\) × 42 = 264 cm.
→ The arc:
The arc length of a circle is given by l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr,
Length of a quarter of a circle, known as quadrant, = \(\frac{\pi r}{2}\)
Length of a semicircle of a circle = πr
For example, the arclength of an arc of the circle whose radius is 7cm and which subtends an angle 90° at the centre is l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr = \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × π × r = \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × π × 42. Using π = \(\frac {22}{7}\), we get the
length of the given arc l = \(\frac{1}{4}\) × 2 × \(\frac{22}{7}\) × 7 = 11 cm.
→ Sector of a Circle:
Sector of a circle is the region enclosed by an arc and two radii connecting the end points of the arc with the centre of the circle.
→ Area of a Circle:
Area of a circle is given by Area of the circle = πr2, where ‘r’ is the radius of the circle. For example, the area of the circle whose radius is 21 cm is = πr2
= π × r × r = π × 21 × 21. Using π = \(\frac {22}{7}\), we get the area of the given circle.
Area = \(\frac {22}{7}\) × 21 × 21 = 1386 cm2.
→ Area of a Triangle:
The area of a triangle is given by A = \(\frac{1}{2}\) × base × height.
→ Heron’s Formula:
Heron’s formula for the area of a triangle in terms of its sides a, b, c is given by
Area = \(\sqrt{s(s-a)(s-b)(s-c)}\), where s is the semi-perimeter of the triangle means half of the
perimeter of the triangle and is calculated by s = \(=\frac{(a+b+c)}{2}\)
→ Area of the triangle circumscribed by the circle of radius ‘R’:
Area of the ∆ABC = \(\frac{a b c}{4 \mathrm{R}}\), where a = BC, b CA, c – AB.
→ Area of the triangle circumscribing the circle of radius V:
→ Area of the Parallelogram = Base × Height.
→ Area of the Square = Side × Side.
→ Area of the Rectangle = Length × Breadth.
→ A median divides the area of a triangle into two equal halves:
In the above figure, BD = DC as median divides the base of a triangle into two equal halves. Height for both the triangles is ‘h’
Height for both the triangles is ‘h’
Area of ∆ABC = \(\frac{1}{2}\) × BD × h
Area of ∆ABC = \(\frac{1}{2}\) × DC × h
Hence, from (i) , (ii) and (iii), we get Area of ∆ABD = Area of ∆ACD.
→ Area of Cyclic 4-gon (Brahmagupta’s Formula):
Area = \(\sqrt{(s-a)(s-b)(s-c)(s-d)}\), where a, b, c, d are the lengths of sides of the Cyclic 4-gon and s = \(\frac{(a+b+c+d)}{2}\)
Note: All squares, all rectangles and all isosceles trapeziums are Cyclic 4-gon. Cyclic 4-gon are quadrilaterals whose all four vertices lie on a circle.
→ Area of a Trapezium:
Area of a trapezium = \(\frac{1}{2}\) × (sum of parallel sides) × h, where h is the distance between the parallel sides.
→ Area of a Rhombus:
Area of a rhombus = \(\frac{1}{2}\) × d1 × d2, where d1 and d2 are its two diagonals.