- Number Systems
- Polynomials
- Factorization of Polynomials
- Linear Equations In Two Variables
- Coordinate Geometry
- Introduction To Euclids Geometry
- Lines And Angles
- Triangles
- Congruence Of Triangles And Inequalities In A Triangle
- Quadrilaterals
- Areas Of Parallelograms And Triangles
- Circles
- Geometrical Constructions
- Areas Of Triangles And Quadrilaterals
- Volume And Surface Area Of Solids
- Mean Median And Mode Of Ungrouped Data
- Probability
RS Aggarwal Solutions Class 9 Mathematics Solutions for Circles Exercise 12A in Chapter 12 - Circles
A chord of length 16cm is drawn in a circle of radius 10cm. Find the distance of the chord from the centre of the circle.
Consider AB as the chord with O as the centre and radius 10cm
So we get
OA = 10 cm and AB = 16cm
Construct OL ⊥ AB
The perpendicular from the centre of a circle to a chord bisects the chord
So we get
AL = ½ × AB
By substituting the values
AL = ½ × 16
So we get
AL = 8 cm
Consider △ OLA
Using the Pythagoras theorem it can be written as
OA^2 = OL^2 + AL^2
By substituting the values we get
10^2 = OL^2 + 8^2
On further calculation
OL^2 = 10^2 - 8^2
So we get
OL^2 = 100 – 64
By subtraction
OL^2 = 36
By taking the square root
OL = √36
So we get
OL = 6cm
Therefore, the distance of the chord from the centre of the circle is 6cm.
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
