questions which are related to nsture in the chapter ares related to circles
Answers
Answer:
Given, k = 0.0248 S cm - 1
c = 0.20 M
Molar conductivity, Am = (k x 1000) / c
= 0.0248 x1000 / 0.20
124 Scm2 mol - 1
Answer:
1. If the radius of a circle is 4.2 cm, compute its area and circumference.
Solution:
Area of a circle = πr2
So, area = π(4.2)2 = 55.44 cm2
Circumference of a circle = 2πr
So, circumference = 2π(4.2) = 26.4 cm
2. What is the area of a circle whose circumference is 44 cm?
Solution:
Circumference of a circle = 2πr
From the question,
2πr = 44
Or, r = 22/π
Now, area of circle = πr2 = π × (22/π)2
So, area of circle = (22×22)/π = 154 cm2
3. Calculate the area of a sector of angle 60°. Given, the circle is having a radius of 6 cm.
Solution:
Given,
The angle of the sector = 60°
Using the formula,
The area of sector = (θ/360°)×π r2
= (60°/360°) × π r2 cm2
Or, area of the sector = 6 × 22/7 cm2 = 132/7 cm2
4. A chord subtends an angle of 90°at the centre of a circle whose chord is 20 cm. Compute the area of the corresponding major segment of the circle.
Solution:
Point to note:
Area of the sector = θ/360 × π × r2
Base and height of the triangle formed will be = radius of the circle
Area of the minor segment = area of the sector – area of the triangle formed
Area of the major segment = area of the circle – area of the minor segment
Now,
Radius of circle = r = 20 cm and
Angle subtended = θ = 90°
Area of the sector = θ/360 × π × r2 = 90/360 × 22/7 × 202
Or, area of the sector = 314.2 cm2
Area of the triangle = ½ × base × height = ½ × 20 × 20 = 200 cm2
Area of the minor segment = 314.2 – 200 = 114.2 cm2
Area of the circle = π × r2 = (22/7) × 202 = 1257.14
Area of the major segment = 1257.14 – 114.2 = 1142 .94 cm2
So, the area of the corresponding major segment of the circle = 1142 .94 cm2
5. A square is inscribed in a circle. Calculate the ratio of the area of the circle and the square.
Solution:
As the square is inscribed in a circle, a diagonal of the square will be = the diameter of the circle.
Let “r” be the radius of the circle and “d” be the length of each diagonal of the square.
We know,
Length of the diagonal of a square = side (s) × √2
So,
d = 2r
And, s × √2 = 2r
Or, s = √2r
We know, the area of the square = s2
Thus, the area of the square = (√2r)2 = 2r2
Now, the area of the circle = π × r2
∴ Area of the circle : area of the square = π × r2 : 2r2 = π : 2
So, the ratio of the area of the circle and the square is π : 2.
6. Find the area of the sector of a circle with radius 4cm and of angle 30°. Also, find the area of the corresponding major sector.
Solution:
Radius = r = 4 cm, θ=30°
Area of sector = [/360]×2
= 30/360×3.14×(4)2
= 1/12×3.14×4×4
= 1/3×3.14×4
= 12.56/3 cm2
= 4.19 cm2
Area of major sector = ((360 − θ)/360)×2
= ((360 − 30))/360×3.14×(4)2
= 330/360×3.14×4×4
= 11/12×3.14×4×4
= 46.05 cm2
7. Calculate the perimeter of an equilateral triangle if it inscribes a circle whose area is 154 cm2
Solution:
Here, as the equilateral triangle inscribed in a circle, the circle is an incircle.
Now, the radius of the incircle is given by,
r = Area of triangle/semi-perimeter
In the question, it is given that area of the incircle = 154 cm2
So, π × r2 = 154
Or, r = 7 cm
Now, assume the length of each arm of the equilateral triangle to be “x” cm
So, the semi-perimeter of the equilateral triangle = (3x/2) cm
And, the area of the equilateral triangle = (√3/4) × x2
We know, r = Area of triangle/semi-perimeter
So, r = [x2(√3/4)/ (3x/2)]
=> 7 = √3x/6
Or, x = 42/√3
Multiply both numerator and denominator by √3
So, x = 42√3/3 = 14√3 cm
Now, the perimeter of an equilateral triangle will be = 3x = 3 × 14√3 = 72.7 cm.
Step-by-step explanation: