Mensuration Chapter 3 Area Of Circle
⇔ Area of circle = πr2
⇔ Area of semi-circle = \(\frac{1}{2}\) πr2
⇔ Area of circular ring = π (R2 – r2) = π (R + r)(R -r)
Read and Learn More WBBSE Solutions For Class 9 Maths
⇔ Sector of a circle = \(\pi r^2 \cdot \frac{\theta}{360}\)
⇔ Area of a square inscribed in a circle = \(\frac{(\text { diagonal })^2}{2}=\frac{(2 r)^2}{2}=2 r^2\)
⇔ Area of a square circumscribing the circle = (2r)2 = 4r2
⇔ Area of an equilateral triangle, when the length of the radius of its in circle = r unit
⇔ Area of the triangle = \(\frac{\sqrt{3}}{4}(2 \sqrt{3} r)^2=3 \sqrt{3} r^2\)
⇔ Area of an equilateral triangle, when the length of the radius of its circumcircle is r unit
⇔ Area of the triangle = \(\frac{\sqrt{3}}{4}(4 \sqrt{3} r)^2=12 \sqrt{3} r^2\)
Mensuration Chapter 3 Area Of Circle Fill In The Blanks
Example 1. If the circumference of a circle be 22 cm, then its area will be ______ sq. cm
Solution: 38.5.
Example 2. The area of a sector making an angle 30° at the centre of a circle of radius 21 cm is ______ sq. cm.
Solution: 115.5.
Example 3. If the ratio of the circumference of two circles be 2: 3, the ratio of their areas will be ______
Solution: 4:9.
Example 4. If x be the area of a circle its circumference is ______
Solution: 2√πx
Example 5. The diameter of a circle inscribed in the square of area 49 sq cm is ______
Solution: 7 cm.
Mensuration Chapter 3 Area Of Circle True Or False
Example 1. If the circumference of a circle is A, then its diameter = \(\frac{A}{\pi}\).
Solution: The statement is true.
Example 2. The radius of a circle is 21 cm, its area is 616 sq. cm.
Solution: The statement is false.
Example 3. Area of a semi-circle of radius 7 cm is 77 cm.
Solution: The statement is true.
Example 4. When a square is inscribed in a circle, its diagonal will be equal to the diameter of the circle.
Solution: The statement is true.
Example 5. When a square circumscribed a circle, its diagonal will be equal to the diameter of the circle.
Solution: The statement is false.
Mensuration Chapter 3 Area Of Circle Short Answer Type Questions
Example 1. If the length of radius of a circular field was increased by 10%, let us write by calculating what % it increase the area.
Solution: Let radius = 100 r unit
Initial Area = 10000 πr2 sq.unit
Increased Area = π(110)2 sq.unit = 12100 πr2 sq.unit
% increase = \(\frac{2100 \pi r^2}{10000 \pi r^2}\) x 100 sq.unit = 21 sq.unit
∴ 21% increased.
Example 2. If the perimeter of a circular field was decreased by 50%. Calculate what percent it decrease the area of circular field.
Solution: Let radius be r unit.
Initial perimeter and area = 2πr unit, πr2sq.unit
Reduced perimeter = \(\frac{2 \pi r}{2}\) unit = πr unit
Reduced radius = \(\frac{r}{2}\) unit
Area decrease = \(=\left\{\pi r^2-\pi\left(\frac{r}{2}\right)^2\right\} \text { sq. unit }=\frac{3 \pi r^2}{4} \text { sq. unit }\)
∴ % decrease = \(\frac{3 \frac{\pi r^2}{4}}{\pi r^2}\) x 100 sq.unit = 75%
Area is decreased by 75%.
Example 3. The length of radius of a circular field is r meter. If the area of other circle is x times of first circle, let us calculate how length of radius of other circle.
Solution: Area = x x πr2
πR2 = πr2
∴ R= r√x unit.
Example 4. Calculate the area of a circular region circumscribe a triangle of which sides are 3 cm, 4 cm, and 5 cm.
Solution: 32+ 42 = 52, It is a right-angled triangle.
Length of circumradius = \(\frac{5}{2}\) cm = 2.5 cm.
Area = \(\frac{22}{7}\) x 2.5 x 2.5 sq.cm = 19\(\frac{9}{14}\) sq.cm.
Example 5. Three circular plates were cut off from a tin plate with equal width if the ratio of the length of diameter of three circles is 3: 5: 7, calculate the ratio of their weight.
Solution: Let weight of per sq. unit tin = x gm
Let diameter be 3R, 5R, 7R units respectively (R > 0)
∴ Ratio weight = \(\pi\left(\frac{3 \mathrm{R}}{2}\right)^2 \cdot x: \pi\left(\frac{5 \mathrm{R}}{2}\right)^2 \cdot x: \pi\left(\frac{7 \mathrm{R}}{2}\right)^2 \cdot x\) = 9: 25: 49
Example 6. Find the area of the shadow region of the figure.
Solution: Area =\(\left\{\frac{90}{360} \pi \cdot(14)^2-\Delta A O B\right\} \mathrm{sq} \cdot \mathrm{cm}\)
= \(\left\{\frac{1}{4} \pi \cdot(14)^2-\frac{1}{2} \cdot 14 \cdot 14\right\} \mathrm{sq} \mathrm{cm}=56 \mathrm{sqcm}\)
Example 7. If the perimeter of a circle = perimeter of an equilateral triangle. Find the ratio of their areas.
Solution: 2πr = 3a ⇒ r = \(\frac{3a}{2 \pi}\)
∴ Ratio = \(\pi\left(\frac{3 a}{2 \pi}\right)^2: \frac{\sqrt{ } 3}{4} a^2=63: 22 \sqrt{3}\)
Example 8. Find The area of the shadow region of
Solution: Area = Area of the square – Area of 4 sectors
= \(\left\{12^2-\frac{90^{\circ}}{360^{\circ}} \times \pi(6)^2 \times 4\right\} \mathrm{sq} \cdot \mathrm{cm}=30 \frac{6}{7} \mathrm{sq} . \mathrm{cm}\)
Example 9. Find the area of shadow region of
Solution: Area Of 4 setors
= \(\left\{\frac{22}{7} \times(3.5)^2\right\}-\left\{\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(3.5)^2\right\} \times 4 \text { sq.cm }=115.5 \text { sq.cm }\)
Example 10. Find the area of the shadow region of
Solution: Area = \(\left(\frac{1}{4} \times \frac{22}{7} \times 12^2-\frac{1}{2} \times 12 \times 12\right) \mathrm{sqcm}\)
= \(\frac{288}{7} \text { sq. }\)
= \(41 \frac{1}{7} \text { sq. } \mathrm{cm}\)
