WBBSE Class 9 Maths Mensuration Chapter 3 Area Of Circle Multiple Choice Questions

WBBSE Class 9 Maths Mensuration Chapter 3 Area Of Circle Multiple Choice Questions

Example 1. If the area of circular field is X sq unit, the perimeter is Y unit and length of the diameter is Z unit, then the value of \(\frac{X}{YZ}\) is

  1. \(\frac{1}{2}\)
  2. \(\frac{1}{4}\)
  3. 1
  4. \(\frac{1}{8}\)

Solution: \(\frac{X}{Y Z}=\frac{\pi r^2}{2 \pi r 2 r}=\frac{1}{4}\)

∴ The correct answer is 2. \(\frac{1}{4}\)

∴ Then the value of \(\frac{X}{YZ}\) is \(\frac{1}{4}\)

Read and Learn More WBBSE Class 9 Maths Multiple Choice Questions

Example 2. The ratio of area of two square circumscribe and inscribe by a circle is

  1. 4:1
  2. 1:4
  3. 2:1
  4. 1:2

Solution: Area of the square inscribed in a circle = 2r2

WBBSE Class 9 Maths Mensuration Chapter 3 Two Square Circumscribe And Inscribe

 

 

 

 

 

 

 

Area of the square circumscribing the circle = 4r2

Ratio = 4r2: 2r2 = 2:1

∴ The correct answer is 3. 2: 1

∴ The ratio of area of two square circumscribe and inscribe by a circle is 2: 1

Example 3. The numerical value of perimeter and area of a circular field in equal. The length of diagonal of square circumscribe by a circle is

  1. 4 unit
  2. 2 unit
  3. 4√2 unit
  4. 2√2 unit

Solution: πr2 ⇒ 2πr  ⇒ r= 2 unit

∴ The correct answer is 2. 2 unit

∴ The length of diagonal of square circumscribe by a circle is 2 unit

WBBSE Class 9 Maths Mensuration Chapter 3 Area Of Circle Multiple Choice Questions

Example 4. The ratio of the areas circumscribed and inscribed circles of an equilateral triangle is

  1. 4:1
  2. 1:4
  3. 2:1
  4. 1:2

Solution: Ratio of length of circumradius and inradius = \(\frac{2}{3}\) x height: \(\frac{1}{3}\) x height = 2:1

Ratio of areas = 22: 12 = 4:1

∴ The correct answer is 1. 4:1

∴ The ratio of the areas circumscribed and inscribed circles of an equilateral triangle is 4:1

Example 5. The inner diameter and external diameter of an iron ring plate are 20 cm, 22 cm. Area of iron plate is

  1. 22 sq. cm
  2. 44 sq. cm
  3. 66 sq. cm
  4. 88 sq. cm

Solution: Area = \(=\pi\left[\left(\frac{22}{2}\right)^2-\left(\frac{20}{2}\right)^2\right]\) sq.cm = 66 sq.cm.

∴ The correct answer is 3. 88 sq. cm

Example 6. The area of a circle is A sq. units and its circumference is C units. The value of \(\frac{\mathbf{A}}{\mathbf{C}^2}\) =

  1. \(\frac{1}{4 \pi}\)
  2. \(\frac{4}{\pi}\)
  3. \(\frac{\pi}{4}\)
  4. None of these

Solution: \(\frac{A}{C^2}=\frac{\pi r^2}{(2 \pi r)^2}=\frac{1}{4 \pi}\)

∴ The correct answer is 1. \(\frac{1}{4 \pi}\)

∴  Area of iron plate is \(\frac{1}{4 \pi}\)

Example 7. The outer and inner circumference of a ring-shaped circular plate are A unit and B unit respectively. If the width is C unit, the value of π is

  1. \(\frac{A – B}{C}\)
  2. \(\frac{A – B}{2C}\)
  3. \(\frac{C}{A – B}\)
  4. \(\frac{A – B}{7C}\)
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Solution: 2π (R − r) = A − B

or, R – r = \(\frac{A – B}{2 \pi}\)

∴ π = \(\frac{A – B}{2C}\)

∴ The correct answer is 2. \(\frac{A – B}{2C}\)

∴ The value of π is \(\frac{A – B}{2C}\)

Example 8. The width of a circular ring shaped is 3.5 cm and its area is 60.5 sq. cm. The sum of inner and outer radii of the plate is

  1. 3.5 cm
  2. 4.5 cm
  3. 5.5 cm
  4. 6.5 cm

Solution: π(R + r) (R − r) = 60.5

or, \(\frac{22}{7}\)(R+r) x 3.5 = 60·5 or, R + r = 5·5

∴ The correct answer is 3. 5.5 cm

∴  The sum of inner and outer radii of the circular ring plate is 5.5 cm

Example 9. The area of a semi-circle is 77 sq. cm. The length of its diameter is

  1. 3.5 cm
  2. 7 cm
  3. 14 cm
  4. 10.5 cm

Solution: \(\frac{\pi r^2}{2}=77\)

or, r = \(\frac{77 \times 7 \times 2}{22}=7^2\)

⇒ R =7, 2R =14

∴ The correct answer is 3. 14 cm

The length of its diameter is

Example 10. The ratio of area of circumscribed and inscribed squares of a circle

  1. 2: 1
  2. 1: 2
  3. 4: 1
  4. 1: 4

Solution: Ratio = \(a^2: \frac{a^2}{2}\) = 2: 1

∴ The correct answer is 1. 2: 1

∴ Ratio of area of circumscribed and inscribed squares of a circle is 2: 1

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