Question 15 - CBSE Class 10 Sample Paper for 2027 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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Question 15 (Method 1) The arc of a circle is of length 6𝜋 cm and the sector it bound has an area of 24𝜋 cm^(2). The radius of the circle is: (A) 4 cm (B) 8 cm (C) 16 cm (D) 18 cm Let the arc subtend an angle 𝜃 at the center Given Length of arc APB = 6𝜋 cm Area of sector AOPB = 24𝜋 cm^(2) We have a formula relating to length of arc and area of sector Area of sector = (𝟏)/(𝟐) × Length of arc × Radius Putting values 24𝜋 = (𝟏)/(𝟐) × 6𝜋 × Radius 24𝜋 = 3𝜋 × Radius (24𝜋)/(3𝜋) = Radius 8 = Radius Radius = 8 cm So, the correct answer is (B) 𝜃 O A B P 6𝜋 cm 24𝜋 = 3𝜋 × Radius (24𝜋)/(3𝜋) = Radius 8 = Radius Radius = 8 cm So, the correct answer is (B) Question 15 (Method 2) The arc of a circle is of length 6𝜋 cm and the sector it bound has an area of 24𝜋 cm^(2). The radius of the circle is: (A) 4 cm (B) 8 cm (C) 16 cm (D) 18 cm Let the arc subtend an angle 𝜃 at the center Given Length of arc APB = 6𝜋 cm Area of sector AOPB = 24𝜋 cm^(2) Now, Length of Arc APB = (𝜃)/(360) × 2𝜋𝑟 6𝜋 = (𝜽)/(𝟑𝟔𝟎°) × 𝟐𝝅𝒓 (6𝜋)/(2𝜋)=(𝜃)/(360°) × 𝑟 𝜃 O A B P 6𝜋 cm 3 = (𝜽)/(𝟑𝟔𝟎°) × 𝒓 (3)/(𝑟) = (𝜽)/(𝟑𝟔𝟎°) (𝜽)/(𝟑𝟔𝟎°)=(𝟑)/(𝒓) Also, Area of sector OAPB= (𝜃)/(360) × 𝜋𝑟^(2) 24𝜋 = (𝜽)/(𝟑𝟔𝟎°) × 𝝅𝒓^(𝟐) Putting (𝜃)/(360°)=(3)/(𝑟) 24𝜋 = (𝟑)/(𝒓) × 𝝅𝒓^(𝟐) 24𝜋 = 3𝜋𝑟 3𝜋r = 24𝜋 r = (24𝜋)/(3𝜋) r = 8 cm 𝜃 O A B P 6𝜋 cm r = (24𝜋)/(3𝜋) r = 8 cm So, the correct answer is (B)