Question 28 (a) - CBSE Class 10 Sample Paper for 2027 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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Question 28 (a) Ridhi drew a polygon with n sides. The smallest exterior angle is 8° and each subsequent exterior angle is 4° more than the previous exterior angle. Find the number of sides of the polygon that Ridhi had drawn. Given that the smallest exterior angle is 8° and each subsequent exterior angle is 4° more than the previous exterior angle Thus, Exterior angles are 8°, 12°, 16°, …. This is an AP with First term = a = 8 Common difference = d = 4 And, Number of sides = n There is a property of exterior angles of polygons Sum of exterior angles = 360° Sum of n terms of AP 8°, 12°, 16°, … = 360° (𝑛)/(2)(2𝑎+(𝑛−1)𝑑)=360° Putting values (𝒏)/(𝟐)(𝟐 × 𝟖+(𝒏−𝟏)𝟒)=𝟑𝟔𝟎 (𝑛)/(2)(16+4𝑛−4)=360 (𝑛)/(2)(12+4𝑛)=360 (𝑛)/(2) × 12+(𝑛)/(2) × 4𝑛=360 5𝑛+2𝑛^(2)=360 2𝑛^(2)+5𝑛−360=0 And, Number of sides = n There is a property of exterior angles of polygons Sum of exterior angles = 360° Sum of n terms of AP 8°, 12°, 16°, … = 360° (𝑛)/(2)(2𝑎+(𝑛−1)𝑑)=360° Putting values (𝒏)/(𝟐)(𝟐 × 𝟖+(𝒏−𝟏)𝟒)=𝟑𝟔𝟎 (𝑛)/(2)(16+4𝑛−4)=360 (𝑛)/(2)(12+4𝑛)=360 (𝑛)/(2) × 12+(𝑛)/(2) × 4𝑛=360 6𝑛+2𝑛^(2)=360 𝟐𝒏^(𝟐)+𝟔𝒏−𝟑𝟔𝟎=𝟎 2(𝑛^(2)+3𝑛−180)=0 𝑛^(2)+3𝑛−180=(0)/(2) 𝒏^(𝟐)+𝟑𝒏−𝟏𝟖𝟎=𝟎 Solving by splitting the middle term 𝑛^(2)+15𝑛−12𝑛−180=0 n(n+15)−12(n+15)=0 (𝐧−𝟏𝟐)(𝐧+𝟏𝟓)=𝟎 Thus, n = 12 and n = –15 But, n cannot be negative as it is number of numbers Thus, n = 12 So, the polygon has 12 sides Splitting the middle term method We need to find two numbers where Sum = 3 Product = 1 × –180 = –180 Since sum is positive and product is negative. One number is positive and other is negative. | Sum | Product | | 3 –180 15 and –12