This question is inspired from 

Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam , your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition , suggests two designs given below.
Observe these carefully

Design I: This design is made with a circle of radius 32cm leaving equilateral triangle ABC in the middle as shown in the given figure.

Design II: This Pookalam is made with 9 circular design each of radius 7cm.

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Refer Design I:

Question 1

The side of equilateral triangle is

(a) 12√3 cm        (b) 32√3 cm       (c) 48 cm       (d) 64 cm

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Question 2

The altitude of the equilateral triangle is

(a) 8 cm           (b) 12 cm      (c) 48 cm       (d) 52 cm

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Refer Design II:

Question 3

The area of square is

(a) 1264 cm 2         (b) 1764 cm 2

(c) 1830 cm 2          (d) 1944 cm 2

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Question 4

Area of each circular design is

(a) 124 cm 2          (b) 132 cm 2

(c) 144 cm 2          (d) 154 cm 2

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Question 5

Area of the remaining portion of the square ABCD is

(a) 378 cm 2                                        (b) 260 cm 2

(c) 340 cm 2                                        (d) 278 cm 2

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  1. Chapter 12 Class 10 Areas related to Circles (Term 1)
  2. Serial order wise

Transcript

Question Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam , your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition , suggests two designs given below. Observe these carefully Design I: This design is made with a circle of radius 32cm leaving equilateral triangle ABC in the middle as shown in the given figure. Design II: This Pookalam is made with 9 circular design each of radius 7cm. Refer Design I: Question 1 The side of equilateral triangle is (a) 12√3 cm (b) 32√3 cm (c) 48 cm (d) 64 cmLet centre of circle be O. Join OA, OB and OC Now, OA = OB = OC = Radius of circle = 32 cm Let’s consider Δ OBC Let’s draw OM ⊥ BC By symmetry, BM = CM = 𝟏/𝟐 BC ∠ BOM = ∠ COM = 𝟏/𝟐 × 120° = 60° In right triangle Δ OMB sin O = (Side opposite to angle O)/Hypotenuse sin 60° = 𝐵𝑀/𝑂𝐵 √3/2=𝐵𝑀/32 √3/2 × 32 = BM 16√3 = BM BM = 16√𝟑 Thus, BC = 2 × BM = 2 × 16 √3 cm = 32√𝟑 cm So, the correct answer is (b) Question 2 The altitude of the equilateral triangle is (a) 8 cm (b) 12 cm (c) 48 cm (d) 52 cmAltitude of equilateral triangle = AM = OA + OM = 32 + OM Let’s find OM Let’s take Δ OMB In right triangle Δ OMB cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 60° = 𝑂𝑀/𝑂𝐵 1/2=𝑂𝑀/32 32/2 = OM 16 = OM OM = 16 Thus, Altitude of equilateral triangle = 32 + OM = 32 + 16 = 48 cm So, the correct answer is (c) Refer Design II: Question 3 The area of square is (a) 1264 cm2 (b) 1764 cm2 (c) 1830 cm2 (d) 1944 cm2Here, Side of square = 3 × Diameter of circle = 3 × 2 × Radius of circle = 6 × Radius = 6 × 7 = 42 cm Now, Area of square = (Side)2 = (42)2 = 1764 cm2 So, the correct answer is (b) Question 4 Area of each circular design is (a) 124 cm2 (b) 132 cm2 (c) 144 cm2 (d) 154 cm2Circular Design is a circle with radius 7 cm Area of circular design = 𝜋𝑟2 = 𝜋(7)^2 = 22/7 × 7 × 7 = 22 × 7 = 154 cm2 So, the correct answer is (d) Question 5 Area of the remaining portion of the square ABCD is (a) 378 cm2 (b) 260 cm2 (c) 340 cm2 (d) 278 cm2Area of remaining portion = Area of square – Area of 9 circles = Area of square – 9 × Area of 1 circle = 1764 − 9 × 154 = 1764 − 1386 = 378 cm2 So, the correct answer is (a)

Chapter 12 Class 10 Areas related to Circles (Term 1)
Serial order wise

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.