Question 2 - Case Based Questions (MCQ) - Chapter 11 Class 10 Areas related to Circles

Transcript

Question A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully. Design A: Brooch A is made with silver wire in the form of a circle with diameter 28mm. The wire used for making 4 diameters which divide the circle into 8 equal parts. Design B: Brooch b is made two colours_Gold and silver. Outer part is made with Gold. The circumference of silver part is 44mm and the gold part is 3mm wide everywhere.Refer to Design A Question 1 The total length of silver wire required is (a) 180 mm (b) 200 mm (c) 250 mm (d) 280 mmTotal wire is used = Length of wire in Circle + Wire used in 4 diameters Now, Diameter of the brooch = 28 mm Radius of brooch = 28/2 = 14 mm Length of wire in circle Length of wire in circle = Circumference of Circle = 2𝜋𝑟 = 2×22/7× 14 = 2 × 22 × 2 = 88 mm Silver wire used in 4 diameters = 4 × Diameter of circle = 4 × 28 = 112 mm Now, Total wire is used = length of wire in circle + Wire used in 5 diameters = 88 + 112 = 200 mm So, the correct answer is (b) Length of wire in circle = Circumference of Circle = 2𝜋𝑟 = 2×22/7× 14 = 2 × 22 × 2 = 88 mm Silver wire used in 4 diameters = 4 × Diameter of circle = 4 × 28 = 112 mm Now, Total wire is used = Length of wire in circle + Wire used in 4 diameters = 88 + 112 = 200 mm So, the correct answer is (b) Question 2 The area of each sector of the brooch is (a) 44 mm2 (b) 52 mm2 (c) 77 mm2 (d) 68 mm2Since the wire divides the circle into 8 equal sectors Area of each sector of the brooch = 1/8× Area of all sectors of the brooch = 𝟏/𝟖× Area of circle = 1/8× 𝜋𝑟2 = 1/8× 22/7×(14)^2 = 1/8× 22/7×14 × 14 = 11 × 1 × 7 = 77 mm2 So, the correct answer is (c) Refer Design B: Question 3 The circumference of outer part (golden) is (a) 48.49 mm (b) 82.2 mm (c) 72.50 mm (d) 62.86 mmGiven that Circumference of silver part is 44mm 2𝜋r = 44 2 × 22/7 × r = 44 44/7 × r = 44 r = 44 × 7/44 r = 7 mm Thus, Radius of golden + sliver part = 7 + 3 = 10 mm Now, Circumference of Golden part = 2𝜋 × Radius of (golden + silver) part = 2 × 22/7 × 10 = 2 × 3.143 × 10 = 6.286 × 10 = 62.86 mm So, the correct answer is (d) Thus, Radius of golden + sliver part = 7 + 3 = 10 mm Now, Circumference of Golden part = 2𝜋 × Radius of (golden + silver) part = 2 × 22/7 × 10 = 2 × 3.143 × 10 = 6.286 × 10 = 62.86 mm So, the correct answer is (d) Question 4 The difference of areas of golden and silver parts is (a) 18 𝜋 (b) 44 𝜋 (c) 51 𝜋 (d) 64 𝜋 Difference b/w areas of golden and silver part = Area of silver − Area of golden = Area of circle of radius 10mm − Area of circle of radius 7mm = 𝜋(10)2 − 𝜋(7)2 = 100𝜋 − 49𝜋 = 51 𝜋 So, the correct answer is (c) Question 5 A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 𝜋 mm? (a) 2 (b) 3 (c) 4 (d) 5Here, Number of revolutions = (𝑻𝒐𝒕𝒂𝒍 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆)/(𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒄𝒐𝒗𝒆𝒓𝒆𝒅 𝒊𝒏 𝟏 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏) Radius of Brooch B = 10 cm Therefore, Distance covered in one revolution = Circumference of wheel = 2𝜋r = 2 × 𝜋 × 10 = 20𝜋 mm Now, Number of revolutions = (𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)/(𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛) = 80𝜋/20π = 4 So, the correct answer is (c)