Chapter 9 Class 8 Mensuration
Serial order wise

Ex 11.1, 3 - The shape of a garden is rectangular in the middle and

Ex 11.1, 3 - Chapter 11 Class 8 Mensuration - Part 2
Ex 11.1, 3 - Chapter 11 Class 8 Mensuration - Part 3
Ex 11.1, 3 - Chapter 11 Class 8 Mensuration - Part 4
Ex 11.1, 3 - Chapter 11 Class 8 Mensuration - Part 5 Ex 11.1, 3 - Chapter 11 Class 8 Mensuration - Part 6

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Transcript

Question 3 The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) meters]. The Garden can be divided into 1 Rectangle & 2 semi circles. Now, Radius of semi circle = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 7/2 = 3.5 cm Also, Length of rectangle = 20 – r – r = 20 – 3.5 – 3.5 = 20 – (3.5 + 3.5) = 20 – 7 = 13 m And, Breadth of rectangle = 7 m Area of Semi-circle Area = 1/2πr2 Putting r = 3.5 Area = 1/2 𝜋(7/2)^2 = 1/2 × 22/7 × 7/2 × 7/2 = 11 × 1/2 × 7/2 = 77/4 Area of rectangle Area = l × b Putting l = 13 m & b = 7 m Area = 13 × 7 = 91 m2 Thus, Area of garden = 2 × Area of Semicircle + Area of rectangle = 2 × 77/4 + 91 = 77/2 + 91 = 38.5 + 91 = 129.5 m2 Now, we need to find Perimeter of garden Perimeter of garden = 2 × Circumference of semi-circle + 2 × Length of rectangle Circumference of semicircle = 1/2 × 2πr = 1/2 × 2 × 22/7 × 7/2 = 11 m Now, we need to find Perimeter of garden Perimeter of garden = 2 × Circumference of semi-circle + 2 × Length of rectangle Circumference of semicircle = 1/2 × 2πr = 1/2 × 2 × 22/7 × 7/2 = 11 m

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo