Following figure depicts a park where two opposite sides are parallel and left and right ends are semi-circular in shape. It has a 7 m wide track for walking

Q 29 - Sample Paper.jpg

Two friends Seema and Meena went to the park. Meena said that area of the track is 4066 m 2 . Is she right? Explain.

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Note : This is similar to Ex 12.3, 8 of NCERT – Chapter 12 Class 10

Check the answer here https://www.teachoo.com/5743/542/Ex-11.1--1---Draw-a-line-segment-of-length-7.6-cm-and-divide-in-ratio/category/Ex-11.1/

  1. Class 10
  2. Solutions of Sample Papers for Class 10 Boards

Transcript

Question 29 Following figure depicts a park where two opposite sides are parallel and left and right ends are semi-circular in shape. It has a 7 m wide track for walking Two friends Seema and Meena went to the park. Meena said that area of the track is 4066 m2. Is she right? Explain. Let us label the figure Area of track = Area ABFE + Area HGCD + (Area of semicircle with diameter AD – Area of semicircle with diameter EH) + (Area of semicircle with diameter BC – Area of semicircle with diameter FG) Area ABFE ABFE is a rectangle with length = 120 m & breadth = 7 m Area of ABFE = Length × Breadth = 120 × 7 = 840 m2 Area HGCD HGCD is a rectangle with length = 120 m & breadth = 7 m Area of HGCD = Length × Breadth = 120 × 7 = 840 m2 Area semicircle with diameter AD Diameter = AD = 70 m Radius = r = 70/2 = 35 m Area semicircle AOD = 1/2×𝜋𝑟2 = 1/2×22/7×35×35 = 1925 m2 Area semicircle with diameter EH Diameter = EH = 70 – 7 – 7 = 56 m Radius = r = 56/2 = 28 m Area semicircle EOH = 1/2×𝜋𝑟2 = 1/2×22/7×28×28 = 1232 m2 Area semicircle with diameter BC Diameter = BC = 70 m Radius = r = 70/2 = 35 m Area semicircle BO’C = 1/2×𝜋𝑟2 = 1/2×22/7×35×35 = 1925 m2 Area semicircle with diameter FG Diameter = FG = 70 – 7 – 7 = 56 m Radius = r = 56/2 = 28 m Area semicircle FO’G = 1/2×𝜋𝑟2 = 1/2×22/7×28×28 = 1232 m2 Area of track = Area ABFE + Area HGCD + (Area of semicircle with diameter AD – Area of semicircle with diameter EH) + (Area of semicircle with diameter BC – Area of semicircle with diameter FG) = 840 + 840 + (1925 – 1232) + (1925 – 1232) = 2 × 840 + 2 (1925 – 1232) = 1680 + 2 (693) = 1680 + 1386 = 3066 m2 Since area of track is not 4066 m2 ∴ Meena was wrong

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.