Question 29 - CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic - Solutions of Sample Papers for Class 10 Boards

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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