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Ex 5.4, 5 (Optional) - A small terrace at football ground of 15 steps

Ex 5.4, 5 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.4, 5 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 3 Ex 5.4, 5 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 4 Ex 5.4, 5 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 5 Ex 5.4, 5 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 6 Ex 5.4, 5 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 7

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Ex 5.4, 5 (Optional) A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = 1/4 × 1/2 × 50 m3] Given Number of steps n = 15 Length of Steps = 50 m Rise in each step = 1/4 m Tread of each other step = 1/2 m Total Volume of concrete required = Sum of the volumes of individual steps. Each step is a cuboid. So, we find Volume of cuboid First step Volume of first step = Length × Breadth × Height = 50 × 1/2 × 1/4 m3 Volume of a cuboid = Base area × Height = Length × Breadth × Height = 50/8 m3 = 6.25 m3 Second step Here, Height = 2 × Height of 1st step Height = 2 × 1/4 m = 1/2 m Volume of Second step = Length × Breadth × Height = 50 × 1/2 × 1/2 m3 = 50/4 = 12.5 m3 Third step Here, Height = 3 × Height of 1st step Height = 3 × 1/4 m = 3/4 m Volume of Third step = Length × Breadth × Height = 50 × 1/2 × 3/4 m3 = 150/8 = 18.75 m3 Now, writing Volumes 6.25, 12.5, 18.75, ….. First Term = a = 6.25 Difference between 1st and 2nd term = 12.5 − 6.25 = 6.25 Difference between 2nd and 3rd term = 18.75 − 12.5 = 6.25 Since, the difference between consecutive terms is equal, ∴ Volume is in AP with First term = a = 6.25 & Common difference = d = 6.25 & Number of steps = n = 15 Now, Total Volume of concrete = Sum of volumes of individual steps = S15 We know that Sum of n terms of an AP is Sn = 𝒏/𝟐 [𝟐𝒂+(𝒏 −𝟏)𝒅] S15 = 15/2 [(2" × 6.25" )+(15−1)(6.25)] = 15/2 [12.5+87.5] = 15/2 × 100 = 15 × 50 = 750 Thus, the total Volume of the concrete required is 750 m3

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.