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 \({{1} \over {4}}\) m and a tread of \({{1} \over {2}}\) 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 = \({{1} \over {4}} × {{1} \over {2}} × 50 m^3\).]

As we can see from the given figure, the first step is \({{1} \over {2}}\) m wide, 2nd step is 1 m wide and 3rd step is \({{3} \over {2}}\) m wide. Thus we can understand that the width of step by \({{1} \over {2}}\) m each time when height is \({{1} \over {4}}\) m. And also, given length of the steps is 50 m all the time. So, the width of steps forms a series AP in such a way that;

\({{1} \over {2}}\) , 1, \({{3} \over {2}}\), 2, ……..
Volume of steps = Volume of Cuboid
= Length × Breadth Height
Now,

Volume of concrete required to build the first step = \({{1} \over {4}} × {{1} \over {2}} ×50 = {{25} \over {4}}\)

Volume of concrete required to build the second step = \({{1} \over {5}} × {{2} \over {2}} × 50 = {{25} \over {2}}\)

Volume of concrete required to build the second step = \({{1} \over {4}} × {{3} \over {2}}×50 = {{75} \over {2}}\)

Now, we can see the volumes of concrete required to build the steps, are in AP series;

\({{25} \over {4}} , {{25} \over {2}} , {{75} \over {2}} \)…..
Thus, applying the AP series concept,

First term, a = \({{25} \over {4}}\)
Common difference, d = \({{25} \over {2}} – {{25} \over {4}} = {{25} \over {4}}\)

As we know, the sum of n terms is;

\(S_n = {{n} \over {2}}[2a+(n-1)d] \)
\(= {{15} \over {2}}(2×({{25} \over {4}} )+({{15} \over {2}} -1){{25} \over {4}})\)

Upon solving, we get,

\(S_n = {{15} \over {2}} (100)\)
\(S_n= 750\)

Hence, the total volume of concrete required to build the terrace is \(750 m^3\).