Q5:

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 $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ 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 = $\frac{1}{4} \times \frac{1}{2} \times 50 m^3$]

Given:

1. Total number of steps ($n$) = $15$.

2. Length of each step ($l$) = $50$ m.

3. Rise of each step ($h$) = $\frac{1}{4}$ m.

4. Tread of each step ($w$) = $\frac{1}{2}$ m.

To Find:

The total volume of concrete required to build the entire terrace.

Step 1: Calculate the volume of individual steps.

The volume of a rectangular prism (each step) is given by the formula: $V = \text{length} \times \text{width} \times \text{height}$.

Volume of the 1st step ($V_1$) = $50 \times \frac{1}{2} \times \frac{1}{4} = \frac{50}{8} = 6.25$ m$^3$.

Volume of the 2nd step ($V_2$) = $50 \times \frac{1}{2} \times (\frac{1}{4} + \frac{1}{4}) = 50 \times \frac{1}{2} \times \frac{2}{4} = 2 \times 6.25 = 12.5$ m$^3$.

Volume of the 3rd step ($V_3$) = $50 \times \frac{1}{2} \times (\frac{1}{4} + \frac{1}{4} + \frac{1}{4}) = 50 \times \frac{1}{2} \times \frac{3}{4} = 3 \times 6.25 = 18.75$ m$^3$.

Step 2: Identify the Arithmetic Progression (AP).

The volumes form an AP where:

First term ($a$) = $6.25$

Common difference ($d$) = $V_2 - V_1 = 12.5 - 6.25 = 6.25$

Number of terms ($n$) = $15$

Step 3: Apply the Sum of an AP formula.

The sum of the first $n$ terms of an AP is given by: $S_n = \frac{n}{2} [2a + (n - 1)d]$.

Substituting the values:

$S_{15} = \frac{15}{2} [2(6.25) + (15 - 1)(6.25)]$

$S_{15} = \frac{15}{2} [12.5 + 14(6.25)]$

$S_{15} = \frac{15}{2} [12.5 + 87.5]$

$S_{15} = \frac{15}{2} [100]$

$S_{15} = 15 \times 50$

$S_{15} = 750$

Final Answer: The total volume of concrete required to build the terrace is 750 m$^3$.