A small terrace at a football ground comprise | vedclass

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(B) From the figure,it can be observed that:$1^{	ext{st}}$ step is $\frac{1}{2} \, m$ wide,$2^{	ext{nd}}$ step is $1 \, m$ wide,$3^{	ext{rd}}$ step

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(B) From the figure,it can be observed that:$1^{	ext{st}}$ step is $\frac{1}{2} \, m$ wide,$2^{	ext{nd}}$ step is $1 \, m$ wide,$3^{	ext{rd}}$ step is $\frac{3}{2} \, m$ wide.Therefore,the width of each step is increasing by $\frac{1}{2} \, m$ each time,whereas their height $\frac{1}{4} \, m$ and length $50 \, m$ remain the same.Thus,the widths of these steps form an arithmetic progression: $\frac{1}{2}, 1, \frac{3}{2}, 2, \dots$Volume of concrete in $1^{	ext{st}}$ step $= 	ext{length} 	imes 	ext{width} 	imes 	ext{height} = 50 	imes \frac{1}{2} 	imes \frac{1}{4} = \frac{25}{4} \, m^3$.Volume of concrete in $2^{	ext{nd}}$ step $= 50 	imes 1 	imes \frac{1}{4} = \frac{25}{2} \, m^3$.Volume of concrete in $3^{	ext{rd}}$ step $= 50 	imes \frac{3}{2} 	imes \frac{1}{4} = \frac{75}{4} \, m^3$.The volumes of concrete in these steps form an $A.P.$ with first term $a = \frac{25}{4}$ and common difference $d = \frac{25}{2} - \frac{25}{4} = \frac{25}{4}$.The total volume for $n = 15$ steps is given by the sum $S_n = \frac{n}{2} [2a + (n - 1)d]$.$S_{15} = \frac{15}{2} \left[ 2 \left( \frac{25}{4} ight) + (15 - 1) \left( \frac{25}{4} ight) ight]$$S_{15} = \frac{15}{2} \left[ \frac{25}{2} + 14 	imes \frac{25}{4} ight] = \frac{15}{2} \left[ \frac{25}{2} + \frac{175}{2} ight]$$S_{15} = \frac{15}{2} \left[ \frac{200}{2} ight] = \frac{15}{2} 	imes 100 = 750$.Thus,the total volume of concrete required is $750 \, m^3$.

$A$ 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 $\frac{1}{2} \, m$. Calculate the total volume of concrete required to build the terrace (in $m^3$).

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