Question

9. Two crossroads, each of width 5 m, ran at right angles through the centre of a rectangular park of length 50 m and breadth 30 m and parallel to its sides. How many square tiles of side 0.5 m will be needed to cover it?​

Answer 1

$\large\underline{\sf{Solution-}}$

Two crossroads, each of width 5 m, ran at right angles through the centre of a rectangular park of length 50 m and breadth 30 m and parallel to its sides.

Length of rectangular park, l = 50 m

Width of rectangular park, b = 30 m

$\rm \: Length \: of \: the \: shorter \: path, \:  l_1 \: = \: 30 \: m$

$\rm \: Width \: of \: the \: shorter \: path, \:  b_1 \: = \: 5 \: m$

$\rm \: Length \: of \: the \: longer \: path, \:  l_2 \: = \: 50 \: m$

$\rm \: Width \: of \: the \: longer \: path, \:  b_2 \: = \: 5 \: m$

Now, Consider

$\rm \: Area \: of \: shorter \: path,  \: A_1=l_1×b_1$

$\rm \: A_1 = 30 \times 5$

$\rm\implies \:A_1 = 150 \: {m}^{2}$

Now, Consider

$\rm \: Area \: of \: longer \: path,  \: A_2=l_2×b_2$

$\rm \: A_2 = 50 \times 5$

$\rm\implies \:A_2 = 250 \: {m}^{2}$

Now,

$\rm \: Area \: of \: common \: path,  \: A_3=5 \times 5$

$\rm\implies \:A_3 \: = \: 25 \: {m}^{2}$

Thus,

$\rm \: Area \: of \: path = A_1 + A_2 - A_3$

$\rm \: = \:150 + 250 - 25$

$\rm \: = \:375 \: {m}^{2}$

$\rm\implies \:\rm \: Area \: of \: path = 375 \: {m}^{2}$

Now, Side of square tile = 0.5 m

$\rm \: Area_{(square tile)} \: = \: {side}^{2}$

$\rm \: Area_{(square tile)} \: = \: {(0.5)}^{2}$

$\rm\implies \:Area_{(square tile)} = 0.25 \: {m}^{2}$

Thus,

$\rm \: Number_{(square tile)} = \dfrac{Area \: of \: path}{Area_{(square tile)}}$

On substituting the values, evaluated above, we get

$\rm \: Number_{(square tile)} = \dfrac{375}{0.25} = 1500$

Thus,

$\rm\implies \: \boxed{\sf{  \:\rm \: \: Number_{(square \: tile)} \: required \: = \: 1500 \: \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$