Question

A rectangular field is 60m wide while its perimeter is 320m. If laying of grass costs rs. 7.50 per square meter;how much would it cost for grass to be laid on the field leaving a bare path 8m wide all round the field for class 6

Answer 1

soulution.
Let the length of rectangle be ' l '.
it's given that,
width =60mt.
perimeter = 320 mt.
A. T. Q,
2(L +B) = perimeter.
2[ L + (60)] =320
2l +120 =320
2l =320-120
2l = 200
L = 100mts.

$2(l + b) = perimeter \: \: of \: the \: rectangle.....$
cost required = 2x 1600 + 2x352=2304 Mts.
therefore, the cost = area X rate
=2304 x7. 5
=17280.0
since, the total cost is rupees is 17280/=

Answer 2

$\rm \: Perimeter \: of \: ABCD = 2AB + 2 BC$

$\sf \implies \: 320 \: m = 2 \times 60 \: m + 2 \: BC$

$\sf \implies \: 2 \: BC = 320 \: m - 120 \: m$

$\sf \implies \: BC = \dfrac{200 \: m}{2} = 100 \: m$

$\rm \bigstar \: PQ = AB - 8 m - 8 m$

$\sf \implies \: PQ = 60 \: m - 16 \: m = \boxed{ \sf \: 44 \: }$

$\rm \bigstar \: QR = BC - 8 m - 8 m$

$\sf \implies \: QR = 100 \: m - 16 m = \boxed{ \sf84 \: m}$

$\rm \bigstar \: Area \: of \: PQRS = PQ \times QR$

$\sf \implies \: 44 \times 84 \: {m}^{2}$

$\implies \boxed{\sf3,696 \: {m}^{2} }$

$\rm \bigstar \: Cost \: of \: lying \: \: grass \: on \: 1 \: {m}^{2}$

$\implies \boxed{ \sf7.50 \: Rs}$

$\rm \bigstar \: Cost \: of \: lying \: \: grass \: on \: \sf3,696 \: {m}^{2}$

$\sf \implies \: 7.50 \times 3,696 \:$

$\sf \implies \boxed{ \bf \: 27,720 \: Rs \: }$

$\begin{gathered}\rule{190pt}{2pt} \\ \end{gathered}$

${ \red{ \mathfrak{Additional\:Information}}}$

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