Question

there is a pentagonal shaped park as shown in the figure

Answer 1

$\pink{\huge{\star Solution \star}}$

$\red{\underline{Jyotis\: Diagram}}\\ CO = \frac{15}{2}\\ CO = 7.5 cm \\\\Area \: of \: Pentagon = 2 (area of trapezium) \\ \implies = 2(\frac{1}{2}(a+b)h)\\\implies \cancel2(\frac{1}{\cancel 2}(30+15)\times 7.5 \\\implies 45\times 7.5 \\\implies 337.5 \: m^{2}$

$\red{ \underline{Kavita's \: Diagram}}\\\ Area\: of\: pentagon = area \:of \:square + area \:of \:triangle\\\implies side^{2} + \frac{1}{2} \times Base \times Altitude \\\implies (15)^{2} + \frac{1}{2} \times 15\times 15 \\\implies 225 + (7.5)(15) \\\implies 225 + 112.5\\\implies 337.5\: m^{2}$

$\huge{\pink{\underline{Answer}}}$

$\red{\boxed{ Area \: of \: pentagon = 337.5\: m^{2} }}$