10.There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.


Find the area of this park using both ways. Can you suggest some other way of finding its area?

(i) From Jyoti’s diagram:


We can observe that, Area of the pentagonal shape = Area of trapezium ABCD + Area of trapezium ADEF

\(= 2 \times Area \;of\; trapezium \;ABCD\)

\(= 2 \times \frac { 1 }{ 2 } \times (a + b) \times h\)

\(= (15 + 30) \times 7.5\)

\(= 45 \times 7.5\)

\(= 337.5 m^2\)

So, the area of the pentagonal shape=337.5\(m^2\)

(ii) From Kavita’s diagram:



We can see that, area of the pentagonal shape = Area of triangle ABE + Area of square BCDE

\(=\frac { 1 }{ 2 }\times b \times h + 15 \times 15\)

\(=\frac { 1 }{ 2 }\times 15 \times 15 + 225\)

\(= 112.5 + 225= 337.5 m^2\)

So, the area of the pentagonal shape =\(337.5 m^2\)

Yes, there is other way to find the area which is as follows:



Required area ABCDEA=Area of rect. ABPQ-2\(\times\)area of triangle CPD

\(=(15\times 30)-(2\times \frac{1}2\times \frac{15}2)\)

\(=(450-\frac{225}2)=\frac{(900-225)}2m^2\)

\(=\frac{675}2m^2=337.5m^2\)