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Answer :
(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
\(=(15\times 30)-(2\times \frac{1}2\times \frac{15}2)\)
\(=(450-\frac{225}2)=\frac{(900-225)}2m^2\)
\(=\frac{675}2m^2=337.5m^2\)