Question

find the area of the adjoining heptagon ABCDEFG Fig. ii) in which AC =6.5 cm , BM = 3.2 cm , CE = 9 cm, DN = 4.5 cm , AG = 10 cm, AP = 4.4 cm.
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Answer 1

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Answer 2

Area of heptagon is 91.35$cm^{2}$.

Step-by-step explanation:

1. Given data

  AC = 6.5 cm

  BM = 3.2 cm

  CE= PF= 9 cm

  AG = 10 cm

  AP = 4.4 cm

  CP = AC-AP= 6.5 - 4.4= 2.1 cm

  ND = 4.5 cm

2. Area of heptagon = area of ΔABC +area of trapezium PAGF + area of rectangle PFEC + area of ΔCED     ...1)

3. Area of ΔABC = $\frac{1}{2}AC\times BM=\frac{1}{2}6.5\times 3.2= 10.4 cm^{2}$         ...2)

4. Area of trapezium PAGF =$\frac{1}{2}(PF+AG)\times AP=\frac{1}{2}(9+10)\times 4.4= 41.8 cm^{2}$       ...3)

5. Area of rectangle PFEC =CE×PC=9×2.1=18.9 $cm^{2}$    ...4)

6. Area of ΔCED =$\frac{1}{2}CE\times ND=\frac{1}{2}9\times 4.5= 20.25 cm^{2}$         ...2)

7. Now from equation 1)

   Area of heptagon = 10.4 +41.8+18.9+20.25 =91.35 $cm^{2}$