Answer :
1) side(a)=38
area of a regular pentagon=[tex] \frac{1}{4} \sqrt{5(5+2 \sqrt{5} )}a ^{2} [/tex]
if a side=38 ,, area ≈2484.37 cm²
2)side(b)=76
area=[tex] \frac{1}{4} \sqrt{5(5+2 \sqrt{5} )}a ^{2} [/tex]
=9937.48 cm²
area of a regular pentagon=[tex] \frac{1}{4} \sqrt{5(5+2 \sqrt{5} )}a ^{2} [/tex]
if a side=38 ,, area ≈2484.37 cm²
2)side(b)=76
area=[tex] \frac{1}{4} \sqrt{5(5+2 \sqrt{5} )}a ^{2} [/tex]
=9937.48 cm²
[tex]Area\ of\ a\ regualr\ pentagon\ of\ side\ 'a'\ is\ given\ by\\ \\A= \frac{1}{4} \sqrt{25+10 \sqrt{5} }\ a^2\\ \\when\ a=38cm,\ A=2484.37\ cm^2\\ when\ a=76cm,\ A=9937.48\ cm^2[/tex]