Question

find the area of regular pentagon whose each side is 5 cm long and the radius of the inscribed circle is 3.5 cm

Answer 1

πr2=5 so 3.14*3.5*3.5=5

Answer 2

Answer:

$\text{The area of regular pentagon is }30.62 cm^2$

Step-by-step explanation:

Given the regular pentagon whose each side is 5 cm long and the radius of the inscribed circle is 3.5 cm.

Side=5 cm

Radius of the inscribed circle=3.5 cm

In ΔOMC, by Pythagoras theorem

$OC^2=OM^2+MC^2$

$(3.5)^2=h^2+(2.5)^2$

$12.25=h^2+6.25$

$6=h^2$

$h=\sqrt6 cm$

$\text{The area of triangle ODC is }\frac{1}{2}\times DC\times OM$

$=\frac{1}{2}\times 5\times \sqrt6 = \frac{5\sqrt6}{2}=6.124 cm^2$

The Pentagon includes 5 triangles

so the area of Pentagon is five times the area of the triangle

$\text{Pentagon's area=}5\times \text{area of one triangle}$

$=5\times 6.124=30.62 cm^2$

$\text{The area of regular pentagon is }30.62 cm^2$