Question

A wire bent in the form of an equilateral triangle encloses an area of 121 under root 3 cm square. If the wire is bent in the form of a circle, find the area enclosed by circle

Answer 1

contact me.
I will give you the full answer

Answer 2

Answer:

346.5 sq. cm.          

Step-by-step explanation:

Area of equilateral triangle = $\frac{\sqrt{3}}{4} a^2$

Where a is the side of the triangle

We are given that area of equilateral triangle is $121\sqrt{3}$

So, $121\sqrt{3}=\frac{\sqrt{3}}{4} a^2$

$121=\frac{a^2}{4}$

$484=a^2$

$22=a$

Perimeter of equilateral triangle = $3 \times Side$

                                                     = $3 \times 22$

                                                     = $66 cm$              

Now  the wire which formed triangle is bent in the form of a circle.

Circumference of circle = $2\pi r$

So,  $66=2\pi r$      

 $66=2 \times  \frac{22}{7} r$            

 $66 \times \frac{7}{44}=r$            

 $6 \times \frac{7}{4}=r$              

$10.5 cm=r$      

Area of circle = $\pi r^{2}$

                       = $\frac{22}{7}(10.5)^{2}$        

                       = $346.5 cm^2$    

Hence the area enclosed by circle is 346.5 sq. cm.