Question

A circle is inscribed in a rhombus whose one angle is 60°. Find the ratio of area of the rhombus to the area of the inscribed circle.

Answer 1

 Best Answer

Rhombus has a 60° angle. Other angles are 120°, 60° and 120°. 
Rhombus has 4 equal sides. 

Each diagonal of a rhombus bisects two angles of the rhombus, and the diagonals bisect each other at right angles. 

Now let longest diagonal bisect 60° angles, and let shorter diagonal bisect 120° angles
Now the two diagonals divide rhombus into four 30°-60°-90° triangles with side lengths x, √3x and 2x. Each triangle, therefore has area = 1/2 x √3x = (√3/2) x² 

Now circle inscribed inside rhombus is tangent to each side of rhombus. If we drop perpendicular from centre of circle (where diagonals intersect) to one side of rhombus (i.e. hypotenuse of smaller triangle), we now have a triangle with base = 2x and height = r (radius of circle). 
Area of triangle = 1/2 2x r = x r 

Now we have 2 formulas for area of triangle 
x r = (√3/2) x² 
r = (√3/2) x 

Area of rhombus = Area of 4 triangles = 4*(√3/2) x² = 2√3 x² 

Area of inscribed circle = π r² = π ((√3/2) x)² = 3/4 π x² 

Ratio of rhombus' area to that of inscribed circle: 
2√3 x² : 3/4 π x² 
8√3 : 3 π 
8 : √3π 

1.47 : 1 
hope I am right

Answer 2

Thank you for asking this question. Here is your answer:

If ∠DAB=∠DCB=60°

This means that DAB  and DCB are equilateral

so ∠ADB=60°

∠POB=θP=(√3/2 cosθ,√3/2sinθ)

A=(0,3–√)B=(1,0)C=(0,−3–√)D=(−1,0)

(Let O the center of the circle

and the radius of the circle is r=AOsin60°

=√3/2 and the distance

AO=√3

If there is any confusion please leave a comment below.