Question

A circle with radius 1 touches the sides of a rhombus. Each of the smaller angle between the sides of the rhombus is 60°. what is the area of the rhombus?

Answer 1

Answer:

Hint:

If ∠DAB=∠DCB=60°∠DAB=∠DCB=60° than The triangles DABDAB and DCBDCB are equilateral, so ∠ADB=60°∠ADB=60°. Let OO the center of the circle, than the radius of the circle is r=AOsin60°=3√2r=AOsin⁡60°=32 and the distance AO=3–√AO=3.

Now you can use a coordinate sistem with center OO, a point on the circle has coordinates P=(rcosθ,rsinθ)P=(rcos⁡θ,rsin⁡θ) and you can find the distances from the vertices of the rhombus.

So you can prove if the sum of th⁸e squares of these distances is constant and find its value.



∠POB=θP=(3–√2cosθ,3–√2sinθ)∠POB=θP=(32cos⁡θ,32sin⁡θ)

A=(0,3–√)