Question

ABCD is a rhombus in which $\angle C={60}^{\circ }$.Then ,AC:BD=?

(a) $\sqrt{3}:1$

(b) $\sqrt{3}:\sqrt{2}$

(c) 3:1

(d) 3:2

Answer 1

(a)√3:1

ABCD is a rhombus. So all of its sides are equal.
Now, BC = DC
⇒ ∠BDC = ∠DBC = x (Angles opposite to equal sides are equal)
Also, ∠BCD = 60°
∴ x + x + 60° = 180°
⇒ 2x = 120°
⇒ x = 60°
i.e., ∠BCD = ∠BDC = ∠DBC = 60°
So, ∆BCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:
AB^2= OA^2 + OB^2
⇒OA^2=AB^2-OB^2 =a^2-a/2^2 =a^2-a^2/4 =3a^2/4
⇒OA^2=3a^2/4
⇒OA=√3a2/4
⇒OA=√3a/2
Now, AC=2×OA=2×√3a/2 =√3a
∴ AC:BD=√3a:a =√3 :1