Question

In a rhombus, each side has length 6. One of the angles of the rhombus is
`$120^\circ.$` Find the area of the rhombus.

Answer 1

Explanation:

Let the rhombus be $ABCD$, where $\angle DAB = 120^\circ$. Then $\angle ABC = 180^\circ - \angle DAB = 180^\circ - 120^\circ = 60^\circ$.

[asy] unitsize(1 cm); pair A, B, C, D; A = (0,1); B = (sqrt(3),0); C = (0,-1); D = (-sqrt(3),0); draw(A--B--C--D--cycle); draw(A--C); label("$A$", A, N); label("$B$", B, E); label("$C$", C, S); label("$D$", D, W); label("$6$", (A + D)/2, NW); [/asy]

Since $AB = BC$, triangle $ABC$ is equilateral. By the same argument, triangle $ACD$ is also equilateral. Each triangle has area

\[\frac{\sqrt{3}}{4} \cdot 6^2 = 9 \sqrt{3},\]so the area of the rhombus is $2 \cdot 9 \sqrt{3} = \boxed{18 \sqrt{3}}$.

Answer 2

31.18 units²

Explanation:

The diagonals bisect the angles.

⇒ Angle at half the rhombus = 120 ÷ 2 = 60°

Both sides of the triangle are 6 units:

⇒ Half the rhombus is also an isosceles triangle.

⇒ the other angle is also 60°.

Both angles are 60°,

⇒ the third angle is 180 - 60 - 60 = 60°

⇒ it is an equilateral triangle.

Area of an equilateral triangle = √3/4 (6)² = 15.59 units²

The rhombus is made up of two of the triangles.

⇒ Area of the rhombus = 15.59 x 2 = 31.18 units²