Question

Find the area of a rhombus with side length 6 and an interior angle with measure $120^\circ$.

Answer 1

Answer:

The area of a rhombus is $18\sqrt{3}$ square units.

Step-by-step explanation:

Side length of rhombus = 6 units.

Interior angle of rhombus = 120°

another Interior angle of rhombus = 180°-120° = 60°.

Draw an altitude.

In a right angled triangle

$\sin \theta=\frac{opposite}{hypotenuse}$

$\sin (60)=\frac{h}{6}$

$\frac{\sqrt{3}}{2}=\frac{h}{6}$

Multiply both sides by 6.

$3\sqrt{3}=h$

The height of the rhombus is $3\sqrt{3}$.

Area of a rhombus is

$Area=base\times height$

$Area=6\times 3\sqrt{3}$

$Area=18\sqrt{3}$

Therefore, the area of a rhombus is $18\sqrt{3}$ square units.

Answer 2

Step-by-step 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 $2 \cdot 9 \sqrt{3} = \boxed{18 \sqrt{3}}$.