The measure of each of the two opposite angles of a rhombus is 60° and the measure of one of its sides is 10 cm. The length of its smaller diagonal is

Options

A. 10 cm
B. `10 sqrt3` cm
C. `10 sqrt(2)` cm
D. `(5)/(2) sqrt2` cm

Answer

To find the length of the smaller diagonal of the rhombus, we can follow these steps:

### Step 1: Understand the properties of the rhombus
A rhombus has opposite angles that are equal, and the sum of the angles in any quadrilateral is 360°. Given that two opposite angles are 60°, the other two angles must also be 120° each (since 360° - 60° - 60° = 240°, and 240°/2 = 120°).

### Step 2: Draw the rhombus and label the angles
Let’s label the rhombus as ABCD, where angle B = angle D = 60° and angle A = angle C = 120°.

### Step 3: Use the properties of triangles
In triangle ABC, we know that AB = BC = 10 cm (since all sides of a rhombus are equal). We can also find the lengths of the diagonals using the properties of triangles.

### Step 4: Use the sine rule or cosine rule
Since we have an isosceles triangle (triangle ABC), we can use the cosine rule to find the length of diagonal AC. The cosine rule states:
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \]
Where:
- \( c \) is the side opposite angle C,
- \( a \) and \( b \) are the other two sides,
- \( C \) is the angle between sides a and b.

In our case:
- \( a = 10 \) cm (AB),
- \( b = 10 \) cm (BC),
- \( C = 60° \).

### Step 5: Calculate the length of diagonal AC
Using the cosine rule:
\[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(60°) \]
\[ AC^2 = 10^2 + 10^2 - 2 \cdot 10 \cdot 10 \cdot \frac{1}{2} \]
\[ AC^2 = 100 + 100 - 100 \]
\[ AC^2 = 100 \]
\[ AC = \sqrt{100} = 10 \text{ cm} \]

### Step 6: Find the smaller diagonal
Since the diagonals of a rhombus bisect each other at right angles, we can find the smaller diagonal BD using the properties of the triangle formed by the diagonals. Since AC is 10 cm, we can use the Pythagorean theorem in triangle AOD (where O is the intersection of the diagonals).

Let \( BD = d \). Then:
\[ AO = \frac{AC}{2} = \frac{10}{2} = 5 \text{ cm} \]
Using the Pythagorean theorem:
\[ AO^2 + OD^2 = AD^2 \]
Where \( AD = 10 \text{ cm} \) (side of the rhombus).
\[ 5^2 + \left(\frac{d}{2}\right)^2 = 10^2 \]
\[ 25 + \frac{d^2}{4} = 100 \]
\[ \frac{d^2}{4} = 100 - 25 \]
\[ \frac{d^2}{4} = 75 \]
\[ d^2 = 300 \]
\[ d = \sqrt{300} = 10\sqrt{3} \text{ cm} \]

### Final Answer
The length of the smaller diagonal BD is \( 10\sqrt{3} \) cm.