In a parallelogram ABCD , AB= 6 cm , BC = 5 cm , and AC= 7 cm . Find the perpendicular distance between `bar(AB) and bar(CD)` .

Options

A. `6sqrt(6)` cm
B. `12sqrt(6) cm`
C. 5 cm
D. `2sqrt(6)` cm

Answer

To find the perpendicular distance between the lines \( \overline{AB} \) and \( \overline{CD} \) in the parallelogram ABCD, we can follow these steps:

### Step 1: Identify the sides and diagonal of the parallelogram
Given:
- \( AB = 6 \, \text{cm} \)
- \( BC = 5 \, \text{cm} \)
- \( AC = 7 \, \text{cm} \)

### Step 2: Calculate the semi-perimeter of triangle \( ABC \)
The semi-perimeter \( s \) is given by the formula:
\[ s = \frac{AB + BC + AC}{2} \]
Substituting the values:
\[ s = \frac{6 + 5 + 7}{2} = \frac{18}{2} = 9 \, \text{cm} \]

### Step 3: Calculate the area of triangle \( ABC \) using Heron's formula
Heron's formula for the area \( A \) of a triangle is:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Where \( a = AB = 6 \, \text{cm} \), \( b = BC = 5 \, \text{cm} \), and \( c = AC = 7 \, \text{cm} \).

Substituting the values:
\[ A = \sqrt{9(9-6)(9-5)(9-7)} = \sqrt{9 \times 3 \times 4 \times 2} \]
Calculating inside the square root:
\[ = \sqrt{216} = 6\sqrt{6} \, \text{cm}^2 \]

### Step 4: Calculate the area of the parallelogram
The area of the parallelogram is twice the area of triangle \( ABC \):
\[ \text{Area of parallelogram} = 2 \times 6\sqrt{6} = 12\sqrt{6} \, \text{cm}^2 \]

### Step 5: Use the area to find the height
The area of a parallelogram can also be expressed as:
\[ \text{Area} = \text{Base} \times \text{Height} \]
Here, the base is \( AB = 6 \, \text{cm} \). Let \( h \) be the height (the perpendicular distance between \( \overline{AB} \) and \( \overline{CD} \)):
\[ 12\sqrt{6} = 6 \times h \]
Solving for \( h \):
\[ h = \frac{12\sqrt{6}}{6} = 2\sqrt{6} \, \text{cm} \]

### Conclusion
The perpendicular distance between \( \overline{AB} \) and \( \overline{CD} \) is \( 2\sqrt{6} \, \text{cm} \).

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