Question

\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

• A. \(\frac{21}{4} \text{ cm}\)
• B. \(\frac{28}{3} \text{ cm}\)
• C. \(\frac{12}{7} \text{ cm}\)
• D. \(5.5 \text{ cm}\)

Hint:

Identify parallel lines and look for the "bow-tie" or "nested" triangle configurations to quickly spot similar triangles in geometry problems.

Answer 1

The Correct Option is B

To find the length of \(FD\) in the parallelogram \(ABCD\), let's use the properties of parallelograms and triangles.

The problem gives us:

• \(AF = 7 \text{ cm}\)
• \(FB = 3 \text{ cm}\)
• \(EF = 4 \text{ cm}\)

From the diagram, you can see that \(F\) is a point on \(AB\), dividing it into segments \(AF\) and \(FB\).

Let's use the property of similar triangles. In parallelogram \(ABCD\), the triangles \(\triangle ADF\) and \(\triangle CBF\) are similar.

Since \(\triangle ADF \sim \triangle CBF\), we have:

\[ \frac{AF}{FB} = \frac{FD}{CF} \]

Given:

\[ AF = 7 \text{ cm}, \quad FB = 3 \text{ cm}, \quad EF = 4 \text{ cm} \]

First, find \(CF\):

\[ CF = EF = 4 \text{ cm} \]

Now, substitute the known values into the ratio:

\[ \frac{7}{3} = \frac{FD}{4} \]

Cross-multiply to solve for \(FD\):

\[ FD = \frac{7 \times 4}{3} = \frac{28}{3} \text{ cm} \]

Therefore, the length of \(FD\) is \(\frac{28}{3} \text{ cm}\).

The correct answer is:

\[ \frac{28}{3} \text{ cm} \]

Let's include the diagram for reference: