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

Question

let mid "“ point of sides of $\Delta$ are $(\frac{5}{2}, 3), (\frac{5}{2}, 7) \, \& \, (4, 5)$. If incentre is $(h, k)$ then value of $3h + k$ is:

Answer 1

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

The problem gives us:

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:

Answer 2