ABCD is a parallelogram such that AF = 7 cm, FB = 3 cm and EF = 4 cm, length FD = equals
Show Hint
Whenever you see a line intersecting two parallel lines (such as the opposite sides of a parallelogram) and forming an "X" shape at an intersection point, the two triangles meeting at that point are always similar.
Setting up the ratio of the corresponding opposite sides of the "X" directly gives you the answer!
Step 1: Identify the similar triangles.
Since $AD \parallel BE$ (as $AD \parallel BC$ in the parallelogram and $E$ lies on $BC$), triangles $ADF$ and $EBF$ are similar by AA. Step 2: Write the similarity as a single ratio k.
Let $k = \frac{FD}{FB} = \frac{AF}{EF}$, the common ratio of corresponding sides. Step 3: Find k using the known sides, then find FD.
\[ k = \frac{AF}{EF} = \frac{7}{4} \]
Since $\frac{FD}{FB} = k$, we get $FD = k \times FB = \frac{7}{4} \times 3 = \frac{21}{4}$ cm.
\[ \boxed{FD = \frac{21}{4}\ \text{cm}} \]
