Calculating Parallelogram Diagonal Length

We are given a parallelogram with adjacent sides of lengths $4a$ and $3a$. The angle between these sides is $60^{\circ}$. We need to find the length of one of its diagonals.

Using the Law of Cosines

Let the adjacent sides of the parallelogram be $p = 4a$ and $q = 3a$. Let the angle between them be $\theta = 60^{\circ}$.

The length of a diagonal ($d_1$) can be found using the Law of Cosines applied to the triangle formed by sides $p$, $q$, and the diagonal $d_1$. The formula is:

$d_1^2 = p^2 + q^2 - 2pq \cos(\theta)$

Applying the Values

Substitute the given values into the formula:

$d_1^2 = (4a)^2 + (3a)^2 - 2(4a)(3a) \cos(60^{\circ})$

Calculate the squares of the sides:

$d_1^2 = 16a^2 + 9a^2 - 2(12a^2) \cos(60^{\circ})$

We know that $\cos(60^{\circ}) = \frac{1}{2}$. Substitute this value:

$d_1^2 = 16a^2 + 9a^2 - 2(12a^2) \left(\frac{1}{2}\right)$

Simplify the expression:

$d_1^2 = 25a^2 - 12a^2$

$d_1^2 = 13a^2$

Finding the Diagonal Length

Take the square root to find the length of the diagonal $d_1$:

$d_1 = \sqrt{13a^2}$

$d_1 = a\sqrt{13}$

This result matches option A.

The other diagonal ($d_2$) would use the angle $180^{\circ} - 60^{\circ} = 120^{\circ}$:

$d_2^2 = (4a)^2 + (3a)^2 - 2(4a)(3a) \cos(120^{\circ})$

$d_2^2 = 16a^2 + 9a^2 - 2(12a^2) \left(-\frac{1}{2}\right)$

$d_2^2 = 25a^2 + 12a^2 = 37a^2$

$d_2 = a\sqrt{37}$

Since $a\sqrt{13}$ is one of the options, it is the required diagonal length.