If a rhombus has area $12 \mathrm{sq} \mathrm{cm}$ and side length $5 \mathrm{~cm}$, then the length, in cm, of its longer diagonal is

We have a rhombus with area = 12 sq cm and side length = 5 cm. We need to find the longer diagonal.

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Area of rhombus = $\dfrac{1}{2} \times d_1 \times d_2$ where $d_1$ and $d_2$ are the lengths of the two diagonals.

In a rhombus, the diagonals bisect each other at right angles, creating four right triangles.

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From the area formula:

$\text{Area} = \dfrac{1}{2} \times d_1 \times d_2 = 12$

$d_1 \times d_2 = 24$ ...(equation 1)

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When diagonals of a rhombus intersect, they create four right triangles.

Each right triangle has hypotenuse = side of rhombus = 5 cm and two legs = $\dfrac{d_1}{2}$ and $\dfrac{d_2}{2}$ (half of each diagonal).

Using the Pythagorean theorem:

$\left(\dfrac{d_1}{2}\right)^2 + \left(\dfrac{d_2}{2}\right)^2 = 5^2$

$\dfrac{d_1^2}{4} + \dfrac{d_2^2}{4} = 25$

$d_1^2 + d_2^2 = 100$ ...(equation 2)

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We have:

$d_1 \times d_2 = 24$ ...(equation 1)

$d_1^2 + d_2^2 = 100$ ...(equation 2)

Using the algebraic identity $(a + b)^2 = a^2 + b^2 + 2ab$:

$(d_1 + d_2)^2 = d_1^2 + d_2^2 + 2d_1d_2$

$(d_1 + d_2)^2 = 100 + 2(24) = 100 + 48 = 148$

Therefore: $d_1 + d_2 = \sqrt{148} = 2\sqrt{37}$ ...(equation 3)

For $(d_1 - d_2)^2$:

$(d_1 - d_2)^2 = d_1^2 + d_2^2 - 2d_1d_2$

$(d_1 - d_2)^2 = 100 - 2(24) = 100 - 48 = 52$

Therefore: $d_1 - d_2 = \sqrt{52} = 2\sqrt{13}$ ...(equation 4)

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Adding equations (3) and (4):

$(d_1 + d_2) + (d_1 - d_2) = 2\sqrt{37} + 2\sqrt{13}$

$2d_1 = 2(\sqrt{37} + \sqrt{13})$

$d_1 = \sqrt{37} + \sqrt{13}$

Subtracting equation (4) from equation (3):

$(d_1 + d_2) - (d_1 - d_2) = 2\sqrt{37} - 2\sqrt{13}$

$2d_2 = 2(\sqrt{37} - \sqrt{13})$

$d_2 = \sqrt{37} - \sqrt{13}$

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Since $\sqrt{37} > \sqrt{13}$, we have $d_1 = \sqrt{37} + \sqrt{13}$ (longer diagonal) and $d_2 = \sqrt{37} - \sqrt{13}$ (shorter diagonal).

Therefore, the length of the longer diagonal is $\sqrt{37} + \sqrt{13}$ cm.