We are given a right triangle $LMN$ with $|LM| = 6$ cm, $\angle LMN = 90^\circ$, $\angle LNM = x$, and $\sin x = \frac{3}{5}$. We want to find the area of triangle $LMN$.

1. Problem Description

We are given a right triangle $LMN$ with $|LM| = 6$ cm, $\angle LMN = 90^\circ$, $\angle LNM = x$, and $\sin x = \frac{3}{5}$. We want to find the area of triangle $LMN$.

2. Solution Steps

Since $\sin x = \frac{3}{5}$, we can form a right triangle with opposite side 3 and hypotenuse

5. Using the Pythagorean theorem, the adjacent side is $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.

Then $\cos x = \frac{4}{5}$ and $\tan x = \frac{3}{4}$.

In triangle $LMN$, we have $\angle LMN = 90^\circ$. Also, $|LM| = 6$ cm. Since $\angle LNM = x$, we can write

$\tan x = \frac{LM}{MN} = \frac{6}{MN}$.

We are given that $\sin x = \frac{3}{5}$, so $\tan x = \frac{3}{4}$. Therefore, we have

$\frac{6}{MN} = \frac{3}{4}$

$MN = \frac{6 \times 4}{3} = \frac{24}{3} = 8$ cm.

The area of triangle $LMN$ is

Area $= \frac{1}{2} \times LM \times MN = \frac{1}{2} \times 6 \times 8 = \frac{48}{2} = 24$ cm$^2$.

3. Final Answer

C. 24 cm$^2$