$\Delta \mathrm{ABC}$ is a right-angled triangle where $m\angle \mathrm{ABC}=90°$. If $m\left(\overline{\mathrm{AB}}\right)=15\text{cm}$, $m\left(\overline{\mathrm{BC}}\right)=20\text{cm}$, and $\overline{\mathrm{BD}}\perp \overline{\mathrm{AC}}$, intersecting $\overline{\mathrm{AC}}$ at the point D, find $m\left(\overline{\mathrm{BD}}\right)$ (in cm).

The correct option is 12.

Step 1: Understand the given information
We are given a right-angled triangle $\Delta \mathrm{ABC}$ where:
• $m\angle \mathrm{ABC}=90°$
• $m\left(\overline{\mathrm{AB}}\right)=15\text{cm}$ (perpendicular side)
• $m\left(\overline{\mathrm{BC}}\right)=20\text{cm}$ (base)
• $\overline{\mathrm{BD}}$ is perpendicular to the hypotenuse $\overline{\mathrm{AC}}$, with $D$ lying on $\overline{\mathrm{AC}}$.

Step 2: Find the length of the hypotenuse $\overline{\mathrm{AC}}$ using the Pythagorean Theorem
In right-angled triangle $\Delta \mathrm{ABC}$:

$A{C}^2=A{B}^2+B{C}^2$

Substitute the given values:

$A{C}^2={15}^2+{20}^2$

$A{C}^2=225+400$

$A{C}^2=625$

$AC=\sqrt{625}=25\text{cm}$

Step 3: Calculate the altitude $m\left(\overline{\mathrm{BD}}\right)$ using the area of the triangle
The area of a triangle can be calculated using two different base-height combinations:

1. Using legs $\mathrm{AB}$ and $\mathrm{BC}$:
$\text{Area}=\frac{1}{2}\times \mathrm{AB}\times \mathrm{BC}=\frac{1}{2}\times 15\times 20=150{\text{cm}}^2$

2. Using hypotenuse $\mathrm{AC}$ and altitude $\mathrm{BD}$:
$\text{Area}=\frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}$

Equating both expressions for area:

$\frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}=\frac{1}{2}\times \mathrm{AB}\times \mathrm{BC}$

$\mathrm{AC}\times \mathrm{BD}=\mathrm{AB}\times \mathrm{BC}$

Substitute the known values:

$25\times \mathrm{BD}=15\times 20$

$25\times \mathrm{BD}=300$

$\mathrm{BD}=\frac{300}{25}=12\text{cm}$

Therefore, $m\left(\overline{\mathrm{BD}}\right)=12\text{cm}$.