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On a triangle $\text{ABC}$, a circle with diameter $\text{BC}$ is drawn, intersecting $\text{AB}$ and $\text{AC}$ at points $\text{P}$ and $\text{Q}$, respectively. If the lengths of $\text{AB, AC}$, and $\text{CP}$ are $30$ cm, $25$ cm, and $20$ cm respectively, then the length of $\text{BQ}$, in cm, is __________

Given that,

$\text{BC}$ is a diameter of circle, the lengths of $\text{AB} = 30 \; \text{cm}, \text{AC} = 25 \; \text{cm},$ and $\text{CP} = 20 \; \text{cm}.$

Now, we can draw the diagram :

The angle inscribed in a semicircle is always a right angle.

So, $\triangle \text{BPC},$ and $\triangle \text{BQC}$ are right angle triangle.

Now, we can draw the $\triangle \text{ABC}$ as :

Area of the $\triangle \text{ABC} = \frac{1}{2} \times \text{Base} \times \text{Height}$

$\quad = \frac{1}{2} \times \text{AB} \times \text{CP}$

$\quad = \frac{1}{2} \times 30 \times 20$

$\quad = 300 \; \text{cm}^{2}$

Again, we can draw the $\triangle \text{ABC}$ as :

Area of the $\triangle \text{ABC} = \frac{1}{2} \times \text{AC} \times \text{BQ}$

$\Rightarrow 300 = \frac{1}{2} \times 25 \times \text{BQ}$

$\Rightarrow 25 \; \text{BQ} = 600$

$\Rightarrow \text{BQ} = \frac{600}{25}$

$\Rightarrow \boxed{\text{BQ} = 24 \; \text{cm}}$

$\therefore$ The length of $\text{BQ}$ is $24 \; \text{cm}.$

Correct Answer $: 24$

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