On a triangle $A B C$, a circle with diameter $B C$ is drawn, intersecting $A B$ and $A C$ at points $P$ and $Q$, respectively. If the lengths of $A B, A C$, and $C P$ are $30 \mathrm{~cm}, 25 \mathrm{~cm}$, and $20 \mathrm{~cm}$ respectively, then the length of $BQ$ , in cm , is

We have triangle ABC with AB = 30 cm, AC = 25 cm, and CP = 20 cm. A circle is drawn with BC as its diameter, intersecting AB at P and AC at Q.

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When a point lies on a circle and we connect it to the endpoints of a diameter, the angle formed is always 90°.

Since BC is the diameter of our circle, point P lies on the circle, so $\angle BPC = 90°$, and point Q lies on the circle, so $\angle BQC = 90°$.

This means PC ⊥ AB (PC is perpendicular to AB) and BQ ⊥ AC (BQ is perpendicular to AC).

Now we have the heights of triangle ABC from two different vertices.

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For any triangle, we can calculate area using different base-height combinations:

Area of triangle ABC = $\tfrac{1}{2} \times \text{base} \times \text{height}$

Using AB as base: Area = $\tfrac{1}{2} \times \text{AB} \times \text{PC}$

Using AC as base: Area = $\tfrac{1}{2} \times \text{AC} \times \text{BQ}$

Since both expressions equal the same area:

$\tfrac{1}{2} \times \text{AB} \times \text{PC} = \tfrac{1}{2} \times \text{AC} \times \text{BQ}$

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$\tfrac{1}{2} \times 30 \times 20 = \tfrac{1}{2} \times 25 \times \text{BQ}$

The $\tfrac{1}{2}$ cancels out from both sides:

$30 \times 20 = 25 \times \text{BQ}$

$600 = 25 \times \text{BQ}$

$\text{BQ} = \dfrac{600}{25} = 24$

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BQ = 24 cm

When you see a circle with a diameter, immediately think about the angle in a semicircle theorem. It often creates perpendicular lines that can be used as heights for area calculations.