Consider the following for the next three (03) items that follow : In the triangle ABC, AB = 6 cm, BC = 8 cm and AC = 10 cm. The perpendicular dropped from B meets the side AC at D. A circle of radius BD (with centre B) cuts AB and BC at P and Q respectively as shown in the figure. 
What is the length of QC ?
3.2 cm
Understanding the Triangle and Circle Problem
The problem describes a triangle ABC with specific side lengths and a perpendicular dropped from one vertex. A circle is then drawn with the vertex as the center and the length of the perpendicular as the radius. We need to find the length of a segment created by the intersection of this circle with one of the triangle's sides.
Analyzing the Given Information
- Triangle ABC with AB = 6 cm, BC = 8 cm, AC = 10 cm.
- BD is perpendicular from B to AC, where D is on AC.
- A circle with center B and radius BD cuts AB at P and BC at Q.
- We need to find the length of QC.
Step-by-Step Solution to Find QC Length
Step 1: Identify the type of triangle ABC
Let's check if the triangle ABC is a right-angled triangle using the Pythagorean theorem. We have side lengths 6 cm, 8 cm, and 10 cm.
Square of the longest side: \(AC^2 = (10)^2 = 100\)
Sum of squares of the other two sides: \(AB^2 + BC^2 = (6)^2 + (8)^2 = 36 + 64 = 100\)
Since \(AB^2 + BC^2 = AC^2\), the triangle ABC is a right-angled triangle, with the right angle at vertex B.
This is important because in a right triangle, the two shorter sides (legs) can be used as base and height to calculate the area.
Step 2: Calculate the area of triangle ABC
Using the legs AB and BC as base and height, the area of triangle ABC is:
Area \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC\)
Area \(= \frac{1}{2} \times 6 \text{ cm} \times 8 \text{ cm} = \frac{1}{2} \times 48 \text{ cm}^2 = 24 \text{ cm}^2\)
Step 3: Calculate the length of the perpendicular BD
The perpendicular BD is the height when AC is taken as the base. The area of the triangle can also be calculated using AC and BD:
Area \(= \frac{1}{2} \times AC \times BD\)
We know the area is 24 cm\(^2\) and AC is 10 cm. Let's substitute these values to find BD:
\(24 \text{ cm}^2 = \frac{1}{2} \times 10 \text{ cm} \times BD\)
\(24 = 5 \times BD\)
\(BD = \frac{24}{5} \text{ cm} = 4.8 \text{ cm}\)
So, the length of the perpendicular BD is 4.8 cm.
Step 4: Determine the radius of the circle
The problem states that a circle of radius BD (with center B) cuts AB and BC at P and Q respectively. Therefore, the radius of the circle is equal to the length of BD.
Radius of circle \(= BD = 4.8 \text{ cm}\).
Step 5: Find the length of BQ
The point Q is on the side BC, and it is one of the points where the circle with center B cuts the triangle's sides. Since B is the center and Q is on the circle, the distance from B to Q is equal to the radius of the circle.
\(BQ = \text{Radius of circle} = BD = 4.8 \text{ cm}\).
Step 6: Calculate the length of QC
The point Q lies on the side BC. The total length of BC is given as 8 cm. The segment BC is divided into two parts by point Q: BQ and QC. Therefore, the length of BC is the sum of the lengths of BQ and QC.
\(BC = BQ + QC\)
We know BC = 8 cm and BQ = 4.8 cm. We can find QC by rearranging the equation:
\(QC = BC - BQ\)
\(QC = 8 \text{ cm} - 4.8 \text{ cm}\)
\(QC = 3.2 \text{ cm}\)
Thus, the length of QC is 3.2 cm.
Revision Table: Key Geometric Concepts
| Concept | Description | Application in Problem |
|---|---|---|
| Pythagorean Theorem | In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (\(a^2 + b^2 = c^2\)). | Used to identify that triangle ABC is right-angled at B. |
| Area of a Triangle | \(\frac{1}{2} \times \text{base} \times \text{height}\). Can be calculated using any side as base and its corresponding altitude as height. | Used to calculate the area of triangle ABC in two ways to find the length of the altitude BD. |
| Circle Properties | All points on the circumference of a circle are equidistant from the center (this distance is the radius). | Used to determine that BQ equals the radius BD, since Q is on the circle and B is the center. |
Additional Information: Right Triangle Properties
Right triangles have many special properties that are useful in geometry problems:
- The altitude to the hypotenuse in a right triangle divides the triangle into two smaller triangles that are similar to the original triangle and to each other.
- The length of the altitude to the hypotenuse (BD in this case) can also be found using the formula \(\frac{AB \times BC}{AC}\) for a right triangle at B. Using the values: \(\frac{6 \times 8}{10} = \frac{48}{10} = 4.8\) cm. This matches our calculation.
- The point D on AC is the foot of the altitude. The lengths of the segments AD and DC can also be calculated using similarity or projection formulas. For example, \(AB^2 = AC \times AD\), so \(6^2 = 10 \times AD\), which means \(AD = 3.6\) cm. Similarly, \(BC^2 = AC \times DC\), so \(8^2 = 10 \times DC\), meaning \(DC = 6.4\) cm. Note that \(AD + DC = 3.6 + 6.4 = 10\) cm, which is AC.
- The points P and Q are located on the sides AB and BC respectively, at a distance equal to the radius (BD = 4.8 cm) from B. BP = 4.8 cm and BQ = 4.8 cm. Since AB = 6 cm, AP = AB - BP = 6 - 4.8 = 1.2 cm. Since BC = 8 cm, QC = BC - BQ = 8 - 4.8 = 3.2 cm. These segments (AP, PB, BQ, QC) can be relevant for other questions related to the same figure.
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