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 radius of the circle ?
4.8 cm
Triangle ABC Geometry Problem Analysis
We are given a triangle ABC with side lengths AB = 6 cm, BC = 8 cm, and AC = 10 cm. A perpendicular line segment BD is dropped from vertex B to the side AC, meeting AC at point D. A circle is drawn with its center at B and radius equal to the length of BD. We need to find the radius of this circle, which is the length of BD.
Identifying the Triangle Type using Pythagorean Theorem
Let's first determine if the triangle ABC is a right-angled triangle by checking if the side lengths satisfy the Pythagorean theorem (\(a^2 + b^2 = c^2\)). The longest side is AC = 10 cm, which would be the hypotenuse if it's a right triangle.
Calculate the square of the lengths of the two shorter sides:
- \(AB^2 = 6^2 = 36\)
- \(BC^2 = 8^2 = 64\)
Sum of the squares of the shorter sides: \(AB^2 + BC^2 = 36 + 64 = 100\).
Calculate the square of the longest side:
\(AC^2 = 10^2 = 100\).
Since \(AB^2 + BC^2 = AC^2\), the triangle ABC is indeed a right-angled triangle, with the right angle located at vertex B.
Calculating the Area of the Triangle
The area of a right-angled triangle can be calculated easily using the two sides that form the right angle as the base and height. In triangle ABC, AB and BC are the legs forming the right angle at B.
Area of triangle ABC \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC\)
Substitute the values of AB and BC:
Area \(= \frac{1}{2} \times 6 \text{ cm} \times 8 \text{ cm} = \frac{1}{2} \times 48 \text{ cm}^2 = 24 \text{ cm}^2\).
Finding the Perpendicular Height BD using Area
The area of any triangle can also be calculated using any side as the base and the corresponding altitude (perpendicular height) from the opposite vertex to that side.
In triangle ABC, we can consider AC as the base and BD as the corresponding height because BD is the perpendicular dropped from B to AC.
Area of triangle ABC \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BD\)
We know the area of triangle ABC is 24 cm² and AC is 10 cm. Substitute these values into the formula:
\(24 = \frac{1}{2} \times 10 \times BD\)
\(24 = 5 \times BD\)
Now, solve for BD:
\(BD = \frac{24}{5}\)
\(BD = 4.8 \text{ cm}\).
Determining the Circle's Radius
The problem states that the circle has a radius equal to the length of BD. Since we found BD = 4.8 cm, the radius of the circle is 4.8 cm.
Triangle Properties Revision Table
| Property | Description | Application in this Problem |
|---|---|---|
| Pythagorean Theorem | For a right triangle with legs a, b and hypotenuse c, \(a^2 + b^2 = c^2\). | Used to confirm ABC is a right triangle (\(6^2 + 8^2 = 10^2\)). |
| Area of a Triangle | General formula: \(\frac{1}{2} \times \text{base} \times \text{height}\). | Used to calculate area in two ways to find BD. |
| Altitude (Perpendicular Height) | A line segment from a vertex perpendicular to the opposite side. | BD is the altitude from B to AC. Its length is the circle's radius. |
Additional Information on Right Triangle Altitudes
In a right-angled triangle, the altitude drawn from the right angle to the hypotenuse has some special properties. The length of this altitude (BD in our case) is the geometric mean of the two segments it divides the hypotenuse into (AD and DC).
Also, there is a direct formula for the altitude from the right angle to the hypotenuse in a right triangle with legs a, b and hypotenuse c: \(h = \frac{a \times b}{c}\).
Using this formula for triangle ABC (a=6, b=8, c=10):
\(BD = \frac{AB \times BC}{AC} = \frac{6 \times 8}{10} = \frac{48}{10} = 4.8 \text{ cm}\).
This confirms our calculation using the area method. The length of the perpendicular BD, which is the radius of the circle, is 4.8 cm.
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Important Questions from Circles, Chords and Tangents
A chord 21 cm long is drawn in a circle of diameter 25 cm. The perpendicular distance of the chord from the centre is:
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