Introduction
This guide explores the method to find the altitude (BD) of a right-angled triangle (ABC) where ∠B 90°. The triangle is defined by the given lengths of AB 6 cm and BC 8 cm. We will demonstrate the solution in a step-by-step manner, harnessing various geometric and trigonometric approaches to ensure a clear understanding for SEO purposes.
Understanding the Problem
Given a right-angled triangle ABC with ∠B 90°, where the lengths of the sides are AB 6 cm and BC 8 cm. We are tasked with finding the length of BD, which is perpendicular to AC.
Method 1: Using the Area Formula
Theorem: The area of a triangle can be expressed as half the product of the base and the height.
Step 1: Calculate the length of the hypotenuse AC.
H2 AB2 BC2 62 82 36 64 100
AC sqrt{100} 10 cm.
Step 2: Use the area formula to determine BD.
[frac{1}{2} cdot AC cdot BD frac{1}{2} cdot AB cdot BC]
[frac{1}{2} cdot 10 cdot BD frac{1}{2} cdot 6 cdot 8]
[10 cdot BD 6 cdot 8]
[BD frac{48}{10} 4.8 text{ cm}]
Method 2: Using Similar Triangles
This approach leverages the similarity of triangles ADB and ABC to derive the relation between BD and the sides of the triangle.
Step 1: Note that ∠ABC 90°, making BD perpendicular to AC.
Step 2: Apply the similarity of triangles ADB and ABC.
[frac{BD}{BC} frac{AB}{AC}]
[frac{BD}{8} frac{6}{10}]
[BD frac{6 cdot 8}{10} frac{48}{10} 4.8 text{ cm}]
Geometric Visualization and Proofs
The diagram below provides a visual representation of the triangle and the altitude BD through the Geogebra tool.
Figure: TriangleABC right-angled at B. BD is perpendicular to ACProving the Problem with Pythagoras' Theorem
Using the Pythagorean theorem, we can validate the hypotenuse AC:
H2 P2 B2
H2 62 82 36 64 100
H sqrt{100} 10 cm.
Applying the ratio of similar triangles ADB and ABC:
[frac{BD}{BC} frac{AB}{AC}]
[frac{BD}{8} frac{6}{10}]
[BD frac{6 cdot 8}{10} 4.8 text{ cm}]
Conclusion
The value of BD in the right-angled triangle where AB 6 cm, BC 8 cm, and ∠B 90° is 4.8 cm. This guide explains the process with a complete geometric proof, utilizing both the area method and the geometric similarity method.