Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]
Given: Equilateral triangle with side length = 6 units.
Step 1: In an equilateral triangle, the altitude bisects the base, creating two equal segments of 3 units each.
Step 2: This forms a right triangle with hypotenuse 6 (the side) and one leg 3 (half the base).
Step 3: Apply Baudhāyana's Theorem to find height h: h² + 3² = 6²
Step 4: Solve: h² + 9 = 36 h² = 27 h = = 3 units
Step 5: Calculate area: Area = (1/2) × base × height = (1/2) × $$6 \times 3$$$\sqrt{3}\sqrt{3}$ square units
Answer: The area is 9 15.59 square units.
Explanation
The chapter's application of Baudhāyana's Theorem extends to finding heights in geometric figures. In an equilateral triangle, the altitude from any vertex bisects the opposite side (by symmetry and congruence). This creates a right triangle with hypotenuse 6 (the side), one leg 3 (half the base), and the other leg being the height h. Using Baudhāyana's Theorem: h² + 3² = 6², so h = = 3. The area is then (1/2) × base × height = (1/2) × $$6 \times 3$$$\sqrt{3}\sqrt{3}$.
Solution Steps
Step 1: In an equilateral triangle, the altitude bisects the base, creating two equal segments of 3 units each.
Step 2: This forms a right triangle with hypotenuse 6 (the side) and one leg 3 (half the base).
Step 3: Apply Baudhāyana's Theorem to find height h: h² + 3² = 6²
Step 4: Solve: h² + 9 = 36, so h² = 27, giving h = = 3 units
Step 5: Calculate area: Area = (1/2) × base × height = (1/2) × $$6 \times 3$$$\sqrt{3}\sqrt{3}$ square units