CBSE Class 8 Mathematics
Question 8 of 8
The Baudhayana-Pythagoras Theorem — Question 9
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Question 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.]
Let ΔABC be an equilateral triangle. AB = BC = CA = 6 cm Let AD be perpendicular to BC. Then ∠1 = ∠2 (each = 90°) AB = AC (each = 6 cm) AD = AD (common) ΔADB ≅ ΔADC (RHS) BD = DC (CPCT) ∴ BD = DC = \(\frac {1}{2}\) × 6 cm = 3 cm In ΔADC, h 2 + 3 2 = 6 2 (Baudhayana’s triple) ⇒ h 2 = 36 – 9 = 27 ⇒ h = \(\sqrt{3 \times 3 \times 3}\) ⇒ h = 3√3 cm Ar ABC = \(\frac {1}{2}\) × BC × AD = \(\frac {1}{2}\) × 6 × 3√3 sq. units = 9√3 sq. units