question 21
please write answer fast tomorrow is my paper
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Since the central angles are equal, the sides AB, BC and CA are also equal.
Let the length of each side of the equilateral triangle be 2x.
Draw AM perpendicular to BC
Since,∆ABC is an equilateral triangle.
Therefore, AM passes through O.
BM = ½BC = ½(2x) = x
Now, In right angled ∆ABM,
We have AM²+BM²=AB²
AM²=AB² - BM² = (2x)²-x² = 4x² - x² = 3x²
AM = √3x
Now,
OM = AM - OA =(√3x-2√3)
Again, In right angled ∆OBM,
(2√3)² = x² + (√3x-2√3)²
12 = x² +3x² - 12x + 12
4x² = 12x
x² = 3x
x = 3
So, each side = 2x = 2×3 = 6
Let the length of each side of the equilateral triangle be 2x.
Draw AM perpendicular to BC
Since,∆ABC is an equilateral triangle.
Therefore, AM passes through O.
BM = ½BC = ½(2x) = x
Now, In right angled ∆ABM,
We have AM²+BM²=AB²
AM²=AB² - BM² = (2x)²-x² = 4x² - x² = 3x²
AM = √3x
Now,
OM = AM - OA =(√3x-2√3)
Again, In right angled ∆OBM,
(2√3)² = x² + (√3x-2√3)²
12 = x² +3x² - 12x + 12
4x² = 12x
x² = 3x
x = 3
So, each side = 2x = 2×3 = 6
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