what is the area of an isosceles triangle
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Area of any triangle ABC with sides a (= BC), b (= AC) and c (= AB), is given by
Δ = √[s(s-a)(s-b)(s-c)
where s = semiperimeter = (a+b+c)/2
Δ = 1/2 b * h
where b = base, h = height or altitude from base to the opposite vertex.
For an isosceles triangle, let b = c (or AC = AB), ∠C = ∠B. The altitude AD = h) divides BC into two. Here a = base.
h² = b² - a²/4 or h² = (4b² - a²)/4
b
Or, s = (a + b + b)/2 = (a+2b)/2
So Δ² = (a+2b)/2 * (2b-a)/2 * a/2 * a/2
Δ = 1/4 * a √ (4b² - a²)
Δ = √[s(s-a)(s-b)(s-c)
where s = semiperimeter = (a+b+c)/2
Δ = 1/2 b * h
where b = base, h = height or altitude from base to the opposite vertex.
For an isosceles triangle, let b = c (or AC = AB), ∠C = ∠B. The altitude AD = h) divides BC into two. Here a = base.
h² = b² - a²/4 or h² = (4b² - a²)/4
b
Or, s = (a + b + b)/2 = (a+2b)/2
So Δ² = (a+2b)/2 * (2b-a)/2 * a/2 * a/2
Δ = 1/4 * a √ (4b² - a²)
kvnmurty:
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