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Question
From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `5sqrt(2)`, then what is the height of ∆ABC?
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Solution
AB = BC ...[Given]
∴ ∠A = ∠C ...[Isosceles triangle theorem]
Let ∠A = ∠C = x ...(i)
In ∆ABC, ∠A + ∠B + ∠C = 180° ...[Sum of the measures of the angles of a triangle is 180°]
∴ x + 90° + x = 180° ...[From (i)]
∴ 2x = 90°
∴ x = `90^circ/2`
∴ x = 45°
∴ ∠A = ∠C = 45°
∴ ∆ABC is a 45° – 45° – 90° triangle.
∴ AB = BC = `1/sqrt(2) xx AC` ...[Side opposite to 45°]
= `1/sqrt(2) xx 5sqrt(2)`
∴ AB = BC = 5 units
∴ The height of ∆ABC is 5 units.
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