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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` .......[From (i)]

∴ 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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