Question

From given figure, In ∆ ABC, AB⊥BC, AB =BC, AC = 5√2 then

what is the height of ∆ ABC ?​

Answer 1

Answer:

Given that

AB=BC and AB is perpendicular to BC

AC=5√2

There fore,

Triangle ABC is Right angle Triangle

Then angle BAC =angle BCA=45°

[Sum of the angles in the triangle=180°]

By pythagoras theorem

AC^2=AB^2+BC^2

as AB=BC

we can write

AC^2=AB^2+AB^2

25x2=2AB^2

25x2/2=AB^2

25=AB^2

√25=AB

AB=5

Ther fore Height of the triangle ABC is 5

Answer 2

Step-by-step explanation:

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