Question

Need geometry help:

Answer 1

Radius of the incircle is r which is given by \(2 \times area\ of\ the \ triangle / perimeter\)

Answer 2

Now area of the triangle = \(4\sqrt2 \times 4\sqrt2 / 2 = 16\)
Length of AC = \(\sqrt{(4\sqrt2)^2+4\sqrt2)^2 } = 8\)
perimeter = \(4\sqrt2 + 4\sqrt2 + 8 = 8(1 + \sqrt2)\)

Answer 3

radius = \(2\times16 / 8(1+\sqrt2) = 4/(1+\sqrt2)\)

Answer 4

BI = \(\sqrt{(r^2 + r^2)} = r \sqrt{2}\)
So.\( BI = \frac{4}{1+\sqrt2} \times \sqrt2\)

Answer 5

Multiplying and dividing by \(1-\sqrt2\)
i.e \( \frac{4\sqrt2}{1+\sqrt2} \times \frac{1-\sqrt2}{1-\sqrt2} = 8 - 4\sqrt2\)

Answer 6

Oh, I get it now, you use the Pythagorean theorem with the inradius, thanks!

Answer 7

Yes