In Pythagoras exercise we have questions which says DB=3CD, or something like that is given to us in the question. And with this we find out the measurements of the sides, how to find those lengths.

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Solution

let CD = x

DB = 3CD = 3x

According to Pythagoras theorem in the triangle BDC

DB should be hypotaneous and CD should be base or lenght

therefore BC = √(DB^2 - CD^2)

BC = √(9x^2 - x^2) = √8x^2 = 2√2x

Hence side BC = 2√2CD, hypotaneous DB = 3CD

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In Pythagoras exercise we have questions which says DB=3CD, or something like that is given to us in the question. And with this we find out the measurements of the sides, how to find those lengths.

let CD = x DB = 3CD = 3x According to Pythagoras theorem in the triangle BDC DB should be hypotaneous and CD should be base or lenght therefore BC = √(DB^2 - CD^2) BC = √(9x^2 - x^2) = √8x^2 = 2√2x Hence side BC = 2√2CD, hypotaneous DB = 3CD

let CD = x

DB = 3CD = 3x

According to Pythagoras theorem in the triangle BDC

DB should be hypotaneous and CD should be base or lenght

therefore BC = √(DB^2 - CD^2)

BC = √(9x^2 - x^2) = √8x^2 = 2√2x

Hence side BC = 2√2CD, hypotaneous DB = 3CD