A square is inscribed in a right isosceles triangle, so that two of its vertices lie on the hypotenuse and two other on the legs. Find the length of the side of the square, if the length of the hypotenuse is 3 in.

The side of the square which is inscribed in a right isosceles triangle = 1 in

Step-by-step explanation:

 The length of each leg = 3/√2

    The area of the given triangle = (1/2) * (3/√2)*(3/√2) = 9/4

So, Area1 + Area2 + Area3 + square area = 9/4

   x²/4 + x²/2 + x²/2 +x² = 9/4

  (9/4) x² = 9/4

  x² = 1

x = 1

The side of the square which is inscribed in a right isosceles triangle = 1 in

The side of the square which is inscribed in a right isosceles triangle = 1 in

Step-by-step explanation:

 The length of each leg = 3/√2

    The area of the given triangle = (1/2) * (3/√2)*(3/√2) = 9/4

So, Area1 + Area2 + Area3 + square area = 9/4

   x²/4 + x²/2 + x²/2 +x² = 9/4

  (9/4) x² = 9/4

  x² = 1

x = 1

The side of the square which is inscribed in a right isosceles triangle = 1 in

goog may have it for you

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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