Question

please find out the answer of this question ​

Answer 1

Answer:

If this is true, then O, the center of that circle, is the intersection of all the perpendiculars dropped on the middles of segments AE, AD, & ED. Let OA = OE=OD = r, and a = the side of the equilateral triangle & also side of the square.

Note that in the isosceles triangle ACD, angle C (in red) is 150 degrees, so the blue angles are all 15 degrees. Then the green angle EDO is 60 degrees. In that little right-angled triangle at the bottom-right, with hypotenuse OD= r, you will have:

cos(60 degrees) = (1/2 of a) divided by r, so then:

1/2 = a/(2r), so r = a, meaning, the radius of that circle is whatever the side of the equilateral triangle is.

Step-by-step explanation:

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Answer 2

If this is true, then O, the center of that circle, is the intersection of all the perpendiculars dropped on the middles of segments AE, AD, & ED. Let OA = OE=OD = r, and a = the side of the equilateral triangle & also side of the square.

Note that in the isosceles triangle ACD, angle C (in red) is 150 degrees, so the blue angles are all 15 degrees. Then the green angle EDO is 60 degrees. In that little right-angled triangle at the bottom-right, with hypotenuse OD= r, you will have:

cos(60 degrees) = (1/2 of a) divided by r, so then:

1/2 = a/(2r), so r = a, meaning, the radius of that circle is whatever the side of the equilateral triangle is.