A Nice Geometry Problem | r=?

Learn how to find the radius of the circle. Important Geometry and skills are also explained: area of a square formula; similar triangles; area of a triangle formula; Pythagorean Theorem; and right triangles. Step-by-step tutorial by FA_Math
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[MonishPratapSingh]

Please make videos of some more tough questions.

[danieljennings]

Here's a much easier way: (I am almost sure that this works)
The 64 square has height of 8, the 49 square has a width of 7, and the 81 square has a height of 9, so if you connect the top right corner of the 64 square to the top left corner of the 81 square then the Pythagorean theorem says that this new line has a length equal to the square root of 50. From this we can construct a square with area 50, and this new square fits perfectly inside the yellow circle (I am not sure how to prove that it fits perfectly, and this might not always be the case, but I can just somehow tell that it fits perfectly as far as this particular problem goes). This means that the yellow circle's radius is equal to half the new square's diagonal. And we know that the diagonal of any square is the side length multiplied by the square root of 2. So multiply the square root of 50 by the square root of 2 to get 10, and then cut it in half to get 5.

[quigonkenny]

Given the areas shown, the side lengths of the three squares are p = √64 = 8, b = √49 = 7, and g = √81 = 9. Let O be the center of the circle and P, B, and G be the points of contact between circle O and the respective squares. Let M and N be the top left and top right vertices of the blue square respectively.

Draw OP, OB, and OG, as well as PB and BG. OP, OB, and OG are all radii of circle O and thus length r. As MN, the top side of the blue square, is tangent to circle O at B, ∠OBM = ∠OBN = 90°.

Drop a perpendicular from P to OB at E. As p-b = 1, OE = r-1. Drop a perpendicular from G to OB at F. As g-b = 2, OF = r-2. Let MB = PE = x, so BN = FG = 7-x.

Triangle ∆PEO:
PE² + OE² = OP²
x² + (r-1)² = r²
x² + r² - 2r + 1 = r²
x² - 2r + 1 = 0
2r = x² + 1

Triangle ∆GFO:
FG² + OF² = OG²
(7-x)² + (r-2)² = r²
49 - 14x + x² + r² - 4r + 4 = r²
53 - 14x + x² - 2(2r) = 0
x² - 14r + 53 - 2(x²+1) = 0
-x² - 14r + 51 = 0
x² + 14r - 51 = 0
(x+17)(x-3) = 0
x = -17 ❌ | x = 3 <== x ≥ 0

2r = (3)² + 1
2r = 9 + 1 = 10
r = 5

[mikegraham]

Are there other ways to get the solution?

[a]

a=8 b=7 c=9
r²=x²+(r+b-a)²=x²+(r-1)²


r²=x²+r²-2r+1
r²=49+x²-14x+r²+4-4r

49-14x+3-2r=0
49+2x²-14x+5-6r=0

x²+14x-51=0
x=-17 x=3, x can be only +
x²-2r+1=0
r=(x²+1)/2=5