Squares inside circle problem | Radius of circle | Advanced math problems | Mathematics

This problem is an application of the properties of circles and chords, and the Pythagoras theorem. This problem may seem to be a little advanced, but a close observation will make it very easy.

Any queries regarding the subject or videos are invited.
Instagram ID : problem_analysis

#maths #mathematics #square #circle #chord #radius #diameter #advancedmathproblems #calculation #problems #geometry

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really cool videos, nice quality, keep it up

[francois]

A= circle contact point – first square on the left
B and C = lower and upper right quarter square contact points respectively
D = point of intersection between the circle and the extension of the vertical side first square on the left
AD and BC are two parallel chords of the circle crossed by the diameter at their midpoint of length respectively
6 and 8, half of AD is 3 and is the height with respect to the hypotenuse (equal to the diameter of the circle) of the right triangle whose vertices are the extremes E and F of the diameter passing through the midpoints of the chords AD and BC and vertex A of the first square on the left
for a theorem of Euclid, we have ... height squared of said triangle = product of the projections of the catheti AE (which we set = x) and AF= 8 or half of AD squared =9=
=8x hence x=9/8
the diameter is the sum of these projections i.e
9/8+8=73/8 and the radius =73/16=4.56

[thorinpalladino]

In determining the length of PC you say symmetry, but where was it established that point P must be the midpoint of OC?

[kevinmorgan]

The fact that the extended topmost side in the topmost square intersects the circle at C is nice... but unnecessary. The centre of the circle lies on the perpendicular bisector of the chord at B. So the centre must be on that line a little over 4 units from that chord. If you label that small distance as x and radius r, you can form two quadratics using Pythagoras. The x^2 term cancel leaving a simple value for r^2. Nice video.

[jimlocke]

Extend the sides of the leftmost square until they intersect the circle, forming 2 chords. Because those chords meet at a right angle, the chord joining those end points is a diameter. We know that the vertical chord, which is also a side of a right triangle, has length 6. If the horizontal chord, which is also the other side of the same right triangle, has length 7 then the diameter chord or hypotenuse, from Pythagorean Theorem, will have length √(85), the correct diameter given in the video. The challenge: how can we prove that the chord has length 7, without working backward (as I did!) from Problem Analysis's solution method?

[PremjitTalwar]

Property 2 is false. Imagine one of the chords is the diameter. The second chord is much shorter, and intersects the diameter at a point close to the perimeter, such that the two segments d and b are equal. It is clear in this case that ab will not equal to cd, since a is much larger than c and b=d.