To find the radius of a circle inscribed in a square with given tangent segments, apply the Pythagorean theorem to a right triangle formed by the radius and tangent segments. Given a circle inscribed in a square with tangent segments of 9 cm and 2 cm, let the radius be r. The perpendicular distance from the center to the tangent is r, and the remaining segment is r-2. The hypotenuse is the full tangent length of 9 cm. Applying Pythagoras: (r-2)² + (9-r)² = r², which simplifies to r² - 4r + 4 + 81 - 18r + r² = r², further simplifying to r² - 22r + 85 = 0. Solving this quadratic equation gives r = 5 or r = 17. Since the radius must be less than the tangent length of 9 cm, the valid solution is r = 5 cm.
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Find the radius of circle. ............... #maths #geometry #olympiad #cds #cat #518 #cbse #ssc #cglAdded:
[clears throat] >> Welcome everyone. This interesting question, can you find the radius of this circle which is inscribed in a square?
Given is the length of this red line segment is 9 cm and length of this green line segment is 2 cm.
Let's check this interesting solution.
Let's start by assuming the radius of circle as r cm. Thus OA length will be r and OC length will be r.
We know that in a circle, radius is perpendicular to the tangent. So angle E is 90. Same logic, angle C is 90°.
Now let's drop a perpendicular from point A on OC.
Now in this right-angle triangle, if we can find this length AF and FO in terms of r, then we can apply Pythagoras theorem to solve it.
Now if you look at this figure ABCF, all the angles are 90, so it's a rectangle. And thus SC length is same as >> [music] >> 2 cm.
How about OF length? That will be full length r minus 2.
Now to get AF length, let's consider this figure OEDC.
It's a rectangle.
So OE is r, thus CD length will also be r.
Now this complete length BD, that's given as 9. So BC length will be 9 minus r.
And AF length, that will be same 9 minus r.
Now solving this triangle by Pythagoras theorem, we'll get r minus 2 squared plus 9 minus r squared.
This is equals to hypotenuse squared r, r squared.
Opening the bracket, applying A minus B whole squared identity, we'll get this as 81 plus r squared minus 18 r.
And this term will be r squared plus 4 minus 4 r. This is equals to r squared.
Solving it, both the side r squared will get cancelled out.
Simplifying it, we'll get a quadratic in terms of r.
Solving it by splitting the middle term, so -22 we'll write it down as -17 and -5.
Now, taking r common and factorizing it, we'll get this as r -17 * r -5 = 0.
So, the radius of circle, it can be 5 or it can be 17.
But, if you look carefully, if BD length is 9 cm, >> [music] >> then the radius has to be less than 9.
So, ignoring this higher value, we get radius of the circle as 5 cm and that's our answer. I hope you enjoyed the solution. I will see you in next video.
Till then, ta-ta, bye-bye.
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