When a square is inscribed in a circle with a chord passing through its center and vertex, the area of the square can be found using the intersecting chord theorem: if the chord is divided into segments of lengths 7 cm and 4 cm, and the square's side length is A, then A² = 7 × 4 = 28 cm², so the area of the square is 28 cm².
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Find area of square. ............ #maths #geometry #circle #olympiad #cds #cat #522 #cbse #ssc #cglAdded:
This amazing question given to us is a circle and a square inside the circle passing through the center of the circle.
This chord which is passing through the vertex of square, its length is given as 7 + 4 cm.
Now exciting part is with this data we have to find the area of square.
Solution is very short and simple.
Let's start by extending this side length CD to meet the circle at point G.
Now this angle D here, this is 90°.
And with respect to circle we have got chord GC is there.
And from center of circle a perpendicular is drawn.
So by theorem, the perpendicular from center of circle on a chord bisects the chord.
So if DC length is A, then GD length will be same A cm.
Now again with respect to circle, we have chord FE is there and GC is there intersecting at point D.
Applying intersecting chord theorem or power of point theorem, we get GD times of DC.
That's equals to ED times of DF.
Putting the values here, we get A times of A is 7 times of 4.
Or A squared is 28 and that's the magic.
A squared is nothing but area of square which you have to find. So area of this given square is 28 cm squared 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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