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Given : point P outsides circle with center O such that PQ = 26 cm & and PQ touches the circle at Q .
To find : Diameter of Circle
Solution:
PQ = 10 cm ( tangent )
PO = 26 cm
OQ = radius
PO² = PQ² + OQ²
=> 26² = 10² + OQ²
=> 676 = 100 + OQ²
=> 576 = OQ²
=> 24² = OQ²
=> 24 = OQ
Radius = 24 cm
Diameter = 2 * radius = 2 * 24 = 48 cm
Diameter of circle = 48 cm
AB = x AC = 32 - x
AC² = AB² + BC²
=> (32 - x)² = x² + 24²
=> 32² + x² - 64x = x² + 24²
=> 64x = 448
=> x = 7
Area of Triangle = (1/2) * 24 * 7
Area of Triangle = (1/2) * (AB + BC + AC) * ( radius of incircle)
AB + BC + AC = 32 + 24 = 56
(1/2) * (56) * ( radius of incircle) = (1/2) * 24 * 7
=> radius of incircle = 3 cm
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REFER TO ATTACHMENT MATE!!!!!!
