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Q7) From a semicircular region OABCD, a triangle ABD in which AB = 3cm and BD = 4cm is removed.Find the perimeter of the remaining figure.
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an triangle formed touches both ends of the diameter , then the triangle is right angle triangle
AO^2 = AB^2 + BD^2
AO^2 = 3^2 + 4^2
AO^2 = 9+ 16
AO^2 = 25 sq.cm
AO = 5 cm
Radius = 5/2 cm
perimeter of remaining figure = pi r + 3cm + 4cm
= 22/7 × 5/2 + 3cm + 4cm
= 55/7 + 3cm + 4cm
= 55 + 21 + 28/7 cm
= 104/7 cm
= 14 6/7 cm
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