From a semicircular region OABCD,A triangle ABCD in which AB=3cm and BD =3cm is removed .Find the area of remaining portion
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Please find the solved answer along with the diagram attached.
We have : AB = 3 CM; BD = 4 CM
We know that, angle in a semi-circle is a right angle
So, ∠ABD = 90°
So, ∆ABD is a right ∆
Now, AD² = BD²+AB² =AD² = 4² + 3² = 16+9 = 25
=AD = 5 cm So, diameter, AD = 5 cm
Now, circumference of semi-circle OABCD = Π x Radius + diameter = 3.14x2.5+5 = 12.85 cm
Now, when the triangle ABD is removed then the figure becomes (attached file)
So, circumference of the remaining figure = 7.85+AB+BD = 7.85+3+4 = 14.85 cm
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