find the area of a regular pentagon whose each side is 5cm.long and the radius of inscribed circle is 3.5cm
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28
Let us find the area of one of the 5 isosceles triangles in the pentagon.
Side of pentagon a = 5cm
Perpendicular distance between centre of pentagon and the side = 3.5 cm
Area of the traingle = (1/2)*5*3.5 = 8.75 cm²
Therefore, area of pentagon = 8.75 * 5 = 43.75 cm²
Side of pentagon a = 5cm
Perpendicular distance between centre of pentagon and the side = 3.5 cm
Area of the traingle = (1/2)*5*3.5 = 8.75 cm²
Therefore, area of pentagon = 8.75 * 5 = 43.75 cm²
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14
side =5cm radius =3.5cm
area of regular Pentagon =n÷2×x×r
=5÷2×5×3.5
=43.75
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