ABCDEF is a regular hexagon with centre O (in the following figure). If the area of triangle OAB is 9 cm², find the area of :
(i)the hexagon and
(ii)the circle in which the hexagon is incribed.
Answers
Answer:
The area of hexagon is 54 cm² and area of circle is 65.324 cm².
Step-by-step explanation:
Let ‘r’ be the radius of a circle and ‘a’ be the side of a equilateral triangle.
Given :
Area of ∆OAB (equilateral) = 9 cm²
Area of hexagon = 6 × Area of equilateral triangle
[A Regular hexagon is made up of 6 equal triangles]
(i) Area of hexagon = 6 × 9 = 54 cm²
Area of hexagon = 54 cm²
(ii) Area of equilateral ∆ = √3/4 × a²
9 = √3/4 × a²
9 × 4 = √3a²
36 = √3a²
a² = 36/√3
Side² = 36/√3 cm
Radius of circle,r = Side of a hexagon
[In regular hexagon inscribed in a circle,its side is equal to the radius of a Circle]
r² = 36/√3 cm
Area of circle,A = πr²
A = 22/7 × 36/√3
A = 22/7 × 36/1.732
A = (22 × 36) /(7 ×1.732)
[√3 = 1.732]
A = 792/12.124
A = 65.324 cm²
Area of circle = 65.324 cm²
Hence, the area of hexagon is 54 cm² and area of circle is 65.324 cm².
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Answer:
The area of hexagon is 54 cm² and area of circle is 65.324 cm².
Step-by-step explanation:
The area of hexagon is 54 cm² and area of circle is 65.324 cm².
Step-by-step explanation:
Let ‘r’ be the radius of a circle and ‘a’ be the side of a equilateral triangle.
Given :
Area of ∆OAB (equilateral) = 9 cm²
Area of hexagon = 6 × Area of equilateral triangle
[A Regular hexagon is made up of 6 equal triangles]
(i) Area of hexagon = 6 × 9 = 54 cm²
Area of hexagon = 54 cm²
(ii) Area of equilateral ∆ = √3/4 × a²
9 = √3/4 × a²
9 × 4 = √3a²
36 = √3a²
a² = 36/√3
Side² = 36/√3 cm
Radius of circle,r = Side of a hexagon
[In regular hexagon inscribed in a circle,its side is equal to the radius of a Circle]
r² = 36/√3 cm
Area of circle,A = πr²
A = 22/7 × 36/√3
A = 22/7 × 36/1.732
A = (22 × 36) /(7 ×1.732)
[√3 = 1.732]
A = 792/12.124
A = 65.324 cm²
Area of circle = 65.324 cm²
Hence, the area of hexagon is 54 cm² and area of circle is 65.324 cm².
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