The perimeter of an equilateral triangle is equal to the perimeter of a regular hexagon. Find the ratio of their areas . Expert help needed! Our coaching asked us this (CatalyseR indore)
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Let each side of the equilateral triangle be a
Let each side of the regular hexagon be r
According to question,
3a=6r
⇒a=2r (eq. i)
Area of regular hexagon= (1/2)(Perimeter of the hexagon)(Perpendicular distance from the centre to any side)
Perpendicular distance from the centre to any side=√3 r/2
(the root is only under 3. Solve this eq. by drawing a circle of r radius and do not adjust the compass but cut the circle by keeping the compass on the circumference and draw arcs to get a hexagon of side r)
Area of eq. triangle/ Area of regular hexagon= {(√3/4)a²} / {(1/2)(Perimeter of the hexagon)(Perpendicular distance from the centre to any side)}
=√3/4*(2r)² / (1/2)(8r)(√3 r/2)
=1:2 is the answer
Let each side of the regular hexagon be r
According to question,
3a=6r
⇒a=2r (eq. i)
Area of regular hexagon= (1/2)(Perimeter of the hexagon)(Perpendicular distance from the centre to any side)
Perpendicular distance from the centre to any side=√3 r/2
(the root is only under 3. Solve this eq. by drawing a circle of r radius and do not adjust the compass but cut the circle by keeping the compass on the circumference and draw arcs to get a hexagon of side r)
Area of eq. triangle/ Area of regular hexagon= {(√3/4)a²} / {(1/2)(Perimeter of the hexagon)(Perpendicular distance from the centre to any side)}
=√3/4*(2r)² / (1/2)(8r)(√3 r/2)
=1:2 is the answer
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