concrete slab in the shape of a regular octagon. The distance from the centerup one of the peaks is 15 cm. Determine the area of a concrete slab

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Each interior angle of octagon= 135 degree.

`(angle s= ((n-2)xx180)/n)`

Draw a perpendicular from O to side of octagon ON .M is one vertex of octagone.

consider triangle OMN , OM=15 cm,

`angle M=(1/2)` of interior angle f octgon

=`135/2=62.5^o`

`angle n=90^o`

`ON=15xxsin(62.5^o)`

`MN= 15xx cos(62.5^o)`

`area of DeltaOMN=(1/2) xxMNxxON`

`=(1/2)(15^2)sin(62.5^o)cos(62.5^o)`

`` area of octagn=`16xx(1/2)xx(15^2) sin(62.5^o) cos(62.5^o)`

`=4xx15^2xxsin(135^o)`

`=636.40 ` sq cm.

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