Karnataka SSLC Karnataka Secondary Education Examination Board

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Areas of Combinations of Plane Figures

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So far, we have calculated the areas of different figures separately. Let us now try to calculate the areas of some combinations of plane figures.
Example 4 : In Fig., two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

For sector ODC, 
∠DOC= θ= 90° (because O is the point of intersection of diagonals of square ABCD
`"Diagonal BD"= "side"sqrt2= 56sqrt2`

therefore, OD=r = `(56sqrt2)/2= 28sqrt2cm`

area of sector ODC= `θ/360 xx πr^2`

                               =`(90 xx 22 xx 28sqrt2 xx 28sqrt2)/360 xx 7`

                               `= 22 xx 2 xx 28`

area of sector ODC= `22 xx 56m^2`

area of ΔDOC= `1/2 xx OD xx OC`

                      = `1/2 xx  28sqrt2 xx 28sqrt 2`

area of ΔDOC= `19 xx 56m^2`

area of flower beds= `2 xx (22 xx 56)- (19 xx 56)`
                               =` 2 xx 56  (22-19)`
                               = `112 xx 8`
area of flower beds= `896m^2`
`"area"   "of"   "lawn"= "side"^2= 56^2`
`"area"  "of"   "lawn"= 3136m^2`
`"Total"   "area" = 3136+896`
`"Total"   "area" = 4032m^2`

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