A SEMICIRCULAR REGION AND A SQUARE REGION HAVE EQUAL PERIMETERS.THE AREA OF THE SQUARE REGION EXCEEDS THAT OF THE SEMICIRCULAR REGION BY 4 CM2 . FIND THE PERIMETRE AND THE AREA OF THE TWO REGIONS.
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As perimeter of semicircular region and a square region is equal
So pie r + 2r = 4a ( r is radius, a is side of square)
Area of square = Area of semi circle + 4
a^2 = pie r^2/2 + 4......(1)
As pie r+ 2r = 4a
a= r( pie +2)/4
put in 1)
r^2( pie +2)^2/16 = pie r^2/2 +4
r^2/16 ( pie ^2 + 4 + 4pie) = pie r^2/2 +4
r^2 pie^2/16 + r^2/4 + pie r^2/4 - pie r^2/2 = 4
r^2 ( pie^2/16 + 1/4 + pie/4 - pie/2 )= 4
r^2( pie^2/16 -pie/4 + 1/4) = 4
r^2 ( pie /4 - 1/2) ^2 = 4
r( pie /4 - 1/2) = 2
r( 11/14 - 1/2) = 2
r( 11 - 7) = 28
r= 28/4 = 7
So perimeter = pie r + 2r = r( pie +2)
= 7( 22/7 +2)
= 22 + 14 = 36
4a = 36
a= 9
Area of semicircular = pie r^2/2 = 22/7 × 7 × 7 × 1/2
= 77
Area of square = a^2 = 81
So pie r + 2r = 4a ( r is radius, a is side of square)
Area of square = Area of semi circle + 4
a^2 = pie r^2/2 + 4......(1)
As pie r+ 2r = 4a
a= r( pie +2)/4
put in 1)
r^2( pie +2)^2/16 = pie r^2/2 +4
r^2/16 ( pie ^2 + 4 + 4pie) = pie r^2/2 +4
r^2 pie^2/16 + r^2/4 + pie r^2/4 - pie r^2/2 = 4
r^2 ( pie^2/16 + 1/4 + pie/4 - pie/2 )= 4
r^2( pie^2/16 -pie/4 + 1/4) = 4
r^2 ( pie /4 - 1/2) ^2 = 4
r( pie /4 - 1/2) = 2
r( 11/14 - 1/2) = 2
r( 11 - 7) = 28
r= 28/4 = 7
So perimeter = pie r + 2r = r( pie +2)
= 7( 22/7 +2)
= 22 + 14 = 36
4a = 36
a= 9
Area of semicircular = pie r^2/2 = 22/7 × 7 × 7 × 1/2
= 77
Area of square = a^2 = 81
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