In the given figure, ABCD is a trapezium with AB II DC, AB = 18 cm, DC = 32 cm and the distance between AB and AC is 14 cm. If arcs of equal radii 7 cm taking A, B, C and D as centres, have been drawn, then find the area of the shaded region.
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i ) In Trapezium ABCD ,
AB // DC ,
AB = 18cm , DC = 32 cm,
Distance between two parallel
lines = ( h ) = 14cm
Area of ABCD = [ ( AB+DC )×h ]/2
= [( 18 + 32 )×14 ]/2
= 40 × 7
= 280 cm² ----( 1 )
ii ) In ABCD Trapezium ,
AB // DC ,
<A + <D = 180° ----( 2 )
Similarly ,
<B + <D = 180° ----( 3 )
Adding
<A + <B + <C + <D = 360° --( 4 )
Arranging 4 sectors with equal radii
become a circle [ see the figure ]
Now ,
Area of the circle = πr²
= ( 22/7 ) × 7²
= 22 × 7
= 154 cm² --- ( 4 )
Therefore ,
Area of the shaded region= ( 1 ) - ( 2 )
= 280 cm² - 154 cm²
= 126 cm²
••••
AB // DC ,
AB = 18cm , DC = 32 cm,
Distance between two parallel
lines = ( h ) = 14cm
Area of ABCD = [ ( AB+DC )×h ]/2
= [( 18 + 32 )×14 ]/2
= 40 × 7
= 280 cm² ----( 1 )
ii ) In ABCD Trapezium ,
AB // DC ,
<A + <D = 180° ----( 2 )
Similarly ,
<B + <D = 180° ----( 3 )
Adding
<A + <B + <C + <D = 360° --( 4 )
Arranging 4 sectors with equal radii
become a circle [ see the figure ]
Now ,
Area of the circle = πr²
= ( 22/7 ) × 7²
= 22 × 7
= 154 cm² --- ( 4 )
Therefore ,
Area of the shaded region= ( 1 ) - ( 2 )
= 280 cm² - 154 cm²
= 126 cm²
••••
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