pls sove my query... question nomber 11. and 12 pls solve it any one
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q. no. 12) i) length of the arc=theta÷360°×2Πr
=90°÷360°×2×22÷7×21
=1÷4×2×22÷7×21,by simplifying
=11÷7×21
=31.571428
therefore,the perimeter=21+21+31.60
=73.60(approximately)
=90°÷360°×2×22÷7×21
=1÷4×2×22÷7×21,by simplifying
=11÷7×21
=31.571428
therefore,the perimeter=21+21+31.60
=73.60(approximately)
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