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Solution: In ∆AOM, Angle M=90° AO^2=AM^2+OM^2 {Pythagoras theorem}
AO^2=(4)^2+(OC-CM)^2
AO^2=16 +(OC-2)^2. {since CM=2}
AO^2= 16+ OC^2+2^2 -2×OC×2
AO^2= 16+ AO^2+4 -4AO {OC=AO=r}
AO^2-AO^2 + 4AO =20
4AO=20
AO=20/4
AO=5 cm=r Answer
AO^2=(4)^2+(OC-CM)^2
AO^2=16 +(OC-2)^2. {since CM=2}
AO^2= 16+ OC^2+2^2 -2×OC×2
AO^2= 16+ AO^2+4 -4AO {OC=AO=r}
AO^2-AO^2 + 4AO =20
4AO=20
AO=20/4
AO=5 cm=r Answer
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