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h=14 cm,
r1+r2=5/2 .......eq(1),
I'm writing here r1=r and r2=R,
then we can write eq(1) as
R+r=5/2 ..........eq(2),
According to the question,
2πRh - 2πrh =44,
2πh(R-r)=44,
2×22/7 ×14 (R-r)=44,
2(R-r)=1,
R-r=1/2 ........eq(3),
now solve eq(2) and eq(3), we get
R=3/2 cm,
r=1 cm
r1+r2=5/2 .......eq(1),
I'm writing here r1=r and r2=R,
then we can write eq(1) as
R+r=5/2 ..........eq(2),
According to the question,
2πRh - 2πrh =44,
2πh(R-r)=44,
2×22/7 ×14 (R-r)=44,
2(R-r)=1,
R-r=1/2 ........eq(3),
now solve eq(2) and eq(3), we get
R=3/2 cm,
r=1 cm
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