The surface

The surface of a truncated rotating cone with side s = 13 cm is S = 510π cm2. Find the radii of the bases when their difference in lengths is 10cm.

Correct answer:

r1 =  15 cm
r2 =  5 cm

Step-by-step explanation:

s=13 cm S=510π=510 3.14161602.2123 cm2  S=π r12+π r22+π (r1+r2) s r1=r2+10  S=π r12+π (r110)2+π (r1+(r110)) s S=πx2+π(x10)2+π(x+(x10))s  1602.21225333=π x2+π (x10)2+π (x+(x10)) 13 6.28318530718x218.85x+1696.46=0 6.28318530718x2+18.85x1696.46=0  a=6.28318530718;b=18.85;c=1696.46 D=b24ac=18.85246.28318530718(1696.46)=42991.9967711 D>0  x1,2=b±D2a=18.85±4299212.5663706144 x1,2=1.5±16.5 x1=15 x2=18   Factored form of the equation:  6.28318530718(x15)(x+18)=0  r1,r2>0 r1=x1=15 cm

Our quadratic equation calculator calculates it.

r2=r110=1510=5 cm   Verifying Solution:  S2=π r12+π r22+π (r1+r2) s=3.1416 152+3.1416 52+3.1416 (15+5) 131602.2123 cm2 S2=S

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