a frustum of right circular core has a diameter of base 20cm, of top 12cm and height 3cm. find the area of its whole surface and volume
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ANSWER
D
1
=20cm,R
1
=10cm
D
2
=12cm,R
2
=6cm
h=3cm
Volume of frustum =
3
1
πh(R
1
2
+R
2
2
+R
1
R
2
)=
3
1
×
7
22
×3[10
2
+6
2
+10×6]
=
7
22
×196=616cm
3
Let slant height, be l. Then,
l=
(R
1
−R
2
)
2
+h
2
=
(10−6)
2
+3
2
=
16+9
=5cm
TSA of frustum of cone =π(R
1
+R
2
)l+π(R
1
2
+R
2
2
)
=
7
22
[10+6]×5+
7
22
(10
2
+6
2
)
=
7
22
[80+1360]=
7
22
×216=678.85cm
2
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