Question

Find CSA and TSA of each following right ciricular cones
(i) op=2cm,ob=3.5cm (ii) op=3.5cm,AB=10cm​

Answer 1

Step-by-step explanation:

CSA of cone=πrl.

TSA of cone=πr(r+l).

(I). Given: OP=2cm,. OB=3.5cm., BP=l=?

BP=√OP²+OB²

=> √2²+(3.5)²

=> √4+12.25

=> √16.25

=>4cm.(approx)

r=OP or OB

Case-1- when r is 2cm.

CSA =22/7×4×2

=>25.14cm².

TSA=22/7×2×(4+2)

=> 22/7×12

=> 37.7cm².

Case-2: when r is 3.5:

CSA=22/7×3.5×4=44cm².

TSA=22/7×3.5(3.5+4)

=> 11(7.5) =82.7cm².

(I).Given: OP=3.5cm,OB=10cm, l=?

PB=√OP²+OB²

=> √(3.5)²+10²

=> √12.25+100

=> √112.25

=> 10.5cm(approx)

r can be 3.5 cm or 10 cm.

Case-1: when r is 3.5cm

CSA=22/7×3.5×10.5

=> 11×10.5

=> 115.5cm²

TSA=22/7×3.5(3.5+10.5)

=> 11×14

=> 154cm².

Case-2: when r is 10cm.

CSA=22/7×10×10.5

=> 22×15=330cm².

TSA=22/7×10×(10+10.5)

=> 220/7×(20.5)

=> 644cm².