CBSE BOARD X, asked by radheyshyamm, 10 months ago

plz answer me this question fast . emergency.

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radheyshyamm: can anyone answer this.....
saurabh2001: hi cutie
shruti6456: it is question of frustum
saurabh2001: yes

Answers

Answered by mohdsuheb01
0
From the figure,

In Δ ABE and Δ ACD,

BE || CD

So, Δ ABE ∼ Δ ACD 

=> AB/AC = BE/CD

=> 4/12 = BE/6

=> 1/3 = BE/6

=> 1 = BE/2

=> BE = 2

Now, AD2 = AC2 + CD2 

=> AD2 = 122 + 62 

=> AD2 = 144 + 36

=> AD2 = 180

=> AD = √180

=> AD = 13.42

Now, total surface are of complete cone = πr(l + r)

                                                        = π*6(6 + 13.42)

                                                        = π*6*19.42

                                                        = 3.14*6*19.42

                                                        =  365.87 

Now, curved surface area of smaller cone of height 4 cm and radius 2 cm = πrl

From the figure,

      l = √(AC2 + CD2 )

=> l = √(42 + 22 )

=> l = √(16 + 4)

=> l = √20

=> l = 4.47

So, the curved surface are = πrl = 3.14*2*4.47 = 28.08

Now, Total surface area of the remaining cone = totalsurface are of bigger cone - curved surface area of smaller cone + area of base of smaller cone

                                                                = 365.87 - 28.08 + π*2*2

                                                                = 365.87 - 28.08 + 3.14*4

                                                                = 365.87 - 28.08 + 12.56

                                                                = 350.35 cm2 

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