CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

From a solid cylinder whose height is $$2.4\ cm$$ and diameter $$1.4\ cm$$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $$cm^{2}$$.


Solution

Given :
radius $$(r) =\dfrac {diameter}{2}= \dfrac {1.4}{2}=0.7 \ cm$$

height $$(h)= 2.4 \ cm$$

Total surface area$$=$$ surface area of cylinder$$+$$ surface area of cone$$-2$$(surface area of above circle).

                              $$=2\pi rh+2\pi r^2+\pi r^2+\pi r\sqrt{h^2+r^2}-2\pi r^2$$

                              $$=2\pi rh+\pi r\sqrt{h^2+r^2}+\pi r^2$$

                              $$=\pi (2rh+r^2+r\sqrt{h^2+r^2})$$

                              $$=\dfrac{22}{7}(2\times 0.7\times 2.4+(0.7)^2+0.7\sqrt{(0.7)^2+(24)^2})$$

                              $$=\dfrac{22}{7}(3.36+0.49+1.75)$$

                             $$=\dfrac{22}{7}(5.6)$$

Total surface area$$=17.6cm^2$$.

Hence, this is the answer.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image