Answer :
Part A r =[tex]6\sqrt{2}[/tex] units & h = 2 units:
Step-by-step explanation:
Here we have , cones A and B both have volume 48(3.14) cubic units but they have different dimensions. Cone A has radius 6 units and height 4 units. We need to find:
Part A
Find one possible radius and height for Cone B.
Since , volume of cone A and B are same so ,
Volume of cone A = [tex]V_1[/tex] , Volume of cone V = [tex]V_2[/tex]
By hit & trial , One possible radius and height for Cone B is r =[tex]6\sqrt{2}[/tex] units & h = 2 units:
⇒ [tex]V_2 = \frac{1}{3}\pi r^2h[/tex]
⇒ [tex]V_2 = \frac{1}{3}(3.14) (6\sqrt{2})^2(2)[/tex]
⇒ [tex]V_2 = 48units^3[/tex]
Part B
Explain how you know Cone B has the same volume as Cone A.
Volume of cone A = [tex]V_1[/tex] :
Cone A has radius 6 units and height 4 units, So
⇒ [tex]V_1 = \frac{1}{3}\pi r^2h[/tex]
⇒ [tex]V_1 = \frac{1}{3}(3.14) (6})^2(4)[/tex]
⇒ [tex]V_2 = 48units^3[/tex]
Volume of cone V = [tex]V_2[/tex]
⇒ [tex]V_2 = \frac{1}{3}\pi r^2h[/tex]
⇒ [tex]V_2 = \frac{1}{3}(3.14) (6\sqrt{2})^2(2)[/tex]
⇒ [tex]V_2 = 48units^3[/tex]
Hence, Volume of both are same!