Question

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.
Part A

Find one possible radius and height for Cone B.
Radius: Height:

Part B
Explain how you know Cone B has the same volume as Cone A. ​

Answer 1

Part A r =$6\sqrt{2}$  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 = $V_1$ , Volume of cone V = $V_2$

By hit & trial , One possible radius and height for Cone B is r =$6\sqrt{2}$  units & h = 2  units:

⇒ $V_2 = \frac{1}{3}\pi r^2h$

⇒ $V_2 = \frac{1}{3}(3.14) (6\sqrt{2})^2(2)$

⇒ $V_2 = 48units^3$

Part B

Explain how you know Cone B has the same volume as Cone A. ​

Volume of cone A = $V_1$ :

Cone A has radius 6 units and height 4 units, So

⇒ $V_1 = \frac{1}{3}\pi r^2h$

⇒ $V_1 = \frac{1}{3}(3.14) (6})^2(4)$

⇒ $V_2 = 48units^3$

Volume of cone V = $V_2$

⇒ $V_2 = \frac{1}{3}\pi r^2h$

⇒ $V_2 = \frac{1}{3}(3.14) (6\sqrt{2})^2(2)$

⇒ $V_2 = 48units^3$

Hence, Volume of both are same!