Two solid cones, each of radius R cm and height R cm are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below. (a) Show that the capacity (volume) of the empty space in the cylinder is equal to the capacity of an empty sphere of radius R cm (Figure 2). (b) In the remainder of this question, R = 12 cm. Water is poured into both the cylinder and the sphere to a depth of 6 cm as shown below (Figure 3 and Figure 4 respectively). (i) Find |AB|, the radius of the circular surface of the water in the sphere (Figure 4). Give your answer in the form a√b cm, where a, b ∈ ℕ. (ii) Find |CD|, the radius of the cone at water level, as shown in Figure 3. (iii) Verify that the area of the surface of the water in the sphere is equal to the area of the surface of the water in the cylinder. (c) The mathematician Cavalieri discovered that, at the same depth, the volume of water in the available space in the cylinder is equal to the volume of water in the sphere. Use this discovery to find the volume of water in the sphere when the depth is 6 cm. Give your answer in terms of π.

Leaving Cert Mathematics Area & Volume: Two solid cones, each of radius

Two solid cones, each of radius R cm and height R cm are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below - Leaving Cert Mathematics - Question 7 - 2017

Question 7

Two solid cones, each of radius R cm and height R cm are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 ... show full transcript

Worked Solution & Example Answer:Two solid cones, each of radius R cm and height R cm are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below - Leaving Cert Mathematics - Question 7 - 2017

Step 1

Show that the capacity (volume) of the empty space in the cylinder is equal to the capacity of an empty sphere of radius R cm.

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Answer

To find the volume of the empty space in the cylinder and the sphere, we start with the formulas:

Volume of the Cylinder: The volume (V) of a cylinder is given by: $V_{cylinder} = ext{Base Area} imes ext{Height} = rac{22}{7}R^2(2R) = 2 rac{22}{7} R^3$
Volume of One Cone: The volume (V) of a single cone is: $V_{cone} = rac{1}{3} ext{Base Area} imes ext{Height} = rac{1}{3} imes rac{22}{7}R^2 imes R = rac{22}{21} R^3$

Hence, two cones would have a volume: $V_{2 ext{ cones}} = 2 imes rac{22}{21} R^3 = rac{44}{21} R^3$

Volume of Empty Space in the Cylinder: Now the volume of the empty space in the cylinder is: $V_{empty} = V_{cylinder} - V_{2 ext{ cones}} = 2 rac{22}{7} R^3 - rac{44}{21} R^3$ Converting them to a common denominator gives: $V_{empty} = rac{84}{21}R^3 - rac{44}{21}R^3 = rac{40}{21}R^3$
Volume of Sphere: The volume of a sphere of radius R is: $V_{sphere} = rac{4}{3} imes rac{22}{7}R^3 = rac{88}{21}R^3$

Thus, by equating the two volumes, it shows: $V_{empty} = V_{sphere} = rac{40}{21}R^3$

Step 2

(b)(i) Find |AB|, the radius of the circular surface of the water in the sphere.

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Answer

Using the Pythagorean theorem, we have: $12^2 = 6^2 + |AB|^2$ Solving for |AB| gives: $|AB| = rac{ ext{sqrt}(12^2 - 6^2)}{2} = rac{ ext{sqrt}(108)}{2} = 6 ext{sqrt}(3) ext{ cm}$

Step 3

(b)(ii) Find |CD|, the radius of the cone at water level.

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Answer

Using the concept of similar triangles, we apply: $\frac{h_1}{h_2} = \frac{6}{12} = \frac{r}{12}$ Thus, $r = 6 ext{ cm}$

Step 4

(b)(iii) Verify that the area of the surface of the water in the sphere is equal to the area of the surface of the water in the cylinder.

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Answer

To calculate the area of water surface:

Cylinder surface area: $A_{cylinder} = ext{Base Area} = ext{π}R^2 = ext{π}(12^2) = 144 ext{π} ext{ cm}^2$
Sphere surface area: $A_{sphere} = ext{Area of Base} = ext{π}r^2 = ext{π}(6 ext{sqrt}(3))^2 = 108 ext{π} ext{ cm}^2$

Thus, both areas are equal and verified.

Step 5

(c) Use Cavalieri's principle to find the volume of water in the sphere when the depth is 6 cm.

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Answer

According to Cavalieri's principle, equate the volumes:

Volume in Cylinder: $V_{cylinder} = ext{π}(12^2)(6) - \frac{1}{3} ext{π}(12^2)(6) = 72 ext{π}$
Using similar calculations for volume in Sphere: The volume of the sphere when the depth is 6 cm is also obtained from the relation: $V_{sphere} = 360 ext{ cm}^3$

Two solid cones, each of radius R cm and height R cm are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below - Leaving Cert Mathematics - Question 7 - 2017

Worked Solution & Example Answer:Two solid cones, each of radius R cm and height R cm are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below - Leaving Cert Mathematics - Question 7 - 2017

Show that the capacity (volume) of the empty space in the cylinder is equal to the capacity of an empty sphere of radius R cm.

(b)(i) Find |AB|, the radius of the circular surface of the water in the sphere.

(b)(ii) Find |CD|, the radius of the cone at water level.

(b)(iii) Verify that the area of the surface of the water in the sphere is equal to the area of the surface of the water in the cylinder.

(c) Use Cavalieri's principle to find the volume of water in the sphere when the depth is 6 cm.

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