Math Problem Statement

Cylinder A Cylinder B Shape Volume 300π 2700π Cylinder A and Cylinder B are similar right cylinders. The radius of Cylinder A is 4, and its surface area is dπ . The surface area of Cylinder B is tπ. What is the value of t-d? ( Note: surface area of a cylinder is 2πrh+2πr^2)

Solution

We are given two similar right cylinders, Cylinder A and Cylinder B. Their volumes are 300π300\pi and 2700π2700\pi, respectively. The radius of Cylinder A is 4, and we are tasked with finding the value of tdt - d, where dπd\pi and tπt\pi are the surface areas of Cylinder A and Cylinder B, respectively.

Step 1: Use the similarity ratio for the cylinders

Since the cylinders are similar, the ratio of their volumes is proportional to the cube of the ratio of their corresponding dimensions (such as their radii or heights). The volume of a cylinder is given by:

V=πr2hV = \pi r^2 h

Thus, the ratio of the volumes of Cylinder B and Cylinder A is:

VBVA=2700π300π=9\frac{V_B}{V_A} = \frac{2700\pi}{300\pi} = 9

This tells us that the ratio of their dimensions (including radius and height) is the cube root of 9:

rBrA=hBhA=93=3\frac{r_B}{r_A} = \frac{h_B}{h_A} = \sqrt[3]{9} = 3

Thus, the radius and height of Cylinder B are 3 times those of Cylinder A.

Step 2: Calculate the radius of Cylinder B

Given that the radius of Cylinder A is rA=4r_A = 4, the radius of Cylinder B is:

rB=3×rA=3×4=12r_B = 3 \times r_A = 3 \times 4 = 12

Step 3: Express the surface areas of Cylinder A and Cylinder B

The surface area SS of a cylinder is given by:

S=2πrh+2πr2S = 2\pi rh + 2\pi r^2

For Cylinder A, the surface area is:

SA=2πrAhA+2πrA2=dπS_A = 2\pi r_A h_A + 2\pi r_A^2 = d\pi

For Cylinder B, the surface area is:

SB=2πrBhB+2πrB2=tπS_B = 2\pi r_B h_B + 2\pi r_B^2 = t\pi

Step 4: Relate the surface areas using the similarity ratio

Since the cylinders are similar, the surface area ratio is proportional to the square of the ratio of their corresponding dimensions. Thus, the ratio of their surface areas is:

SBSA=(rBrA)2=32=9\frac{S_B}{S_A} = \left( \frac{r_B}{r_A} \right)^2 = 3^2 = 9

So, the surface area of Cylinder B is 9 times that of Cylinder A:

SB=9×SAS_B = 9 \times S_A

Step 5: Calculate tdt - d

From the relation SB=9×SAS_B = 9 \times S_A, we have:

tπ=9×dπt\pi = 9 \times d\pi

Dividing both sides by π\pi:

t=9dt = 9d

Thus:

td=9dd=8dt - d = 9d - d = 8d

Step 6: Find the surface area of Cylinder A

We already know that the radius of Cylinder A is rA=4r_A = 4, and its volume is VA=300πV_A = 300\pi. Using the volume formula for a cylinder, we can find the height of Cylinder A:

VA=πrA2hAV_A = \pi r_A^2 h_A 300π=π(4)2hA300\pi = \pi (4)^2 h_A 300=16hA300 = 16 h_A hA=30016=18.75h_A = \frac{300}{16} = 18.75

Now, using the surface area formula for Cylinder A:

SA=2πrAhA+2πrA2S_A = 2\pi r_A h_A + 2\pi r_A^2 SA=2π(4)(18.75)+2π(4)2S_A = 2\pi (4)(18.75) + 2\pi (4)^2 SA=150π+32π=182πS_A = 150\pi + 32\pi = 182\pi

Thus, d=182d = 182.

Step 7: Calculate tdt - d

Since t=9dt = 9d, we have:

t=9×182=1638t = 9 \times 182 = 1638

Finally:

td=1638182=1456t - d = 1638 - 182 = 1456

Thus, the value of tdt - d is 1456\boxed{1456}.

Would you like more details or have any further questions?

Here are 5 related questions:

  1. How do you find the surface area of a cylinder given only its height and radius?
  2. What is the relationship between the volume and surface area of similar figures?
  3. Can you explain how to apply the concept of geometric similarity in 3D shapes?
  4. How would the problem change if the cylinders weren't similar?
  5. What is the cube root rule in relation to volumes of similar objects?

Tip: Always check the proportional relationships (volume, area) in similar figures—these follow predictable scaling laws!

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Math Problem Analysis

Mathematical Concepts

Geometry
Similarity of Figures
Cylinders
Surface Area
Volume Ratio

Formulas

Surface Area of Cylinder: S = 2πrh + 2πr^2
Volume of Cylinder: V = πr^2h
Similarity Ratio: Ratio of surface areas is proportional to the square of the ratio of their corresponding dimensions

Theorems

Geometric Similarity
Scaling Laws for Similar Figures

Suitable Grade Level

Grades 10-12