Math Problem Statement

A rectangular prism is 12 inches wide, 5 inches long, and 6 inches tall. Tamara's work to find the length of its longest diagonal is incorrect. Identify the mistake and find the correct length. Additionally, what is the height of the square-based pyramid shown, and what is the shortest side of a right triangle with given sides?

Solution

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Question 5

Problem Statement: A rectangular prism is 12 inches wide, 5 inches long, and 6 inches tall. Tamara made a mistake when calculating the length of the prism’s longest diagonal. We need to identify her mistake and calculate the correct length of the diagonal.

Tamara's work (visible in the image):

  • She calculated 122+52+62=d212^2 + 5^2 + 6^2 = d^2.
  • Solved for dd, finding 23=4.8in\sqrt{23} = 4.8 \, \text{in}.
  • Her result: "The longest diagonal is about 4.8 in."

Mistake Analysis: Tamara incorrectly calculated 122+52+6212^2 + 5^2 + 6^2. The correct sum of squares should be: 122+52+62=144+25+36=205.12^2 + 5^2 + 6^2 = 144 + 25 + 36 = 205. Thus, the length of the diagonal should be: d=20514.32in.d = \sqrt{205} \approx 14.32 \, \text{in}.

So, the longest diagonal is approximately 14.32 inches.


Question 6

Problem Statement: What is the height of the square-based pyramid shown?

  • The base is a square with side lengths of 10.4 inches.
  • The slant height is given as 8 inches.

To find the height hh of the pyramid, we can use the Pythagorean theorem. The slant height, the height of the pyramid, and half the base form a right triangle. The relationship is: h2+(10.42)2=82.h^2 + \left(\frac{10.4}{2}\right)^2 = 8^2. Simplifying: h2+5.22=64,h^2 + 5.2^2 = 64, h2+27.04=64,h^2 + 27.04 = 64, h2=6427.04=36.96,h^2 = 64 - 27.04 = 36.96, h=36.966.08in.h = \sqrt{36.96} \approx 6.08 \, \text{in}.

So, the height of the pyramid is approximately 6.08 inches.


Question 7

Problem Statement: Two sides of a right triangle have lengths 6 and 828\sqrt{2}. There are two possible lengths for the third side, and the shortest possible side needs to be found.

We apply the Pythagorean theorem:

  1. If 66 and 828\sqrt{2} are the legs, the hypotenuse cc is: c=62+(82)2=36+128=164.c = \sqrt{6^2 + (8\sqrt{2})^2} = \sqrt{36 + 128} = \sqrt{164}. Simplifying: c=241.c = 2\sqrt{41}.

  2. If one of them is the hypotenuse (say 828\sqrt{2}), we solve for the missing leg bb: (82)2=62+b2,(8\sqrt{2})^2 = 6^2 + b^2, 128=36+b2,128 = 36 + b^2, b2=92,b^2 = 92, b=92=223.b = \sqrt{92} = 2\sqrt{23}.

The shortest side is 223\boxed{2\sqrt{23}}.


Do you have any questions or would you like further details? Here are five related questions to explore:

  1. How would the length of the diagonal in Question 5 change if one dimension doubled?
  2. What would be the volume of the pyramid from Question 6?
  3. How can you generalize the formula for the diagonal of any rectangular prism?
  4. How does the Pythagorean theorem apply to three-dimensional figures?
  5. What are the real-world applications of square-based pyramids in architecture?

Tip: For problems involving the Pythagorean theorem, always check whether you are working in 2D or 3D to apply the correct formula.

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Math Problem Analysis

Mathematical Concepts

Geometry
Pythagorean Theorem
3D Shapes
Right Triangles

Formulas

Diagonal of rectangular prism: d = sqrt(l^2 + w^2 + h^2)
Pythagorean Theorem: a^2 + b^2 = c^2

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 9-10