Math Problem Statement

Find the missing side or hypotenuse of a right triangle given one leg as 10 and multiple-choice options.

Solution

The image shows a right triangle with the hypotenuse labeled as aa and one of the legs labeled as bb. The length of the other leg is given as 10. The question appears to be asking for either aa or bb using the Pythagorean Theorem.

Step-by-Step Solution:

Let's denote:

  • The length of the given leg as 1010.
  • The other leg as bb.
  • The hypotenuse as aa.

By the Pythagorean Theorem: a2=102+b2a^2 = 10^2 + b^2 a2=100+b2a^2 = 100 + b^2

If we are to solve for bb, we need more information. However, assuming that aa is what we need to find, we can solve the problem once we know bb.

Considerations for Choices:

The options provided are:

  • 626\sqrt{2}
  • 2\sqrt{2}
  • 10210\sqrt{2}
  • 20220\sqrt{2}

We are likely dealing with a multiple-choice problem where calculating bb and then substituting to find aa would lead to matching one of these choices. Let’s assume the triangle is a special 45-45-90 triangle:

  • If it is, then bb would also be 1010.
  • The hypotenuse aa would then be 10210\sqrt{2}.

Conclusion:

The correct answer appears to be: Option C: 10210\sqrt{2}.

Would you like more details, or do you have any specific questions?

Follow-up Questions:

  1. How do you identify if a triangle is a 45-45-90 triangle?
  2. What is the Pythagorean Theorem, and how is it applied to right triangles?
  3. Can the hypotenuse of a right triangle ever be shorter than any of its legs?
  4. How do you rationalize square roots?
  5. What other properties apply to special triangles like 30-60-90 and 45-45-90 triangles?

Tip:

Remember that a 45-45-90 triangle always has legs of equal length, and the hypotenuse is 2\sqrt{2} times the length of one leg.

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Math Problem Analysis

Mathematical Concepts

Geometry
Right Triangles
Special Triangles
Square Roots

Formulas

Pythagorean Theorem: a^2 = b^2 + c^2

Theorems

Pythagorean Theorem
Properties of 45-45-90 Triangles

Suitable Grade Level

Grades 8-10