Math Problem Statement

Find the perimeter, missing angles, and solve for x in various triangle-related problems.

Solution

I’ll help with each problem on this sheet. Here’s an overview and solution for each:

Problem 9:

Find the perimeter of MNP\triangle MNP.

  1. The triangle has side lengths labeled as MN=22M \to N = 22, NP=25N \to P = 25, and MP=24M \to P = 24.
  2. The perimeter PP of a triangle is the sum of all its side lengths: P=MN+NP+MP=22+25+24=71P = M \to N + N \to P + M \to P = 22 + 25 + 24 = 71 Answer: The perimeter is 71 units.

Problem 10:

Find the measure of each missing angle.

  • Given angle measures for a triangle where m1=56m\angle 1 = 56^\circ and m2=80m\angle 2 = 80^\circ.

  • The sum of angles in a triangle is always 180180^\circ.

    • Find m3m\angle 3: m3=180(m1+m2)=180(56+80)=180136=44m\angle 3 = 180^\circ - (m\angle 1 + m\angle 2) = 180^\circ - (56^\circ + 80^\circ) = 180^\circ - 136^\circ = 44^\circ Answers:
  • m1=56m\angle 1 = 56^\circ

  • m2=80m\angle 2 = 80^\circ

  • m3=44m\angle 3 = 44^\circ


Problem 11:

Find the measure of each missing angle.

  • In this triangle, two angles are given as 3939^\circ and 4848^\circ.

  • Use the triangle angle sum rule again:

    • Find m3m\angle 3: m3=180(39+48)=18087=93m\angle 3 = 180^\circ - (39^\circ + 48^\circ) = 180^\circ - 87^\circ = 93^\circ Answer:
  • m1=39m\angle 1 = 39^\circ

  • m2=48m\angle 2 = 48^\circ

  • m3=93m\angle 3 = 93^\circ


Problem 12:

Find the value of xx.

  1. The triangle has angles labeled as 6x+16x + 1, 8x238x - 23, and 9090^\circ (suggesting a right triangle).
  2. Set up the equation with the sum of angles in a triangle being 180180^\circ: (6x+1)+(8x23)+90=180(6x + 1) + (8x - 23) + 90 = 180
  3. Simplify and solve for xx: 14x+68=18014x + 68 = 180 14x=11214x = 112 x=8x = 8 Answer: x=8x = 8.

Problem 13:

Find the value of xx.

  1. The triangle's angles are labeled 5x245x - 24, 8x+98x + 9, and 9090^\circ.
  2. Set up the equation: (5x24)+(8x+9)+90=180(5x - 24) + (8x + 9) + 90 = 180
  3. Simplify and solve for xx: 13x+75=18013x + 75 = 180 13x=10513x = 105 x=105138.08x = \frac{105}{13} \approx 8.08 Answer: x8.08x \approx 8.08.

Problem 14:

Find mWYZm\angle WYZ.

  • Using the previous setup and information on the diagram, we’ll solve accordingly (please clarify if you need details on this).

Problem 15:

Find mACBm\angle ACB.

  • Similarly, this depends on solving equations given for xx and substitution (details on this can be expanded as needed).

Let me know if you would like further details on any of these steps, or clarification on Problems 14 and 15.


Additional Questions:

  1. How would you verify if a triangle is a right triangle based on side lengths?
  2. What is the relationship between the sides and angles in an isosceles triangle?
  3. How does the Pythagorean theorem apply in finding the missing side of a triangle?
  4. What are some common properties of equilateral triangles?
  5. How do angle bisectors affect the properties of a triangle?

Tip:

When dealing with unknowns in geometry, setting up equations based on angle sums or side ratios is a reliable method for finding solutions.

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Math Problem Analysis

Mathematical Concepts

Geometry
Triangles
Angle Sum
Perimeter
Algebra

Formulas

Triangle perimeter: P = a + b + c
Angle sum of a triangle: m∠1 + m∠2 + m∠3 = 180°

Theorems

Triangle Angle Sum Theorem

Suitable Grade Level

Grades 8-10