Pythagoras Theorem Worksheet for 15-Year-Olds | Math & Geometry Practice

Instructions

1. Read the conceptual explanations and examples carefully before starting the exercises.
2. Work through each section sequentially, as the difficulty level increases progressively.
3. Write your step-by-step calculations in the spaces provided. Do not write down only the final answer.
4. Use a standard scientific calculator for square roots, and round your answers to the nearest tenth where necessary.
5. Complete the final challenge section to test your spatial reasoning skills!

Section 1: Core Concept & Identification

The Pythagorean Theorem states that in any right-angled triangle, the square of the length of the hypotenuse ($c$) is equal to the sum of the squares of the lengths of the other two sides ($a$ and $b$).

$$\mathbf{a^2 + b^2 = c^2}$$

• The Hypotenuse ($c$): Always the longest side, directly opposite the $90^\circ$ right angle.
• The Legs ($a$ and $b$): The two shorter sides that form the right angle.

Quick Identification Exercise: For each of the triangles described below, identify which side length is the hypotenuse ($c$):

1. Triangle XYZ with sides $x = 7\text{ cm}$, $y = 24\text{ cm}$, and $z = 25\text{ cm}$.

   • Hypotenuse is side: __

2. Triangle PQR where the right angle is at vertex Q.

   • Hypotenuse is side: __

Section 2: The Calculation Station

Complete the table below by finding the missing side of each right-angled triangle.

• Hint: If you are finding the hypotenuse ($c$), use: $c = \sqrt{a^2 + b^2}$
• Hint: If you are finding a leg ($a$ or $b$), use: $a = \sqrt{c^2 - b^2}$

Section 3: Real-World Applications

Solve these practical problems. Sketch a simple diagram first to help you visualize the triangle!

Problem 1: The Phone Upgrade You are buying a new smartphone. The manufacturer's specifications state that the screen has a width of $3\text{ inches}$ and a height of $6\text{ inches}$. Phone screen sizes are measured diagonally from corner to corner.

• Sketch & Setup:
  
  
  
  
• Calculation:
  
  
  
  
• Final Answer (rounded to the nearest tenth): ___

Problem 2: The Sidewalk Shortcut Every day on her walk home from school, Clara walks along two perpendicular sidewalks around a corner lot: $120\text{ meters}$ due East, and then $50\text{ meters}$ due North. Today, she decides to take a shortcut directly across the grassy lot from the start point to the end point.

• How many meters does Clara save by taking the diagonal path instead of the sidewalks?
  
  
  
  
• Calculation:
  
  
  
  
• Final Answer: ___

Section 4: The Converse Check (Is it Right?)

The Converse of the Pythagorean Theorem states that if the side lengths of a triangle satisfy $a^2 + b^2 = c^2$, then the triangle must be a right-angled triangle.

Determine if the following structural frames are perfectly square (contain a $90^\circ$ angle):

1. Frame A: A triangular bracing timber with side lengths of $9\text{ inches}$, $12\text{ inches}$, and $15\text{ inches}$.

   • Calculation:
     
     
   • Is it a right triangle? (Yes/No): _____

2. Frame B: A triangular shade sail with side lengths of $5\text{ feet}$, $8\text{ feet}$, and $10\text{ feet}$.

   • Calculation:
     
     
   • Is it a right triangle? (Yes/No): _____

Section 5: The 3D Space Challenge (Extension)

Ready to level up? Let's apply Pythagoras to three dimensions!

You have a rectangular shipping box that measures $12\text{ inches}$ long, $4\text{ inches}$ wide, and $3\text{ inches}$ high. You want to fit a thin metal rod inside the box diagonally from the bottom-left-front corner to the top-right-back corner.

• Step 1: Find the diagonal of the bottom face of the box ($d$) using the length ($12\text{ in}$) and width ($4\text{ in}$).

• Step 2: Use that diagonal length ($d$) and the height of the box ($3\text{ in}$) to find the 3D space diagonal ($D$).

• Your Working Out:
  
  
  
  
  
  

• What is the maximum length of the rod that can fit? ___

Answer Key

Section 1

1. Hypotenuse is side $z$ (since $25$ is the longest side).
2. Hypotenuse is side $PR$ (opposite vertex Q which contains the right angle).

Section 2

• Row 2: $5^2 + 12^2 = c^2 \rightarrow 25 + 144 = 169 \rightarrow c = \sqrt{169} \rightarrow$ $c = 13$
• Row 3: $8^2 + b^2 = 10^2 \rightarrow 64 + b^2 = 100 \rightarrow b^2 = 36 \rightarrow b = \sqrt{36} \rightarrow$ $b = 6$
• Row 4: $a^2 + 15^2 = 17^2 \rightarrow a^2 + 225 = 289 \rightarrow a^2 = 64 \rightarrow a = \sqrt{64} \rightarrow$ $a = 8$
• Row 5: $6^2 + 8^2 = c^2 \rightarrow 36 + 64 = 100 \rightarrow c = \sqrt{100} \rightarrow$ $c = 10$
• Row 6: $7^2 + 24^2 = c^2 \rightarrow 49 + 576 = 625 \rightarrow c = \sqrt{625} \rightarrow$ $c = 25$

Section 3

• Problem 1:
  • $3^2 + 6^2 = c^2 \rightarrow 9 + 36 = 45 \rightarrow c = \sqrt{45} \approx 6.7\text{ inches}$.
• Problem 2:
  • Sidewalk distance = $120\text{ m} + 50\text{ m} = 170\text{ m}$.
  • Shortcut distance ($c$) = $\sqrt{120^2 + 50^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130\text{ m}$.
  • Distance saved = $170\text{ m} - 130\text{ m} = 40\text{ meters}$.

Section 4

1. Frame A: $9^2 + 12^2 = 81 + 144 = 225$. Since $15^2 = 225$, the equation holds true. Yes, it is a right triangle.
2. Frame B: $5^2 + 8^2 = 25 + 64 = 89$. Since $10^2 = 100$ and $89 \neq 100$, the equation does not hold true. No, it is not a right triangle.

Section 5

• Step 1 (Bottom diagonal): $d^2 = 12^2 + 4^2 = 144 + 16 = 160$.
• Step 2 (3D Space diagonal): $D^2 = d^2 + \text{height}^2 \rightarrow D^2 = 160 + 3^2 \rightarrow D^2 = 160 + 9 = 169 \rightarrow D = \sqrt{169} = 13\text{ inches}$.
• Answer: The maximum length of the rod is $13\text{ inches}$.