7.2 The Pythagorean Relationship & Right-Triangle Problems

7.2 The Pythagorean Relationship & Right-Triangle Problems

Quick Summary: The Pythagorean Theorem is a cornerstone of geometry, measurement, and spatial reasoning on the Ontario Mathematics Proficiency Test (MPT). This section covers right-triangle properties, leg and hypotenuse identification, formula applications ($a^2 + b^2 = c^2$), common Pythagorean triples, non-calculator square root estimation, the converse theorem for triangle classification, and real-world distance applications.

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Anatomy of a Right Triangle & The Pythagorean Relationship

A right triangle (or right-angled triangle) is a triangle containing one $90^\circ$ right angle. The sides of a right triangle are classified into two categories:

• Legs ($a$ and $b$): The two shorter sides that meet at a $90^\circ$ angle to form the right-angle vertex.
• Hypotenuse ($c$): The longest side of the right triangle, located directly across (opposite) from the $90^\circ$ right angle.

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The Pythagorean Theorem

In any right-angled triangle, the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the two legs:

$$a^2 + b^2 = c^2$$

Where:

• $a$ and $b$ are the lengths of the legs.
• $c$ is the length of the hypotenuse.

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Solving for Unknown Sides: Hypotenuse vs. Leg

Depending on which side length is missing, rearrange the Pythagorean formula:

1. Solving for the Hypotenuse ($c$)

When both legs $a$ and $b$ are known:

$$c^2 = a^2 + b^2 \implies c = \sqrt{a^2 + b^2}$$

Example: If leg $a = 6\text{ cm}$ and leg $b = 8\text{ cm}$:

$$c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$

2. Solving for an Unknown Leg ($a$ or $b$)

When the hypotenuse $c$ and one leg are known:

$$a^2 = c^2 - b^2 \implies a = \sqrt{c^2 - b^2}$$ $$b^2 = c^2 - a^2 \implies b = \sqrt{c^2 - a^2}$$

Example: If hypotenuse $c = 13\text{ m}$ and leg $a = 5\text{ m}$:

$$b = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ m}$$

Critical Error Warning: Candidates often mistakenly add squares when solving for a missing leg. Remember: ADD to find the hypotenuse ($c = \sqrt{a^2 + b^2}$), but SUBTRACT to find a leg ($a = \sqrt{c^2 - b^2}$)!

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Common Pythagorean Triples & Scaled Multiples

A Pythagorean triple is a set of three positive integers $(a, b, c)$ that satisfy $a^2 + b^2 = c^2$. Memorizing fundamental triples and their scaled multiples saves time on the MPT, especially on non-calculator sections.

Fundamental Pythagorean Triples:

1. $3\text{--}4\text{--}5$ Triple: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$
2. $5\text{--}12\text{--}13$ Triple: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$
3. $8\text{--}15\text{--}17$ Triple: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$
4. $7\text{--}24\text{--}25$ Triple: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$

Scaled Families of Triples

Multiplying any triple by a positive scale factor $k$ creates a new valid right triangle: $(k \cdot a, k \cdot b, k \cdot c)$.

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Non-Calculator Square Root Estimation

The non-calculator section of the MPT frequently requires candidates to evaluate non-perfect square roots (e.g., $\sqrt{50}$ or $\sqrt{75}$).

Step-by-Step Bounding & Estimation Method

1. Identify Perfect Square Bounds: Locate the closest perfect square below ($n^2$) and above ($(n+1)^2$) the given radicand $N$. $$n^2 < N < (n+1)^2 \implies n < \sqrt{N} < n+1$$
2. Linear Interpolation: Estimate the fractional decimal portion: $$\sqrt{N} \approx n + \frac{N - n^2}{(n+1)^2 - n^2}$$

Perfect Squares Reference List ($1$ to $20$):

$$\begin{array}{ccccc} 1^2 = 1 & 2^2 = 4 & 3^2 = 9 & 4^2 = 16 & 5^2 = 25 \\ 6^2 = 36 & 7^2 = 49 & 8^2 = 64 & 9^2 = 81 & 10^2 = 100 \\ 11^2 = 121 & 12^2 = 144 & 13^2 = 169 & 14^2 = 196 & 15^2 = 225 \\ 16^2 = 256 & 17^2 = 289 & 18^2 = 324 & 19^2 = 361 & 20^2 = 400 \end{array}$$

Estimation Example: Estimate $\sqrt{75}$

• Step 1: Perfect squares around $75$ are $64$ ($8^2$) and $81$ ($9^2$). So $8 < \sqrt{75} < 9$.
• Step 2: Difference from lower bound: $75 - 64 = 11$. Interval width: $81 - 64 = 17$.
• Step 3: Fractional part: $\frac{11}{17} \approx 0.65$.
• Result: $\sqrt{75} \approx 8.65 \approx 8.7$.

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The Converse of the Pythagorean Theorem

The Converse of the Pythagorean Theorem allows you to test whether a given triangle with side lengths $a, b, c$ (where $c$ is the longest side) is right, acute, or obtuse.

• Right Triangle: If $a^2 + b^2 = c^2$, the angle opposite $c$ is exactly $90^\circ$.
• Acute Triangle: If $a^2 + b^2 > c^2$ (or $c^2 < a^2 + b^2$), all three interior angles are less than $90^\circ$.
• Obtuse Triangle: If $a^2 + b^2 < c^2$ (or $c^2 > a^2 + b^2$), the angle opposite $c$ is greater than $90^\circ$.

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Real-World Applications & 3D Distance

1. Ladder Safety & Wall Reach

A ladder leaning against a vertical wall forms a right triangle with the ground and wall. Safety regulations often specify maximum base distance to prevent slipping.

2. Field & Screen Diagonals

Television screen sizes represent diagonal lengths. A rectangular screen with width $w$ and height $h$ has a diagonal size of $d = \sqrt{w^2 + h^2}$.

3. 3D Spatial Diagonal of a Rectangular Box

To find the direct straight-line distance across opposite corners of a 3D rectangular box with length $l$, width $w$, and height $h$:

$$d_{\text{3D}} = \sqrt{l^2 + w^2 + h^2}$$

Derivation: First find base diagonal $d_{\text{base}} = \sqrt{l^2 + w^2}$. Then apply Pythagoras to the vertical height: $d_{\text{3D}} = \sqrt{d_{\text{base}}^2 + h^2} = \sqrt{l^2 + w^2 + h^2}$.

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Multi-Step Worked Problem Examples

Example 1: Ladder Positioning & Heights

Problem: A maintenance technician must reach a window $16\text{ metres}$ above the ground. Safety regulations state that the distance from the base of the wall to the ladder base must be at least $12\text{ metres}$.

1. What is the minimum length of ladder required to reach the window?
2. If the technician uses a $20\text{-metre}$ ladder placed $12\text{ metres}$ from the wall, how high up the wall will the ladder reach?

Solution:

1. Step 1: Calculate Minimum Ladder Length The ladder is the hypotenuse $c$. Legs are $a = 12\text{ m}$ and $b = 16\text{ m}$.

   $$c = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ metres}$$

   (Note: This is a $3\text{--}4\text{--}5$ triple scaled by $k=4$, i.e., $12\text{--}16\text{--}20$.)

2. Step 2: Calculate Height Reachable with 20m Ladder at 12m Distance Here hypotenuse $c = 20\text{ m}$ and base leg $a = 12\text{ m}$. Find wall leg $b$:

   $$b = \sqrt{c^2 - a^2} = \sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} = 16\text{ metres}$$

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Example 2: 3D Box Spatial Diagonal

Problem: A shipping crate has internal dimensions: length $l = 12\text{ cm}$, width $w = 9\text{ cm}$, and height $h = 8\text{ cm}$. Can a rigid metal rod of length $17.5\text{ cm}$ lay flat inside the crate along its 3D spatial diagonal?

Solution:

1. Step 1: Calculate 2D Base Diagonal

   $$d_{\text{base}} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ cm}$$

   (Scaled $3\text{--}4\text{--}5$ triple: $9\text{--}12\text{--}15$.)

2. Step 2: Calculate 3D Spatial Diagonal

   $$d_{\text{3D}} = \sqrt{d_{\text{base}}^2 + h^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{ cm}$$

   (This forms the $8\text{--}15\text{--}17$ Pythagorean triple!)

3. Step 3: Compare Rod Length to Spatial Diagonal The maximum internal clearance is $17\text{ cm}$. A rigid rod of $17.5\text{ cm}$ exceeds $17\text{ cm}$ and cannot lay flat inside the crate.