Pythagoras' theorem

Unit 6 Pythagoras' theorem

Key unit competence: Solve right-angled triangles in real-life situations using Pythagoras' theorem.

Key ideas

• In a right-angled triangle, the hypotenuse is the side opposite the right angle; it is the longest side. The other two sides are the legs.
• Pythagoras' theorem: the square of the hypotenuse equals the sum of the squares of the other two sides: $c^2=a^2+b^2$.
• To find a leg: $a^2=c^2-b^2$.
• Converse: if $c^2=a^2+b^2$ for the sides of a triangle, then the triangle is right-angled (opposite $c$).
• Pythagorean triples are whole numbers $(a,b,c)$ with $a^2+b^2=c^2$: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), and multiples like (6, 8, 10).
• Applications: ladders, diagonals of rectangles ($d=\sqrt{l^2+w^2}$), heights of isosceles triangles, distances between points: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$.

Formulas to know

Common misconceptions

• "$c=a+b$." No: $c^2=a^2+b^2$; for 3 and 4, $c=5$, not 7.
• "To find a leg, add the squares." No: subtract from the square of the hypotenuse.
• "Pythagoras works in every triangle." Only in right-angled triangles.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find the hypotenuse of a right triangle with legs 9 cm and 40 cm.

   Show answer

   $\sqrt{81+1600}=\sqrt{1681}=41$ cm.

   Common mistake: Adding the legs: 49 cm.

2. A pole is supported by a wire from its top to a point 5 m from its foot. The wire is 13 m long. How tall is the pole?

   Show answer

   $\sqrt{169-25}=12$ m.

   Common mistake: Thinking the pole is the hypotenuse.

3. A rectangular field is 60 m by 80 m. How much shorter is walking along the diagonal than along two sides?

   Show answer

   Diagonal $=\sqrt{3600+6400}=100$ m; two sides $=140$ m; saving 40 m.

   Common mistake: Comparing with one side only.

4. The height of an isosceles triangle is 12 cm and its base is 10 cm. Find the equal sides and the perimeter.

   Show answer

   $\sqrt{12^2+5^2}=13$ cm; perimeter $=13+13+10=36$ cm.

   Common mistake: Using the full base 10 instead of half (5).

5. Determine whether the triangle with sides 10, 24, 26 is right-angled.

   Show answer

   $100+576=676=26^2$. Yes.

   Common mistake: Squaring the wrong side as the hypotenuse.

6. Generate a Pythagorean triple from (5, 12, 13) and check it.

   Show answer

   Multiply by 3: (15, 36, 39): $225+1296=1521=39^2$. ✓

   Common mistake: Adding 3 to each number (that does not keep the property).

7. A 6.5 m ladder reaches a window 6 m high. How far is the foot of the ladder from the wall?

   Show answer

   $\sqrt{42.25-36}=\sqrt{6.25}=2.5$ m.

   Common mistake: Adding the squares.

8. Find the distance between $P(-3,-2)$ and $Q(5,4)$.

   Show answer

   $\sqrt{8^2+6^2}=10$.

   Common mistake: Sign errors: $5-(-3)=8$, not 2.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Which is the missing number in the triple (8, 15, ?)?	a. 17 cm
   2. Find the diagonal of a rectangle 8 cm by 15 cm.	b. 17
   3. The legs of a right triangle are $x$ and $x+7$, and the hypotenuse is 13. Find $x$.	c. 5
   4. Find the space diagonal of a cuboid 3 cm by 4 cm by 12 cm.	d. 13 cm

   Show answer

   1 → b (17) · 2 → a (17 cm) · 3 → c (5) · 4 → d (13 cm)

   • $64+225=289=17^2$.
   • $\sqrt{64+225}=\sqrt{289}=17$.
   • Step 1. $x^2+(x+7{)}^2=169\Rightarrow 2x^2+14x-120=0\Rightarrow x^2+7x-60=0$.
     Step 2. $(x+12)(x-5)=0$; $x>0$, so $x=5$.
   • Step 1. Base diagonal: $\sqrt{3^2+4^2}=5$ cm.
     Step 2. Space diagonal: $\sqrt{5^2+12^2}=13$ cm.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   14  ·  10  ·  $5\sqrt{2}$ cm  ·  right-angled
   1. The distance between $(-2,1)$ and $(4,9)$ is .
   2. The sides of a triangle are 20, 21 and 29. The triangle is .
   3. A square has side 5 cm. Its diagonal is .
   Show answer

   1: 10 · 2: right-angled · 3: $5\sqrt{2}$ cm

   • $\sqrt{6^2+8^2}=10$.
   • $400+441=841=29^2$.
   • $\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. A rhombus has diagonals 10 cm and 24 cm. Find the length of a side.	a. 8 cm
   2. An isosceles triangle has equal sides 10 cm and base 12 cm. Find its height.	b. 13 cm
   3. Find $x$ if $x^2+9^2=15^2$.	c. 168 cm²
   4. A rectangle has diagonal 25 cm and one side 7 cm. Find its area.	d. 12

   Show answer

   1 → b (13 cm) · 2 → a (8 cm) · 3 → d (12) · 4 → c (168 cm²)

   • Step 1. The diagonals bisect each other at right angles: half-diagonals 5 and 12.
     Step 2. Side $=\sqrt{5^2+12^2}=13$ cm.
   • Half the base is 6: $\sqrt{100-36}=8$.
   • $x^2=225-81=144$.
   • Step 1. Other side: $\sqrt{25^2-7^2}=\sqrt{576}=24$ cm.
     Step 2. Area $=7\times 24=168$ cm².

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\sqrt{7}$  ·  5  ·  6 m  ·  13 km
   1. Find the distance between $(1,2)$ and $(4,6)$. →
   2. A 10 m ladder reaches 8 m up a wall. How far is its foot from the wall? →
   3. A man walks 12 km east and then 5 km north. How far is he from his start? →
   Show answer

   1: 5 · 2: 6 m · 3: 13 km

   • $\sqrt{3^2+4^2}=5$.
   • $\sqrt{100-64}=6$.
   • $\sqrt{144+25}=13$.

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