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Pythagoras’ theorem

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visibility 94update 6 months agobookmarkshare

🎯 In this topic you will

Understand the relationship between the three sides of a right-angled triangle (Pythagoras’ theorem).
Use two side lengths of a right-angled triangle to calculate the third side.

🧠 Key Words

hypotenuse
Pythagoras’ theorem

Show Definitions

hypotenuse: The longest side of a right-angled triangle, opposite the right angle.
Pythagoras’ theorem: In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: $c^2=a^2 b^2$.

A builder has built a wall. He wants to build a second wall at right angles to the first wall. How can he do this? One way is to use the $3\!-\!4\!-\!5$ rule.

The builder measures out lengths of $3$ metres, $4$ metres and $5$ metres on the ground in a triangle with the $3$-metre side along the first wall as shown. They build the second wall along the $4$-metre side. The triangle is right-angled. The right angle is between the $3$ m and $4$ m sides. The longest side of a right-angled triangle is called the hypotenuse. In this example, the hypotenuse is $5$ m long.

$3^2=9$, $4^2=16$ and $5^2=25$. Hence $3^2 + 4^2 = 5^2$.

Tip

$9 + 16 = 25$

The square of the hypotenuse is equal to the sum of the squares of the other two sides.

This result is true for any right-angled triangle: $a^2 + b^2 = c^2$.

This result is called Pythagoras’ theorem. Pythagoras was born on the Greek island of Samos and wrote a proof of this theorem over $2500$ years ago.

📘 Worked example

ABCD is a rectangle. Calculate the length of the diagonal AC.

Answer:

ADC is a right-angled triangle and AC is the hypotenuse. Let $AC=x$ cm. By Pythagoras’ theorem,

$x^2 = 7.5^2 + 11.3^2 = 56.25 + 127.69 = 183.94$
$x = \sqrt{183.94} \approx 13.6$ cm (to 1 d.p.).

In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: $c^2 = a^2 + b^2$. Here, the legs are $7.5$ cm and $11.3$ cm, so add their squares and then take the square root to find $AC$. Finally round to one decimal place.

The square root of $183.94$ is irrational, so its decimal does not terminate or repeat. Give the answer to an appropriate degree of accuracy (here, 1 d.p.).

📘 Worked example

Ari says: $a^2 = 6.0^2 + 3.5^2 = 48.25$, and so $a=\sqrt{48.25}=6.9$.

a. Show that Ari is incorrect.

b. Work out the correct value of $a$.

Answer:

a. Ari is incorrect because in the triangle the hypotenuse is the 6.0 cm side (it lies opposite the right angle), not the side of length $a$. So $a$ is a shorter leg, and you must subtract its square from the hypotenuse’s square.

b. Using Pythagoras’ theorem with hypotenuse $6.0$ cm and other leg $3.5$ cm: $a^2 3.5^2 = 6.0^2$ so $a^2 = 6.0^2 - 3.5^2 = 36 - 12.25 = 23.75$, and $a = \sqrt{23.75} \approx 4.9$ cm to 1 d.p.

Key idea: The hypotenuse is the side opposite the right angle. In a right-angled triangle, $c^2 = a^2 + b^2$ with $c$ the hypotenuse. If you’re solving for a leg, rearrange to $\text{leg}^2 = \text{hypotenuse}^2 - \text{other leg}^2$.

🧠 PROBLEM-SOLVING Strategy

Pythagoras’ Theorem: sides, diagonals & quick checks

Use this for Exercises $5.5$–$6$, $11$–$14$: find unknown sides, test for right angles, and handle diagonals of rectangles/TVs.

Spot the right angle and label the hypotenuse. The hypotenuse is opposite the right angle and is always the longest side.
Pick the correct form of Pythagoras.
- Hypotenuse unknown: $c=\sqrt{a^{2}+b^{2}}$.
- A leg unknown: $a=\sqrt{c^{2}-b^{2}}$ (or $b=\sqrt{c^{2}-a^{2}}$).
Diagonals are hypotenuses. For a rectangle/TV of width $w$ and height $h$, diagonal $d=\sqrt{w^{2} h^{2}}$. Screen “size” is this diagonal length.
Converse test (is it right-angled?). If the three sides satisfy $a^{2} b^{2}=c^{2}$ (with $c$ the largest), then the included angle opposite $c$ is a right angle.
Keep exact values until the end. Simplify surds (e.g., $\sqrt{98}=7\sqrt2$), then round only at the final step (e.g., to $1$ d.p.).
Units & context. Convert where needed (e.g., mm ↔ cm ↔ m). For ladders/walls and pitch diagonals, identify the horizontal/vertical legs before calculating.
Handy patterns.
- Pythagorean spiral lengths from a fixed point are $\sqrt{1},\sqrt{2},\sqrt{3},\dots$.
- Common integer triples help you check work fast.

Situation	Formula	Example result
Hypotenuse	$c=\sqrt{a^{2} b^{2}}$	$\sqrt{8^{2} 6^{2}}=10$
Missing leg	$a=\sqrt{c^{2}-b^{2}}$	$\sqrt{15^{2}-9^{2}}=12$
Rectangle/TV diagonal	$d=\sqrt{w^{2}+h^{2}}$	$\sqrt{70^{2}+39^{2}}\approx 80$ (cm)
Converse check	$a^{2}+b^{2}\stackrel{?}{=}c^{2}$	$5.1^{2}+6.8^{2}=8.5^{2}$

Common triples:$(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$.

❓ EXERCISE

1. Calculate the length of the hypotenuse in each of these triangles.

👀 Show answers

a. Legs: 8 cm and 6 cm. Hypotenuse = $\sqrt{8^2+6^2}=\sqrt{64 36}=\sqrt{100}=10$ cm.

b. Legs: 12 cm and 5 cm. Hypotenuse = $\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$ cm.

c. Right angle is between the sides labelled 8 cm and 15 cm, so these are the legs. Hypotenuse = $\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17$ cm.

❓ EXERCISES

2. Calculate the length of the hypotenuse in each of these triangles. Round your answers to $1$ decimal place.

👀 Show answer

a. Legs $2.5$ cm and $3.5$ cm → hypotenuse $=\sqrt{2.5^2+3.5^2}=\sqrt{6.25+ 12.25}=\sqrt{18.5}\approx 4.3$ cm.

b. Legs $7.4$ cm and $9.6$ cm → hypotenuse $=\sqrt{7.4^2+9.6^2}=\sqrt{54.76 92.16}=\sqrt{146.92}\approx 12.1$ cm.

c. Legs $13.5$ cm and $4.5$ cm → hypotenuse $=\sqrt{13.5^2+4.5^2}=\sqrt{182.25+ 20.25}=\sqrt{202.5}\approx 14.2$ cm.

3. Calculate the length of the missing side in each of these triangles.

👀 Show answer

a. Hypotenuse $15$ cm, other leg $9$ cm → missing side $=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12$ cm.

b. Hypotenuse $8.0$ m, other leg $6.4$ m → missing side $=\sqrt{8.0^2-6.4^2}=\sqrt{64-40.96}=\sqrt{23.04}=4.8$ m.

c. Hypotenuse $85$ mm, one leg $40$ mm → missing side $=\sqrt{85^2-40^2}=\sqrt{7225-1600}=\sqrt{5625}=75$ mm.

4. Calculate the length of the missing side in each of these triangles. Round your answers to $1$ decimal place.

👀 Show answer

a. Hypotenuse $8.0$ cm, other leg $4.5$ cm → missing side $=\sqrt{8.0^2-4.5^2}=\sqrt{64-20.25}=\sqrt{43.75}\approx 6.6$ cm.

b. Hypotenuse $10.2$ cm, other leg $8.9$ cm → missing side $=\sqrt{10.2^2-8.9^2}=\sqrt{104.04-79.21}=\sqrt{24.83}\approx 5.0$ cm.

c. Legs $12.6$ m and $4.8$ m → hypotenuse $=\sqrt{12.6^2+4.8^2}=\sqrt{158.76+ 23.04}=\sqrt{181.8}\approx 13.5$ m.

5. In this question, you can leave your answers as square roots. $AB=BC=CD=DE=1$ unit

a. Calculate the length of $AC$.

b. Calculate the length of $AD$.

c. Calculate the length of $AE$.

d. Show how the pattern continues.

e. If you continue the pattern, will there be a line with length $3$? Or with length $4$? Explain your answer.

👀 Show answer

a. Right triangle with legs $AB=1$ and $BC=1$ → $AC=\sqrt{1^2+1^2}=\sqrt{2}$.

b. Next right triangle has legs $AC=\sqrt{2}$ and $CD=1$ → $AD=\sqrt{(\sqrt{2})^2+ 1^2}=\sqrt{3}$.

c. Next step: legs $AD=\sqrt{3}$ and $DE=1$ → $AE=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{4}=2$.

d. Each new segment from $A$ to the next point has length $\sqrt{n}$ for $n=1,2,3,4,\dots$ (a spiral of right triangles, each time adding a unit side at right angles).

e. Yes. When $n=9$, the length is $\sqrt{9}=3$; when $n=16$, the length is $\sqrt{16}=4$. So both lengths occur in the pattern.

❓ EXERCISES

6. The size of a TV screen is the length of the diagonal.

a. A TV screen is 70 cm wide and 39 cm high. Show that this is an 80 cm screen.

👀 Show answer

Diagonal = √(70²+39²) = √(4900+1521) = √6421 ≈ 80.1 cm. Rounded to the nearest centimetre, it is an 80 cm screen.

b. Another TV screen is 105 cm wide and 58 cm high. Calculate the size of this screen.

👀 Show answer

Diagonal = √(105²+58²) = √(11025+3364) = √14389 ≈ 119.95 cm. So it is a 120 cm screen (to the nearest cm).

7. A ladder is leaning against a wall. The length of the ladder is 3.50 m. The foot of the ladder is 0.91 m from the foot of the wall. How far up the wall is the top of the ladder? Round your answer to the nearest centimetre.

👀 Show answer

Height = √(3.50² − 0.91²) = √(12.25 − 0.8281) = √11.4219 ≈ 3.3796 m = 338 cm (nearest cm). (≈ 3.38 m)

a. Construct a triangle with sides 5.1 cm, 6.8 cm and 8.5 cm.

b. Use a calculation to show that one of the angles must be a right angle.

c. Check that the triangle you have drawn is right-angled.

👀 Show answers

b. Check Pythagoras: 5.1²+6.8² = 26.01 46.24 = 72.25, and 8.5² = 72.25. Since sum of the smaller two squares equals the largest square, the angle opposite the 8.5 cm side is a right angle.

c. On your construction, the right angle is between the 5.1 cm and 6.8 cm sides (opposite 8.5 cm). A set square/protractor confirms it.

9. Two sides of a right-angled triangle are 15 cm and 20 cm. Work out the possible lengths of the third side. Give reasons for your answers.

👀 Show answer

If 15 cm and 20 cm are the legs, the hypotenuse is √(15²+20²) = √625 = 25 cm.
If 20 cm is the hypotenuse and 15 cm a leg, the other leg is √(20² − 15²) = √175 = 5√7 ≈ 13.2 cm.

(15 cm cannot be the hypotenuse since the hypotenuse must be the longest side.)

10. A soccer pitch is 90 m long and 40 m wide. Zena walks from A to B round the edge of the pitch.

a. How far does Zena walk?

b. How much shorter is the distance if Zena walks from A to B in a straight line?

👀 Show answers

a. Along two edges: 90 m 40 m = 130 m.

b. Straight-line distance (diagonal): √(90²+40²) = √9700 ≈ 98.5 m. The route round the edge is longer by 130 − 98.5 ≈ 31.5 m.

❓ EXERCISES

11. Here are a square and a rectangle.

a. Show that the square and rectangle have the same perimeter.

b. Calculate the length of the diagonal of each shape.

c. Sketch another rectangle with the same perimeter. Calculate its diagonal.

d. Sofia says:

What evidence can you find to support Sofia’s conjecture?

👀 Show answers

a. Square side $=25\text{ mm}$, so perimeter $=4\times25=100\text{ mm}$. Rectangle sides $30\text{ mm}$ and $20\text{ mm}$, so perimeter $=2(30+20)=100\text{ mm}$. The perimeters are equal.

b. Square diagonal $=25\sqrt2\text{ mm}\approx35.36\text{ mm}$. Rectangle diagonal $=\sqrt{30^2+20^2}= \sqrt{1300}=10\sqrt{13}\text{ mm}\approx36.06\text{ mm}$.

c. Example with same perimeter: $40\text{ mm}\times10\text{ mm}$ (since $2(40+10)=100$). Its diagonal $=\sqrt{40^2+10^2}=\sqrt{1700}=10\sqrt{17}\text{ mm}\approx41.23\text{ mm}$.

d. Numerical evidence: with equal perimeter $100\text{ mm}$, the square’s diagonal $25\sqrt2\approx35.36$ mm is smaller than the rectangle’s $10\sqrt{13}\approx36.06$ mm, and also smaller than the $40\times10$ rectangle’s $10\sqrt{17}\approx41.23$ mm. Reason: for fixed perimeter $P$, if the rectangle has sides $a$ and $b$ with $a b=\tfrac P2$, then the diagonal is $d=\sqrt{a^2+b^2}=\sqrt{(a b)^2-2ab}=\sqrt{(P/2)^2-2ab}$. This is minimized when $ab$ is maximized, which occurs at $a=b$ (a square). So the square gives the smallest diagonal for a given perimeter.

12. One side of a right-angled triangle is $\sqrt{17}$ units. The other two sides are integers. How long are the other two sides? Is there more than one possible answer?

👀 Show answer

There are two possibilities:

Case 1 (hypotenuse $=\sqrt{17}$): Need integers $x,y$ with $x^2 y^2=17$. One solution is $1^2+4^2=17$. So sides are $1$, $4$, and $\sqrt{17}$.
Case 2 (leg $=\sqrt{17}$): Need integers $k,h$ with $h^2-k^2=17$. Factor $(h-k)(h k)=17\Rightarrow h-k=1,\ h k=17$, giving $h=9$, $k=8$. So sides are $\sqrt{17}$, $8$, and $9$.

Therefore there is more than one possible answer: $(1,4,\sqrt{17})$ or $(\sqrt{17},8,9)$.

🧠 Think like a Mathematician

Theme: Diagonals of rectangles with Pythagoras’ theorem

Equipment: Ruler, pencil, calculator

Method (solo):

Sketch each rectangle with the given side lengths.
Draw the diagonal. This forms a right-angled triangle with the sides as the legs.
Use Pythagoras: for sides $a$ and $b$, diagonal $d$ satisfies $d=\sqrt{a^{2} b^{2}}$.
Simplify the surd if possible and state the final expression/length.

Tasks:

a. A rectangle has sides $7.5$ units and $5.5$ units. Show that the diagonal has length $\sqrt{86.5}$ units.

b. A rectangle has sides $x$ units and $5.5$ units. Show that the diagonal has length $\sqrt{x^{2} 30.25}$ units.

c. A rectangle has sides $x$ and $y$. Let the diagonal be $d$. Find a formula for $d$ in terms of $x$ and $y$.

Show Answers

a.$d=\sqrt{7.5^{2}+5.5^{2}}=\sqrt{56.25+30.25}=\sqrt{86.5}$ units.
b.$d=\sqrt{x^{2}+5.5^{2}}=\sqrt{x^{2}+30.25}$ units.
c. By Pythagoras, $d=\sqrt{x^{2}+y^{2}}$.

❓ EXERCISES

14.

a. Each side of this square is $7$ units.

Show that:

i the length of the diagonal is $\sqrt{98}$

ii $\sqrt{98}=7\sqrt{2}$

👀 Show answer

Since the diagonal of a square makes a right-angled triangle with the sides, by Pythagoras, $d^2=7^2+7^2=49+49=98\;\Rightarrow\;d=\sqrt{98}$. Also, $\sqrt{98}=\sqrt{49\cdot2}=7\sqrt{2}$.

b. Each side of this square is $x$.

Show that the length of the diagonal is $x\sqrt{2}$.

👀 Show answer

Let the diagonal be $d$. Then $d^2=x^2+x^2=2x^2$, so $d=\sqrt{2x^2}=x\sqrt{2}$ (for $x>0$).

📘 What we've learned

Pythagoras’ theorem (right-angled triangles): $c^2=a^2 + b^2$ where $c$ is the hypotenuse (opposite the right angle).
Finding sides: hypotenuse $c=\sqrt{a^2+b^2}$; a leg $a=\sqrt{c^2-b^2}$.
Converse test for a right angle: if the largest side $c$ satisfies $a^2 b^2=c^2$, the triangle is right-angled.
Rectangle/TV diagonal: $d=\sqrt{w^2+h^2}$ (screen “size” is the diagonal).
Exact values & surds: keep in surd form then simplify (e.g., $\sqrt{98}=7\sqrt2$) and round only at the end when asked.
Real contexts (ladder/wall, soccer pitch): choose horizontal/vertical legs, apply the theorem, and keep units consistent.
Useful checks with Pythagorean triples: $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$.
Spiral/pattern idea: successive lengths can be $\sqrt1,\sqrt2,\sqrt3,\dots$, giving exact lengths like $3=\sqrt9$ and $4=\sqrt{16}$.

Pythagoras’ theorem

🎯 In this topic you will

🧠 Key Words

🧠 PROBLEM-SOLVING Strategy

Pythagoras’ Theorem: sides, diagonals & quick checks

❓ EXERCISE

❓ EXERCISES

❓ EXERCISES

❓ EXERCISES

🧠 Think like a Mathematician

❓ EXERCISES

📘 What we've learned

Related Past Papers

Related Tutorials

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