Finding the Hypotenuse
Use Pythagoras' theorem to calculate the hypotenuse when both shorter sides are known. Substitute, sum the squares and take the square root.
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You know a right triangle has legs of 5 cm and 12 cm. Estimate the hypotenuse first (longer or shorter than 12?), then calculate it using Pythagoras.
To find the hypotenuse, square both legs, add them, then square-root the total. The square root undoes the squaring at the end.
Start with $c^2 = a^2 + b^2$. Substitute the leg values, add the squares, then take $\sqrt{}$ of both sides to isolate $c$. Always round to 2 decimal places unless told otherwise — and include the unit (cm, m, km).
Know
- The formula $c^2 = a^2 + b^2$ for the hypotenuse
- To take the square root only at the end
- To round to 2 decimal places by default
Understand
- Why the hypotenuse is always longer than either leg
- Why $\sqrt{a^2 + b^2} \neq a + b$
- How rounding errors accumulate when rounding mid-calculation
Can Do
- Substitute leg values into $c^2 = a^2 + b^2$
- Use a calculator to find $\sqrt{\text{sum}}$
- Present answers with correct units and rounding
Wrong: “$c = a + b$.” No — you must square first, sum, then square-root.
Right: Use $c = \sqrt{a^2 + b^2}$. Square the legs, add, square-root.
Wrong: “$\sqrt{9 + 16} = 3 + 4$.” Wrong. $\sqrt{25} = 5$, not 7.
Right: The square root acts on the WHOLE sum, not term-by-term.
Solving for the hypotenuse follows a reliable four-step pattern. Get into the habit so it becomes automatic.
Step 1: Label the sides (legs $a$, $b$; hyp $c$). Step 2: Substitute into $c^2 = a^2 + b^2$. Step 3: Compute $a^2 + b^2$. Step 4: $c = \sqrt{\text{sum}}$, rounded to 2 d.p.
Calculator order matters. A wrong key press gives a wildly wrong answer. Practice the key sequence below.
| To find | Key sequence |
|---|---|
| $c$ from $a=5$, $b=12$ | $\sqrt{}$ ( 5 $x^2$ + 12 $x^2$ ) = |
| Result | 13 (exact in this case) |
Watch Me Solve It · 3 examples
- 1Set up$c^2 = 5^2 + 12^2$Substitute leg values into Pythagoras.
- 2Compute$c^2 = 25 + 144 = 169$
- 3Square-root$c = \sqrt{169} = 13$ cm169 is a perfect square — no rounding needed.
- 1Identify hypLadder = hypotenuse (longest, slanted).The ladder is opposite the right angle at the wall/ground corner.
- 2Substitute$c^2 = 1.8^2 + 3.2^2 = 3.24 + 10.24 = 13.48$
- 3Root + round$c = \sqrt{13.48} \approx 3.67$ m
- 1Set updiagonal$^2 = 56.7^2 + 31.9^2$The diagonal of any rectangle is the hypotenuse of two right triangles.
- 2Compute$= 3214.89 + 1017.61 = 4232.5$
- 3Rootdiagonal $= \sqrt{4232.5} \approx 65.06$ inches
Common Pitfalls
Method
- Label legs $a$, $b$; hyp $c$
- $c^2 = a^2 + b^2$
- $c = \sqrt{a^2 + b^2}$
- Round to 2 d.p.
Calculator
- $\sqrt{}$ ( $a$ $x^2$ + $b$ $x^2$ ) =
- Brackets matter
- Use $x^2$ key
Sanity check
- $c$ must be longer than each leg
- Common triples skip rounding
- Watch units
Example
- Legs 5, 12 → $c = 13$
- Legs 1.8, 3.2 → $c \approx 3.67$
- Legs 7, 24 → $c = 25$
How are you completing this lesson?
Brain Trainer · 4 problems
Four quick drills to lock in today's skill. Try each, then reveal the answer.
-
1 Legs 6 and 8 — find $c$.
$6^2+8^2=36+64=100$.$c = 10$ -
2 Legs 9 and 12 — find $c$.
$81+144=225$, $\sqrt{225}=15$.$c = 15$ -
3 Legs 1 and 1 — find $c$ to 2 d.p.
$1+1=2$, $\sqrt{2} \approx 1.41$.$c \approx 1.41$ -
4 Legs 2.5 and 6 — find $c$ to 2 d.p.
$6.25+36=42.25$, $\sqrt{42.25}=6.5$.$c = 6.5$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Find the hypotenuse of a right triangle with legs (a) 9 cm and 40 cm, (b) 2 m and 5 m (2 d.p.).
Q7. A ladder leans against a wall. The foot is 1.5 m from the wall, the top reaches 3.6 m up. How long is the ladder?
Q8. A rectangular soccer field is 100 m by 64 m. A player runs along the diagonal. Find the diagonal length and explain why it is longer than running 100 m + 64 m / 2.
Quick Check
1. B — $\sqrt{64+225}=\sqrt{289}=17$.
2. C — $\sqrt{9+16}=5$ cm.
3. A — $c=\sqrt{a^2+b^2}$.
4. D — $\sqrt{13.48}\approx 3.67$.
5. B — $c^2$ exceeds either $a^2$ or $b^2$ individually.
Show Your Working Model Answers
Q6 (3 marks): (a) $c^2=81+1600=1681$, $c=41$ cm [2]. (b) $c^2=4+25=29$, $c=\sqrt{29}\approx 5.39$ m [1].
Q7 (2 marks): $c^2=2.25+12.96=15.21$ [1]. $c=\sqrt{15.21}=3.9$ m [1].
Q8 (4 marks): Diagonal$^2 = 100^2+64^2 = 10000+4096 = 14096$ [1]. $d=\sqrt{14096}\approx 118.73$ m [1]. The diagonal is shorter than 100+64=164 m because the triangle inequality ensures the third side is less than the sum of the other two [1]. The diagonal is the shortest path between opposite corners through the interior — straight-line distance is always less than going around the perimeter [1].
Stacking two right triangles
A right triangle has legs 3 and 4. Glued onto its hypotenuse is another right triangle, whose other leg is 12 (perpendicular to the shared hypotenuse). Find the longest side of the second triangle.
Reveal solution
First hypotenuse: $\sqrt{9+16}=5$. Second hypotenuse: $\sqrt{5^2+12^2}=\sqrt{169}=13$.
Formula
$c = \sqrt{a^2+b^2}$
Method
Label, substitute, sum, root
Round
2 d.p. unless told otherwise
Common triples
3-4-5, 5-12-13, 8-15-17
Sanity check
$c$ > each leg
Calculator
$\sqrt{}$ before bracket
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