Geometry

high importance~4 Q in Tier 137 formulas⚡ 19 shortcuts6 subtopics

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Pythagoras theorem and triplets

⏱ 4 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

In a right triangle, the square of the longest side (the hypotenuse) equals the sum of squares of the other two sides: $h^2=p^2+b^2$ . Learn the triplet families: $(3,4,5)$ , $(5,12,13)$ , $(7,24,25)$ , $(8,15,17)$ , $(9,40,41)$ and their multiples. The test also works backwards: if $a^2+b^2=c^2$ holds, the triangle is right-angled. The median to the hypotenuse equals half of it.

The rule and where it hides

In a right triangle with right angle at $C$ :

a^2+b^2=c^2

$c$ is the hypotenuse: the side opposite the right angle, always the longest.

Every ladder, pole-and-wire or walk-north-then-east question is this equation with a story wrapped around it. Draw the right triangle first. Label the hypotenuse before anything else. A 17 m ladder with its foot 8 m from a wall reaches $\sqrt{17^2-8^2}=\sqrt{225}=15$ m up. Same rule, story skin only.

Rule: The hypotenuse is opposite the right angle. Ladder, wire and shortest-path lengths are hypotenuses.

Learn the triplet families

Family	Multiples you will meet
3-4-5	6-8-10, 9-12-15, 12-16-20, 15-20-25
5-12-13	10-24-26, 15-36-39, 25-60-65
7-24-25	14-48-50
8-15-17	16-30-34
9-40-41, 12-35-37, 20-21-29	as they are

See two sides, recall the third. Hypotenuse 26 with one side 10 is 10-24-26 ( $5$ - $12$ - $13$ times 2). Diagonal 65 with one side 25 gives 60 instantly. Two sides 20 and 21? That is 20-21-29, straight from the table.

Tip: Check the leg ratios $3:4$ , $5:12$ , $8:15$ , $7:24$ first. Exam numbers come from these families.

Missing side: add or subtract?

Missing leg: subtract. $\text{leg}=\sqrt{h^2-a^2}$ .
Missing hypotenuse: add. $h=\sqrt{a^2+b^2}$ .

Legs $9$ and $12$ : $h=\sqrt{81+144}=\sqrt{225}=15$ , the 3-4-5 family times 3. Hypotenuse $25$ with leg $7$ : other leg $=\sqrt{625-49}=24$ .

Watch: Writing $\sqrt{h^2+a^2}$ for a missing leg is the most common rush error. Subtract for legs, add for the hypotenuse.

Is the triangle right-angled?

The rule works both ways. Square the two shorter sides, add, and compare with the square of the longest:

equal: right-angled
sum smaller ( $6,8,11$ : $100<121$ ): obtuse
sum larger ( $6,8,9$ : $100>81$ ): acute

One comparison answers the whole question. For $6,8,11$ : $36+64=100$ against $121$ , so obtuse. For $6,8,9$ : $100>81$ , so acute.

Two free results

Median to the hypotenuse $=\dfrac{c}{2}$ . Hypotenuse $13$ gives median $6.5$ .
Diagonals: rectangle $l\times b$ has diagonal $\sqrt{l^2+b^2}$ ; a square of side $a$ has diagonal $a\sqrt2$ . A square of side $5$ has diagonal $5\sqrt2$ . A rope from the top of a 12 m pole to a point 5 m out is the 5-12-13 hypotenuse, 13 m.

Special right triangles: 45-45-90 has sides $x$ , $x$ , $x\sqrt2$ ; 30-60-90 has $x$ , $x\sqrt3$ , $2x$ .

Example: Diagonal 65, one side 25: other side $=\sqrt{4225-625}=\sqrt{3600}=60$ . The 5-12-13 family, times 5.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Ladder and two-leg stories

How to spot it:

A ladder against a wall, a wire on a pole, walking north then east: a right-triangle story with one side missing.

a^2+b^2=c^2

Method

Draw the right triangle. The ladder, wire or return path is the hypotenuse.
Mark the two known sides.
Subtract for a missing leg, add for a missing hypotenuse.
Check the answer against a triplet family.

Why it works:

Every story is the same equation; only the names of the sides change.

Try this

A 17 m ladder leans against a wall with its foot 8 m from the base. How high up the wall does it reach?

Show solution

Height $=\sqrt{17^2-8^2}$ .
$=\sqrt{289-64}=\sqrt{225}$ .
$=15$ m (the 8-15-17 family).

Answer

15 m

Type 2common2 practice Q

Classify the triangle

How to spot it:

Three sides given; the triangle must be named right-angled, obtuse or acute.

a^2+b^2\lessgtr c^2

Method

Square the two shorter sides and add.
Compare with the square of the longest side.
Equal means right, smaller sum means obtuse, larger means acute.

Why it works:

The rule works both ways, so the comparison classifies the biggest angle.

Try this

What type of triangle has sides 9 cm, 12 cm and 15 cm?

Show solution

$9^2+12^2=81+144=225$ .
$15^2=225$ .
Equal, so it is right-angled (3-4-5 times 3).

Answer

Right-angled

Type 3common2 practice Q

Median to the hypotenuse

How to spot it:

A median from the right angle (or to the hypotenuse) appears; its length or the hypotenuse is asked.

m=\dfrac{c}{2}

Method

Confirm it is a right triangle and the median ends on the hypotenuse.
Median $=$ half the hypotenuse.
Double the median if the hypotenuse is asked.

Why it works:

The right-angle corner sits on the circle whose diameter is the hypotenuse, so the midpoint is equidistant from all three corners.

Try this

In a right triangle, the median to the hypotenuse is 6.5 cm. Find the hypotenuse.

Show solution

$c=2\times6.5$ .
$c=13$ cm.

Answer

13 cm

Type 4very common2 practice Q

Rectangle and square diagonals

How to spot it:

A diagonal of a rectangle, field or room with one dimension missing, or a square's side from its diagonal.

d=\sqrt{l^2+b^2},\quad a=\dfrac{d}{\sqrt2}

Method

The diagonal splits the rectangle into two right triangles.
Missing side $=\sqrt{d^2-\text{known}^2}$ .
For a square, side $=\dfrac{d}{\sqrt2}$ .

Why it works:

Every corner of a rectangle is a right angle, so the diagonal is a hypotenuse.

Try this

The diagonal of a rectangular field is 65 m and one side is 25 m. Find the other side.

Show solution

$\sqrt{65^2-25^2}$ .
$=\sqrt{4225-625}=\sqrt{3600}$ .
$=60$ m (25-60-65 is 5-12-13 times 5).

Answer

60 m

Type 5common2 practice Q

Triplet families in disguise

How to spot it:

Legs in a ratio like $3:4$ or $5:12$ with the hypotenuse given, or hypotenuse plus one leg with perimeter or area asked.

(3k,\ 4k,\ 5k)

Method

Spot the family in the ratio or the given pair.
Find the multiplier $k$ .
Write all three sides.
Answer what is asked: perimeter, area or the missing side.

Why it works:

Exam numbers come from these families, so recognition replaces calculation.

Try this

The legs of a right triangle are in the ratio $3:4$ and the hypotenuse is 20 cm. Find its perimeter.

Show solution

Legs $3k$ , $4k$ , so hypotenuse $5k$ .
$5k=20$ , so $k=4$ .
Sides $12$ , $16$ , $20$ : perimeter $=48$ cm.

Answer

48 cm

Formula sheet

Pythagoras

h^2=p^2+b^2

h = hypotenuse (longest side, opposite the right angle).

Missing leg

p=\sqrt{h^2-b^2}

Subtract when a leg is missing.

Median to hypotenuse

m=\dfrac{h}{2}

The median drawn from the right-angle corner.

Rectangle diagonal

d=\sqrt{l^2+b^2}

The corner angles of a rectangle are right angles.

Triangle type test

a^2\lessgtr b^2+c^2

a = longest side; equal means right, greater means obtuse, smaller means acute.

Shortcuts that save time

⚡ See two numbers, recall the third

12 with 13 gives 5. 24 with 25 gives 7. 15 with 17 gives 8. Multiples scale: 10-24-26 is 5-12-13 times 2.

Example

A 25 m ladder stands with its foot 7 m from a wall. How high up the wall does it reach?

Show solution

7-24-25 triplet.
Ladder 25 and foot 7 are given.
Height $=24$ m.

Answer

24 m

⚡ Median to the hypotenuse is half of it

In any right triangle, the median from the right angle equals half the hypotenuse.

Example

In a right triangle with hypotenuse 13 cm, find the median to the hypotenuse.

Show solution

$m=\dfrac{13}{2}$ .
$m=6.5$ cm.

Answer

6.5 cm

⚡ Classify: square the longest side only

To name the triangle type, compare the longest side's square with the sum of the other two squares.

Example

Which of these is right-angled: (i) 6, 8, 11 (ii) 6, 8, 10 (iii) 6, 8, 9?

Show solution

$6^2+8^2=100$ for all three.
(ii) $10^2=100$ : right-angled.
(i) $121>100$ obtuse; (iii) $81<100$ acute.

Answer

(ii) 6, 8, 10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding squares when a leg is missing.

A missing leg subtracts: $\sqrt{h^2-a^2}$ . Only the hypotenuse adds.

Mistake 02

Treating a leg as the hypotenuse because it looks longest.

The hypotenuse is opposite the right angle. Check the angle, not the drawing.

Mistake 03

Calling any near-triplet right-angled without checking.

Test $a^2+b^2=c^2$ on the squares. 6, 8, 11 fails ( $100\ne121$ ).

Mistake 04

Confusing the median to the hypotenuse with half a leg.

Half the hypotenuse, always. The right-angle corner is what matters.

Mistake 05

Using $a\sqrt2$ for a rectangle diagonal.

$a\sqrt2$ is only for squares. Rectangles need $\sqrt{l^2+b^2}$ .

Quick revision

Read this the night before the exam.

$h^2=p^2+b^2$ ; the hypotenuse is opposite the right angle.
Triplets: 3-4-5, 5-12-13, 7-24-25, 8-15-17, 9-40-41 and multiples. Spot, don't compute.
Missing leg: subtract squares. Missing hypotenuse: add squares.
Converse: equal squares means right; longer longest side means obtuse; shorter means acute.
Median to the hypotenuse $=$ half the hypotenuse.
Rectangle diagonal $\sqrt{l^2+b^2}$ ; square diagonal $a\sqrt2$ .

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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