Geometry

high importance~4 Q in Tier 136 formulas⚡ 19 shortcuts6 subtopics

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Subtopic 4 of 6·← Congruence, similarity and BPT·Quadrilaterals and polygons →

Pythagoras theorem and triplets

In a right triangle with hypotenuse $h$ : $h^2=p^2+b^2$ . The converse also holds — it is how CGL tests "which triangle is right-angled". Memorise the standard triplets (and their multiples): $(3,4,5)\ (5,12,13)\ (7,24,25)\ (8,15,17)\ (9,40,41)\ (11,60,61)\ (12,35,37)\ (20,21,29)$ Useful extras: median to the hypotenuse $=\frac{h}{2}$ ; the diagonal of a rectangle is $\sqrt{l^2+b^2}$ ; in an obtuse triangle $a^2>b^2+c^2$ (acute if $a^2<b^2+c^2$ ) where $a$ is the longest side.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The theorem and how questions hide it

In a right triangle with the right angle at $C$ : $a^2 + b^2 = c^2$ , where $c$ is the hypotenuse (opposite the right angle — always the longest side). Every "ladder", "pole and wire", "man walks north then east" question is this equation with a story wrapped around it. Draw the right triangle, label the hypotenuse first, then substitute.

The triplet arsenal (memorise cold)

Family	Multiples you will meet
3-4-5	(6,8,10), (9,12,15), (12,16,20), (15,20,25), (18,24,30), (21,28,35)
5-12-13	(10,24,26), (15,36,39), (20,48,52)
8-15-17	(16,30,34)
7-24-25	(14,48,50)
9-40-41, 20-21-29, 12-35-37, 28-45-53	as-is

A question saying "hypotenuse 26, one side 10" is really (10, 24, 26) — recognise the family and the third side appears without any squaring. Check leg ratios like $\frac{3}{4}$ , $\frac{5}{12}$ , $\frac{8}{15}$ for the same reason. Micro-example: diagonal 65 with one side 25 is the 5-12-13 family times 5, so the other side is 60 instantly. When the sides don't fit a family, square honestly — but always subtract the smaller square from the larger for a missing leg, and add them for a missing hypotenuse; writing the wrong direction is the most common slip in a rush.

The converse (and the obtuse/acute test)

If $a^2 + b^2 = c^2$ , the triangle is right-angled at the side facing $c$ . If $a^2 + b^2 < c^2$ , it is obtuse; if $a^2+b^2 > c^2$ , acute. Classification questions are one comparison away — compute both squares, compare, name the triangle.

Two formulas built on Pythagoras

Median to the hypotenuse $= \frac{c}{2}$ — the right-angle vertex sits on the circle with the hypotenuse as diameter, so the hypotenuse midpoint is equidistant from all three vertices. Asked constantly, almost free.
Rectangle/square diagonals: diagonal of $l \times b$ is $\sqrt{l^2+b^2}$ ; diagonal of a square of side $a$ is $a\sqrt{2}$ , and diagonal $d$ gives side $= \frac{d}{\sqrt2}$ .

Standard question shells

Ladder: length is the hypotenuse; wall height and ground distance are the legs.
Two poles with a rope/crow between tops: horizontal gap is one leg, difference of heights is the other.
Displacement: north-then-east legs; shortest distance back is the hypotenuse.
Diagonal of a field/room: one leg known, diagonal known → other leg.

Quick revision

$a^2+b^2=c^2$ ; hypotenuse is opposite the right angle.
Triplet families 3-4-5, 5-12-13, 8-15-17, 7-24-25 (+ multiples) — spot, don't compute.
Converse: equality → right; $a^2+b^2<c^2$ → obtuse; $>$ → acute.
Median to hypotenuse $= \frac{c}{2}$ .
Rectangle diagonal $\sqrt{l^2+b^2}$ ; square diagonal $a\sqrt2$ .

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Ladder / two-leg Pythagoras storiesvery common3 practice Q

How to spot it:

A ladder against a wall, a pole with a guy-wire, walking north then east — any right-triangle story where the hypotenuse (or a leg) is missing.

a^2 + b^2 = c^2

Draw the right triangle; the ladder/wire/displacement is the hypotenuse.
Identify the two knowns; the unknown is a leg ( $\sqrt{c^2-a^2}$ ) or the hypotenuse ( $\sqrt{a^2+b^2}$ ).
Match against a triplet family to skip the arithmetic.

Why: every shell is the same equation; only the story changes which side is which.

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

Leg $= \sqrt{17^2 - 8^2} = \sqrt{225} = 15$ m — the 8-15-17 family.

Type 2: Converse: classify the trianglecommon2 practice Q

How to spot it:

Three sides are given and the triangle must be identified as right/obtuse/acute, or 'is it right-angled?' is asked.

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

Square the two smaller sides; add.
Compare with the square of the largest side.
Equal → right; smaller → obtuse; larger → acute.

Why: the theorem is an if-and-only-if, so the inequality direction classifies the angle facing the longest side.

Example: Sides 9 cm, 12 cm and 15 cm: what type of triangle is formed?

$9^2 + 12^2 = 81 + 144 = 225 = 15^2$ → right-angled (3-4-5 $\times 3$ ).

Type 3: Median to the hypotenusecommon2 practice Q

How to spot it:

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

m_c = \frac{c}{2}

Right triangle + median to hypotenuse → the median is half the hypotenuse.
From the median, double for the hypotenuse; then Pythagoras gives any leg.

Why: the right-angle vertex lies on the circle whose diameter is the hypotenuse, and the circumcentre (hypotenuse midpoint) is equidistant from all three vertices.

Example: In a right triangle the median drawn to the hypotenuse is $6.5$ cm. The hypotenuse is:

$c = 2 \times 6.5 = 13$ cm (a 5-12-13 hypotenuse, in fact).

Type 4: Rectangle and square diagonalsvery common2 practice Q

How to spot it:

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

d = \sqrt{l^2 + b^2};\quad a = \frac{d}{\sqrt{2}}

The diagonal splits the rectangle into two right triangles — apply $a^2+b^2=c^2$ .
Missing dimension: $\sqrt{d^2 - \text{known}^2}$ .
Square: side $= \frac{d}{\sqrt2}$ , or use the 1-1- $\sqrt2$ shape.

Why: the corner angles of a rectangle are right angles, so the diagonal is literally a hypotenuse.

Example: The diagonal of a rectangular field is $65$ m and one side is $25$ m. The other side is:

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

Type 5: Triplet families in disguisecommon2 practice Q

How to spot it:

A ratio of legs ( $3:4$ , $5:12$ …) with the hypotenuse given, or hypotenuse + one leg with area/perimeter asked.

(3k, 4k, 5k)

Spot the family in the ratio or the given pair (e.g. 10 & 26 → 5-12-13 ×2).
Find $k$ ; write all three sides.
Answer the actual ask — perimeter, area, or the missing side.

Why: exam numbers almost always come from these families; recognising them turns a 60-second calculation into a 10-second recall.

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

Legs $3k, 4k$ with $5k = 20 \Rightarrow k = 4$ : legs $12$ and $16$ ; perimeter $= 12+16+20 = 48$ cm.

Formulas

Pythagoras

h^2=p^2+b^2

Median to hypotenuse

m_{\text{hyp}}=\frac{h}{2}

Rectangle diagonal

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

Triangle type test

a\ \text{longest}:\ a^2\gtrless b^2+c^2\Rightarrow \text{obtuse}/\text{acute}

Shortcut tricks

⚡ Triplet recognition

See two numbers, recall the third: 12 and 13 → 5; 24 and 25 → 7; 15 and 17 → 8. Multiples scale: 6-8-10, 9-12-15, 10-24-26 are all 3-4-5 families.

Example: A ladder 25 m long stands with its foot 7 m from a wall. How high does it reach?

7-24-25 triplet → height $=24$ m.

⚡ Median to the hypotenuse is half of it

In any right triangle the median from the right angle equals half the hypotenuse (three equal pieces: the two half-hypotenuses and the median).

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

$\frac{13}{2}=6.5$ cm.

⚡ Converse: check $a^2$ vs $b^2+c^2$

To identify the right/obtuse/acute triangle, square only the longest side and compare.

Example: Which of the triangles with sides (i) 6, 8, 11 (ii) 6, 8, 10 (iii) 6, 8, 9 is right-angled?

(ii) $10^2=6^2+8^2$ . (i) is obtuse ( $121>100$ ), (iii) is acute ( $81<100$ ).

Where students lose marks

Adding instead of subtracting: the unknown leg is $\sqrt{h^2-b^2}$ , not $\sqrt{h^2+b^2}$ .
Calling any big triangle right-angled without checking a triplet or the squares.
Half-hypotenuse median confused with the median to a leg.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

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