CDS Mathematics · Quadrilaterals

Any Quadrilateral: Diagonals, Triangles and Areas

A diagonal splits any quadrilateral into two triangles, so triangle rules — the triangle inequality, Pythagoras, area on a common base — answer most questions.

All Quadrilaterals notes CDS Mathematics strategy

Why this matters

Eleven PYQs, four of them HARD. None needs a special shape: draw a diagonal and work with the two triangles. Bounding a diagonal is the triangle inequality twice; area questions compare triangles that share the diagonal as a base.

Concept 1 of 2: Lengths through a diagonal

A diagonal is a side of two triangles at once. Each triangle limits it, so its possible lengths are the overlap of two ranges. A right angle on each side makes a chain of Pythagorean triples.

Definition

Sides $a, b$ around a diagonal $x$ : $|a - b| < x < a + b$ . Apply it in BOTH triangles and intersect the ranges.
The perimeter always exceeds the sum of the diagonals.
Right angles at $B$ and at $C$ (in $\angle ACD$ ): $AB^2 + BC^2 = AC^2$ and $AC^2 + CD^2 = AD^2$ , two triples sharing $AC$ .
A rectangle's diagonals are equal and bisect each other.

Range of a diagonal

|a - b| < x < a + b

Worked example

ABCD

AB = 5

BC = 14

CD = 8

and

DA = 7

. Find the range of

BD

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2024 · CDS (I) 2024 — Elementary Mathematics · Q48Moderate

Example 1 · Quadrilaterals · General Quadrilaterals and their Diagonals

In a quadrilateral ABCD, AB = 6 cm, BC = 18 cm, CD = 6 cm and DA = 10 cm. If diagonal BD = x, then which one of the following is correct ?

One triangle is not enough

Each triangle gives its own range for the diagonal. The answer is where the two ranges OVERLAP; either range alone is too wide and matches a wrong option.

Drill 6 more on lengths through a diagonal

Concept 2 of 2: Rules that hold in every quadrilateral

Triangles on the same base have areas in the ratio of their heights. The two triangles on a diagonal have heights in the ratio in which the other diagonal cuts it.

Definition

Diagonals meet at $O$ with $AO : OC = m : n$ : $[ABD] : [CBD] = m : n$ .
The four triangles at $O$ satisfy $[AOB]\cdot[COD] = [AOD]\cdot[BOC]$ .
A point $O$ inside a rectangle $ABCD$ : $OA^2 + OC^2 = OB^2 + OD^2$ .
If a diagonal bisects the angles at both its ends, the figure is a kite: the two triangles are congruent.
A quadrilateral with an inscribed circle: $AB + CD = BC + DA$ .

Area on a shared diagonal

\dfrac{[CBD]}{[ABD]} = \dfrac{OC}{AO}

Worked example

The diagonals of

ABCD

meet at

O

with

AO : OC = 3 : 5

, and

[ABD] = 24

. Find

[CBD]

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (I) 2026 — Elementary Mathematics · Q53Hard

Example 2 · Quadrilaterals · General Quadrilaterals and their Diagonals

In a quadrilateral ABCD, the diagonals intersect at O. Let the area of the triangle ABD be p. If

AO : OC = m : n

, then what is the area of the triangle BCD ?

Heights, not squares

Triangles on a common base compare by HEIGHT, a straight ratio. Squaring the ratio belongs to similar triangles, which these are not.

Drill 3 more on rules that hold in every quadrilateral

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Lengths through a diagonal
Range of a diagonal
$|a - b| < x < a + b$
Rules that hold in every quadrilateral
Area on a shared diagonal
$\dfrac{[CBD]}{[ABD]} = \dfrac{OC}{AO}$

Watch out for (2)

One triangle is not enough
→ Lengths through a diagonal
Heights, not squares
→ Rules that hold in every quadrilateral

Test yourself on Quadrilaterals

15 past CDS questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Quadrilaterals chapter test Drill all 11 questions