Properties of each quadrilateral — diagonals, angles, symmetry: Exercises

Intuition How to read this page (notation legend)

A few conventions used throughout — none of them are "extra maths," they are just reading aids:

Coloured admonition boxes like this one are pedagogical labels, not formal notation: [!intuition] = the big idea in plain words, [!definition] = a precise meaning, [!recall]- = a collapsible box (click to open — here it hides a full worked solution so you can test yourself), [!mistake] = a common trap and how to escape it.
Highlighted words (shown with a coloured marker) simply mark a key term worth memorising — the == is a styling highlight only and carries no mathematical meaning.
The arrow $\Rightarrow$ is read "therefore / this forces": whatever is written after it is a logical consequence of what came before. So "all sides equal $\Rightarrow$ rhombus" means "the fact that all sides are equal forces the shape to be a rhombus."
Convexity caveat: every quadrilateral on this page is assumed convex — no dented-in corners, so both diagonals lie inside the shape and always cross at an interior point $O$ . (Concave "arrowhead" quadrilaterals behave differently and are out of scope here.)

Intuition How to use this page

Each problem below is a locked box: try it yourself first, then open the collapsible Solution to check your reasoning — not just the number, but the why of every step. Difficulty climbs in five steps:

L1 Recognition — name the shape / recall the fingerprint property.
L2 Application — plug one property into a calculation.
L3 Analysis — combine two or three properties.
L4 Synthesis — build a proof or a multi-shape argument.
L5 Mastery — coordinate / limiting / edge-case reasoning.

Everything here rests on the parent note the topic overview plus these prerequisites: 1.2.01-Types-of-quadrilaterals, 1.2.07-Angle-sum-property-of-quadrilaterals, 1.12-Triangle-congruence-criteria, 1.3.05-Line-symmetry, 1.3.06-Rotational-symmetry, 2.1.08-Pythagorean-theorem, 2.2.04-Area-of-quadrilaterals.

Before we start, one shared picture. A diagonal is a straight line joining two corners that are not next to each other. A quadrilateral has exactly two diagonals. The figure below labels the corners $A, B, C, D$ (always going around the shape, never criss-cross) and the point $O$ where the diagonals cross.

Figure — Properties of each quadrilateral — diagonals, angles, symmetry

What to look for in the figure above: the corners are named $A, B, C, D$ going around the outline (never criss-cross); the coral line is diagonal $A C$ , the mint line is diagonal $B D$ , and the black dot where they cross is $O$ — note it sits inside the shape because the quadrilateral is convex.

Keep this labelling in your head for every problem: sides are the edges $A B, B C, C D, D A$ ; diagonals are $A C$ and $B D$ ; $O$ is their crossing point.

L1 · Recognition

Problem 1.1

A quadrilateral has both pairs of opposite sides parallel, all four sides equal, but its angles are not all $90°$ . Name it, and state how many lines of symmetry it has.

Recall Solution

WHAT the clues say: "opposite sides parallel" $\Rightarrow$ it is at least a parallelogram. "All four sides equal" $\Rightarrow$ it is a rhombus. "Angles not all $90°$ " rules out the square (a square is a rhombus with right angles).

Symmetry: a rhombus has 2 lines of symmetry — its two diagonals — and rotational symmetry of order $2$ . Answer: rhombus, 2 lines of symmetry.

Problem 1.2

Which quadrilateral has diagonals that are equal in length and bisect each other and meet at right angles? Name the only shape that has all three at once.

Recall Solution

Read the three fingerprints one at a time:

"diagonals equal" $\Rightarrow$ rectangle-family (rectangle or square).
"bisect each other" $\Rightarrow$ parallelogram-family (true for both).
"meet at right angles" $\Rightarrow$ rhombus-family (rhombus or square).

The only shape sitting in all three families is the square. It is the meeting point of rectangle and rhombus.

L2 · Application

Problem 2.1

In parallelogram $A B C D$ the diagonals meet at $O$ . Given $A O = 6$ cm and $B O = 4.5$ cm, find the full lengths of both diagonals $A C$ and $B D$ .

Recall Solution

WHY we may double: in a parallelogram the diagonals bisect each other, so $O$ is the midpoint of each diagonal. That means $O$ splits $A C$ into two equal halves $A O = O C$ , and likewise $B D$ into $B O = O D$ . $A C = 2 \times A O = 2 \times 6 = 12 cm$ $B D = 2 \times B O = 2 \times 4.5 = 9 cm$ Note the diagonals are not equal ( $12 \neq = 9$ ) — and that's fine, a general parallelogram has unequal diagonals.

Problem 2.2

A rhombus has diagonals of length $10$ cm and $24$ cm. Find (a) its side length, and (b) its area.

Recall Solution

WHY Pythagoras applies: a rhombus's diagonals are perpendicular and bisect each other, so they cut the rhombus into four right-angled triangles. The two legs of each triangle are the half-diagonals.

Look at the picture below: each half-diagonal is $\frac{10}{2} = 5$ and $\frac{24}{2} = 12$ .

What to look for in the figure above: the two half-diagonals (coral

= 5

, lavender

= 12

) meet at the centre

O

at a right angle (little square marker); one shaded quarter is a right triangle whose hypotenuse (dark line) is a side of the rhombus.

(a) The side of the rhombus is the hypotenuse of one such triangle. By 2.1.08-Pythagorean-theorem: $s = 5^{2} + 1 2^{2} = 25 + 144 = 169 = 13 cm$

(b) Using the diagonal area rule (see 2.2.04-Area-of-quadrilaterals): $Area = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 10 \times 24 = 120 cm^{2}$

L3 · Analysis

Problem 3.1

Trapezium $P QR S$ has $P Q ∥ S R$ . Given $∠ P = 72°$ and $∠ Q = 68°$ , find $∠ R$ and $∠ S$ .

Recall Solution

WHY the angles pair up: when two parallel lines are crossed by a slanted line (a transversal), the two angles trapped on the same side, between the parallels are co-interior and add to $180°$ . Here the legs $P S$ and $QR$ are the transversals.

$P S$ crosses $P Q$ and $S R$ : so $∠ P + ∠ S = 180°$ $∠ S = 180° - 72° = 108°$
$QR$ crosses $P Q$ and $S R$ : so $∠ Q + ∠ R = 180°$ $∠ R = 180° - 68° = 112°$ Check with the angle-sum property (1.2.07-Angle-sum-property-of-quadrilaterals): $72° + 68° + 112° + 108° = 360° ✓$

Problem 3.2

An isosceles trapezium has parallel sides (bases) $a = 14$ cm and $b = 6$ cm, and height $h = 8$ cm. Find the length of each diagonal. (Recall: $h$ is the perpendicular distance between the bases — see the height definition above.)

Recall Solution

WHY coordinates help: placing the shape on a grid turns "length of a slanted diagonal" into a plain horizontal-run/vertical-rise problem we can feed to Pythagoras. Because the height is the perpendicular gap between the bases, the top base sits exactly $h$ units straight up — that is why $h$ becomes the vertical coordinate.

Put the long base on the x-axis and centre the short base by symmetry: $A = (0, 0), B = (14, 0), C = (\frac{a + b}{2}, h) = (10, 8), D = (\frac{a - b}{2}, h) = (4, 8)$ The diagonal $A C$ runs from the outer corner $A$ to the opposite inner corner $C$ :

horizontal run $= \frac{a + b}{2} = \frac{14 + 6}{2} = 10$
vertical rise $= h = 8$ $d = (\frac{a + b}{2})^{2} + h^{2} = 1 0^{2} + 8^{2} = 100 + 64 = 164 = 241 \approx 12.81 cm$ Both diagonals are equal — that is the fingerprint of an isosceles trapezium, guaranteed by its left–right line of symmetry.

L4 · Synthesis

Problem 4.1

Prove that in any parallelogram $A B C D$ the diagonals bisect each other. (Use triangle congruence.)

Recall Solution

Goal: show that where the diagonals cross at $O$ , we get $A O = O C$ and $B O = O D$ .

Compare $△ A O B$ and $△ C O D$ (see figure):

What to look for in the figure above: the two shaded triangles

△ A O B

(coral) and

△ C O D

(mint) are mirror-matched across

O

; the equal alternate angles at

A

–

C

and at

B

–

D

, together with the equal side

A B = C D

, are exactly the three ingredients ASA needs.

$A B ∥ C D$ , and $A C$ is a transversal $\Rightarrow$ $∠ O A B = ∠ O C D$ (alternate angles).
$A B ∥ C D$ , and $B D$ is a transversal $\Rightarrow$ $∠ O B A = ∠ O D C$ (alternate angles).
$A B = C D$ (opposite sides of a parallelogram are equal).

Two angles and the included side match, so by ASA (1.12-Triangle-congruence-criteria): $△ A O B ≅ △ C O D$ Corresponding parts of congruent triangles are equal, therefore $A O = O C$ and $B O = O D$ . The diagonals cut each other exactly in half. $■$

Problem 4.2

Prove that a rectangle's diagonals are equal in length.

Recall Solution

Goal: show $A C = B D$ in rectangle $A B C D$ .

Compare $△ A B C$ and $△ B A D$ :

$A B = B A$ (the same side, shared).
$B C = A D$ (opposite sides of a rectangle — it is a parallelogram — are equal).
$∠ A B C = ∠ B A D = 90°$ (every angle of a rectangle is a right angle).

Two sides and the included angle match, so by SAS: $△ A B C ≅ △ B A D$ Hence the third sides are equal: $A C = B D$ . The diagonals of a rectangle are equal. $■$ (Note we did not claim they are perpendicular — that only happens in the square.)

L5 · Mastery

Problem 5.1 (Coordinate reasoning)

A quadrilateral has vertices $A = (0, 0)$ , $B = (4, 0)$ , $C = (6, 3)$ , $D = (2, 3)$ . Classify it as precisely as possible, and confirm using diagonal midpoints.

Recall Solution

Step 1 — check side vectors. A vector is just "run then rise" from one corner to the next. $A B = (4, 0), D C = (6 - 2, 3 - 3) = (4, 0)$ $A D = (2, 3), B C = (6 - 4, 3 - 0) = (2, 3)$ Both pairs of opposite sides are equal and point the same way $\Rightarrow$ both pairs are parallel $\Rightarrow$ it is a parallelogram.

Step 2 — are all sides equal (rhombus)? $∣ A B ∣ = 4^{2} + 0^{2} = 4, ∣ A D ∣ = 2^{2} + 3^{2} = 13 \approx 3.61$ Not equal, so it is not a rhombus.

Step 3 — are the angles $90°$ (rectangle)? Two sides are perpendicular only if their run/rise vectors give a zero dot product (the dot product $=$ run $\times$ run $+$ rise $\times$ rise measures how much two directions agree; $0$ means a perfect right angle): $A B \cdot A D = (4) (2) + (0) (3) = 8 \neq = 0$ Not perpendicular, so not a rectangle.

Conclusion: it is a plain parallelogram (neither rhombus nor rectangle).

Confirm bisection: the midpoint of a diagonal is the average of its two endpoints. $mid A C = (\frac{0 + 6}{2}, \frac{0 + 3}{2}) = (3, 1.5), mid B D = (\frac{4 + 2}{2}, \frac{0 + 3}{2}) = (3, 1.5)$ Same point $\Rightarrow$ the diagonals share a midpoint $\Rightarrow$ they bisect each other, exactly as a parallelogram demands. $✓$

Problem 5.2 (Limiting / edge case)

Start with an isosceles trapezium of bases $a$ (long) and $b$ (short) and height $h$ , whose diagonal is $d = (\frac{a + b}{2})^{2} + h^{2}$ . What shape and what diagonal length do you get as $b \to a$ ? Interpret the answer geometrically.

Recall Solution

WHAT " $b \to a$ " means: we slowly stretch the short base until it is as long as the long base. When $b = a$ , the top and bottom edges have the same length and are parallel — the trapezium has become a parallelogram.

WHY it is specifically a rectangle (not just any parallelogram): recall that the height $h$ is measured perpendicular to the bases. In our coordinate model the legs run from $(0, 0)$ to $(\frac{a - b}{2}, h)$ — a horizontal shift of $\frac{a - b}{2}$ . As $b \to a$ , that horizontal shift $\frac{a - b}{2} \to 0$ , so each leg becomes purely vertical, i.e. perpendicular to the bases. Four sides meeting at $90°$ is exactly a rectangle.

Substitute $b = a$ into the diagonal formula: $d = (\frac{a + a}{2})^{2} + h^{2} = a^{2} + h^{2}$ Geometric check: the resulting rectangle has width $a$ and height $h$ , and its diagonal — by Pythagoras — is exactly $a^{2} + h^{2}$ . The formula degenerates smoothly into the rectangle's diagonal. No contradiction, no jump.

The other edge, $b \to 0$ : the short base shrinks to a point and the trapezium collapses into a triangle of base $a$ and height $h$ ; the "diagonal" becomes $(a /2)^{2} + h^{2}$ , which is the distance from a bottom corner to the apex — again perfectly sensible.

Recall Quick self-quiz (reveal after answering)

How many lines of symmetry does a rhombus have? ::: 2 (its two diagonals) Diagonals equal AND perpendicular AND bisecting — which shape? ::: The square Isosceles-trapezium diagonal with $a = 14, b = 6, h = 8$ ? ::: $241 \approx 12.81$ cm Rhombus with diagonals $10$ and $24$ : side length? ::: $13$ cm Parallelogram diagonal test in coordinates? ::: The two diagonals share the same midpoint