Diagonals of special quadrilaterals

This is part of the ICSE Class 8 chapter "Special Types of Quadrilaterals," which classifies trapeziums, kites, parallelograms, rectangles, rhombi and squares and studies their diagonal properties by verification and measurement, not by formal proof — that deductive step, using congruent triangles or the mid-point theorem, belongs to Class 9. For a quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ meet at $O$, measurement tells us the following for each shape.

"Diagonals bisect each other" means $AO = OC$ and $BO = OD$ — the point $O$ is the midpoint of both diagonals, so each diagonal is cut into two equal halves.

Why the properties stack up. Every rectangle, rhombus and square is a parallelogram, so all three automatically inherit "diagonals bisect each other" — the base-level property. A rectangle adds the condition that all four angles are $90^\circ$; measuring rectangles of every size shows their two diagonals always come out equal, on top of bisecting each other. A rhombus instead adds the condition that all four sides are equal; measuring the angle at $O$ in any rhombus always gives $90^\circ$, so the diagonals are perpendicular, again on top of bisecting each other. A square is simultaneously a rectangle and a rhombus, so it collects all three properties at once: bisecting, equal, and perpendicular diagonals together.

A rhombus's perpendicular, bisecting diagonals split it into four right-angled triangles, which is what makes a simple area shortcut work: if the diagonals have lengths $d_1$ and $d_2$,

where: $d_1$ and $d_2$ are the two diagonal lengths, measured along the full diagonal, not the half-diagonals. This is a Class 7 recall formula, restated here because it depends directly on the rhombus diagonal properties above.

Problem: $PQRS$ is a rhombus with diagonals $PR = 16$ cm and $QS = 12$ cm, meeting at $O$. Find (a) the length of side $PQ$, and (b) the area of the rhombus. As a check, also find $BO$ and $OD$ in a rectangle $ABCD$ whose diagonal $AC = 18$ cm.

Step 1. Since the diagonals of a rhombus bisect each other at right angles, $PO = PR/2 = 16/2 = 8$ cm and $QO = QS/2 = 12/2 = 6$ cm, with $\angle POQ = 90^\circ$.

Step 2. Triangle $POQ$ is right-angled at $O$, so by the Pythagoras theorem, $PQ^2 = PO^2 + QO^2 = 8^2 + 6^2 = 64 + 36 = 100$, giving $PQ = \sqrt{100} = 10$ cm. Since all four sides of a rhombus are equal, every side of $PQRS$ is $10$ cm.

Step 3. Using the area formula derived above, $\text{Area} = \tfrac12\, d_1 d_2 = \tfrac12 \times 16 \times 12 = 96\ \text{cm}^2$. So $PQ = 10$ cm and the area of rhombus $PQRS$ is $96\ \text{cm}^2$.

Competitive stretch

Step 1. For rectangle $ABCD$ with diagonals meeting at $O$ and $AC = 18$ cm: since a rectangle's diagonals are equal, $BD = AC = 18$ cm.

Step 2. Since the diagonals also bisect each other, $O$ is the midpoint of $BD$, so $BO = OD = BD/2 = 18/2 = 9$ cm. Answer: $BO = OD = 9$ cm. This needed two separate properties — "equal diagonals" to get $BD$ from $AC$, then "bisecting diagonals" to split $BD$ into equal halves.

• Confusing "bisect" with "equal." A plain parallelogram and a rhombus both have bisecting diagonals, but neither has equal diagonals in general — bisection and equality are independent properties.
• Assuming perpendicular diagonals force a square. A rhombus has perpendicular diagonals without being a square, since its diagonals are unequal and its angles are not $90^\circ$.
• Using full diagonals instead of half-diagonals in Pythagoras. The legs of the right triangle at $O$ are $d_1/2$ and $d_2/2$, not $d_1$ and $d_2$.
• Forgetting a square combines every property. Students often quote only "diagonals are equal" for a square and forget it is also perpendicular and bisecting, all three together.

Ask the tutor: If the difference between bisecting, equal and perpendicular diagonals still feels confusing, or you want more practice problems on rectangles and rhombi, ask the AI tutor for a guided walkthrough.