Exercises — Quadrilaterals — square, rectangle, parallelogram, rhombus, trapezium, kite

Walk the definitions in order:

• Both pairs of opposite sides parallel → it is a parallelogram.
• All four sides equal → among parallelograms, "all sides equal" is the extra rule for a rhombus.
• Angles are not $90^{\circ }$ → it is not a square (a square needs all sides equal and all angles $90^{\circ }$).

Answer: a rhombus.

Why not "square"? A square is a rhombus with the bonus of right angles. Missing that bonus means we stop one rung lower.

"Exactly one pair of parallel sides" is the defining line of a trapezium (US: trapezoid).

• The two parallel sides are the bases.
• The two non-parallel sides are the legs.

Answer: a trapezium; the parallel sides are its bases.

The word exactly matters: if both pairs were parallel it would be a parallelogram, not a trapezium.

Equal sides that sit next to each other (rather than opposite) is the signature of a kite.

• In a rhombus all four sides are equal and opposite sides are parallel.
• In a kite only adjacent sides match ($5,5$ meet at one vertex; $8,8$ meet at the opposite vertex), and the sides are generally not parallel.

Answer: a kite.

Step 1 — opposite angles equal (WHAT & WHY). Call the parallelogram $ABCD$ with $\angle A=48^{\circ }$. Why is the opposite angle $\angle C$ also $48^{\circ }$? Draw the diagonal $AC$. It is a transversal across the parallel pairs, so by alternate interior angles $\angle BAC=\angle DCA$ and $\angle BCA=\angle DAC$ (see Basic Geometry - Parallel Lines). Adding each pair back together gives $\angle A=\angle BAC+\angle DAC=\angle DCA+\angle BCA=\angle C$. So the angle across from $48^{\circ }$ is also $48^{\circ }$. Step 2 — consecutive angles supplementary (WHY). $\angle A$ and $\angle B$ are co-interior angles on the same side of the transversal $AB$ cutting the parallel pair $AD\parallel BC$, and co-interior angles add to $180^{\circ }$: $\angle B=180^{\circ }-48^{\circ }=132^{\circ }.$ Step 3 — opposite of $\angle B$. By the same opposite-angle argument as Step 1, $\angle D=\angle B=132^{\circ }$.

Answer: $48^{\circ },132^{\circ },48^{\circ },132^{\circ }$. Check: $48+132+48+132=360^{\circ }$ ✓ (matches the interior-angle sum).

Step 1 — pick the formula and say why. Area of a trapezium is the average of the two bases times the height: $A=\frac{1}{2}(a+b)h.$ We average because two copies of the trapezium tile into a parallelogram of base $(a+b)$. Step 2 — substitute. $A=\frac{1}{2}(6+14)\times 9=\frac{1}{2}\times 20\times 9=90{\text{cm}}^2.$

Answer: $90{\text{cm}}^2$.

Step 1 — why the diagonals cross at right angles. In a rhombus all four sides are equal. The two triangles either side of a diagonal are therefore congruent isosceles triangles, and a diagonal of a rhombus bisects the vertex angles. Where the diagonals meet at $O$, consider triangles $AOB$ and $AOD$: they share side $AO$, have $AB=AD$ (equal sides), and $OB=OD$ (diagonals of a parallelogram bisect each other). By SSS they are congruent, so the two angles at $O$ along a straight line are equal — and two equal angles that sum to $180^{\circ }$ must each be $90^{\circ }$. Hence the diagonals are perpendicular, slicing the rhombus into four right triangles whose areas sum to $\frac{1}{2}{d}_1{d}_2$. Step 2 — substitute. $A=\frac{1}{2}{d}_1{d}_2=\frac{1}{2}\times 16\times 30=240{\text{cm}}^2.$

Answer: $240{\text{cm}}^2$.

Step 1 — half the diagonals (WHAT & WHY). As justified in Exercise 2.3, the diagonals of a rhombus bisect each other at right angles. So each side of the rhombus is the hypotenuse of a right triangle whose two legs are the half-diagonals: $\frac{18}{2}=9,\frac{24}{2}=12.$ Step 2 — Pythagoras (WHY this tool). We know two legs of a right triangle and want the hypotenuse — that is exactly the question Pythagoras answers: $s=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\text{cm}.$ Step 3 — perimeter (WHY). A rhombus has all four sides equal, so $P=4s=60\text{cm}$.

Answer: side $=15\text{cm}$, perimeter $=60\text{cm}$.

Step 1 — find the horizontal overhang (WHY). Because the legs are equal, the top base sits centred over the bottom base. The extra bottom length is $10-4=6$, split equally on the two ends, so each end sticks out by $6/2=3\text{cm}$. Step 2 — right triangle at the leg. Drop a perpendicular from a top corner. It forms a right triangle with horizontal leg $3$, hypotenuse (the leg) $5$, and vertical leg $h$. Use Pythagoras to undo the hypotenuse: $h=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4\text{cm}.$ Step 3 — area. $A=\frac{1}{2}(10+4)\times 4=\frac{1}{2}\times 14\times 4=28{\text{cm}}^2.$

Answer: height $=4\text{cm}$, area $=28{\text{cm}}^2$.

Step 1 — the diagonal is a hypotenuse (WHY). A rectangle's corners are right angles, so a diagonal splits it into two right triangles with legs $8$ and $6$: $d=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{cm}.$ Step 2 — why both diagonals match. By symmetry the other diagonal cuts the same $8$-by-$6$ right triangle, so it too measures $10\text{cm}$. (Equal diagonals are a defining perk of rectangles that ordinary parallelograms lack.)

Answer: each diagonal $=10\text{cm}$.

Step 1 — side lengths (WHY). The distance formula $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$ measures each side directly from coordinates:

• $AB=\sqrt{(5-1{)}^2+(1-1{)}^2}=\sqrt{16}=4$.
• $CD=\sqrt{(2-6{)}^2+(4-4{)}^2}=\sqrt{16}=4$.
• $BC=\sqrt{(6-5{)}^2+(4-1{)}^2}=\sqrt{1+9}=\sqrt{10}$.
• $AD=\sqrt{(2-1{)}^2+(4-1{)}^2}=\sqrt{1+9}=\sqrt{10}$. Opposite sides are equal ($AB=CD$, $BC=AD$) → parallelogram. Step 2 — is it a rectangle? (WHY it matters). $AB$ is horizontal (both endpoints have $y=1$). $BC$ goes right-and-up, not straight vertical, so the corner at $B$ is not $90^{\circ }$. Hence it is a plain parallelogram, not a rectangle. Step 3 — area by base × height. Take $AB=4$ as base (it lies on the line $y=1$). The opposite side $DC$ lies on the line $y=4$, so the perpendicular height is $4-1=3$. $A=b\times h=4\times 3=12.$

Answer: a parallelogram, area $=12$ square units.

Step 1 — why the diagonals meet at $90^{\circ }$ (WHY). A kite has a mirror line: the long diagonal $d_1$. Reflecting the kite across $d_1$ maps the short diagonal $d_2$ onto itself but flips its two halves, so $d_1$ must cut $d_2$ into two equal halves and cross it at a right angle (a mirror line always meets what it reflects at $90^{\circ }$). Note the asymmetry: $d_1$ bisects $d_2$, but $d_2$ does not bisect $d_1$ — that is why $d_1$ is split unequally into $6$ and $14$. Step 2 — Pythagoras for half of $d_2$. The short side ($10$) is the hypotenuse of a right triangle whose legs are the $6\text{cm}$ piece of $d_1$ and one half of $d_2$: $\frac{d_2}{2}=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8.$ So $d_2=16\text{cm}$. Step 3 — area. $A=\frac{1}{2}{d}_1{d}_2=\frac{1}{2}\times 20\times 16=160{\text{cm}}^2.$

Answer: $d_2=16\text{cm}$, area $=160{\text{cm}}^2$.

Proof. Let $\angle P=90^{\circ }$.

• Consecutive angles of a parallelogram are supplementary (co-interior angles across parallel sides), so $\angle Q=180^{\circ }-90^{\circ }=90^{\circ }$.
• Opposite angles are equal, so $\angle R=\angle P=90^{\circ }$ and $\angle S=\angle Q=90^{\circ }$.
• All four angles are $90^{\circ }$ → by definition it is a rectangle. ∎ Application. With right angle at $Q$ between the sides $PQ$ and $QR$, the diagonal $PR$ is the hypotenuse of right triangle $PQR$ with legs $PQ=9$, $QR=12$: $PR=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15.$

Answer: it must be a rectangle; $PR=15$.

Step 1 — name the pieces. Let the diagonals be $d_1=3k$ and $d_2=4k$ for some positive scale $k$. Half-diagonals are $\frac{3k}{2}$ and $\frac{4k}{2}=2k$. Step 2 — side by Pythagoras (WHY). Diagonals of a rhombus meet at right angles (Exercise 2.3), so the side is the hypotenuse of legs $\frac{3k}{2}$ and $2k$: $s=\sqrt{{\left(\frac{3k}{2}\right)}^2+(2k{)}^2}=\sqrt{\frac{9k^2}{4}+4k^2}=\sqrt{\frac{25k^2}{4}}=\frac{5k}{2}.$ Step 3 — area and perimeter. $A=\frac{1}{2}{d}_1{d}_2=\frac{1}{2}(3k)(4k)=6k^2,P=4s=4\cdot \frac{5k}{2}=10k.$ Step 4 — set them equal and solve. $6k^2=10k\Rightarrow 6k=10\Rightarrow k=\frac{5}{3}.$ Step 5 — read off the shape. $d_1=3k=5,d_2=4k=\frac{20}{3},s=\frac{5k}{2}=\frac{25}{6}.$ Check: $A=6k^2=6\cdot \frac{25}{9}=\frac{50}{3}$, and $P=10k=\frac{50}{3}$ ✓.

Answer: $d_1=5$, $d_2=\frac{20}{3}$, side $=\frac{25}{6}$ (area $=$ perimeter $=\frac{50}{3}$).

Step 1 — midpoints (WHY). A midpoint is found by averaging the two endpoint coordinates:

• $P=\left(\frac{0+6}{2},\frac{0+0}{2}\right)=(3,0)$
• $Q=\left(\frac{6+8}{2},\frac{0+5}{2}\right)=(7,2.5)$
• $R=\left(\frac{8+2}{2},\frac{5+7}{2}\right)=(5,6)$
• $S=\left(\frac{2+0}{2},\frac{7+0}{2}\right)=(1,3.5)$ Step 2 — opposite side lengths via distance formula.
• $PQ=\sqrt{(7-3{)}^2+(2.5-0{)}^2}=\sqrt{16+6.25}=\sqrt{22.25}$
• $RS=\sqrt{(1-5{)}^2+(3.5-6{)}^2}=\sqrt{16+6.25}=\sqrt{22.25}$ → equal ✓
• $QR=\sqrt{(5-7{)}^2+(6-2.5{)}^2}=\sqrt{4+12.25}=\sqrt{16.25}$
• $SP=\sqrt{(3-1{)}^2+(0-3.5{)}^2}=\sqrt{4+12.25}=\sqrt{16.25}$ → equal ✓ Step 3 — conclude. Both pairs of opposite sides are equal, and a quadrilateral with both pairs of opposite sides equal is a parallelogram, so $PQRS$ is a parallelogram. General theorem (Varignon): the midpoints of any quadrilateral's four sides always form a parallelogram. Why: each side of $PQRS$ is a midsegment of one of the triangles cut by a diagonal of $ABCD$, so it is parallel to that diagonal and half its length — giving two pairs of equal, parallel sides.

Answer: $PQRS$ is a parallelogram ($PQ=RS=\sqrt{22.25}$, $QR=SP=\sqrt{16.25}$).

Which level is each skill? ::: L1 name shapes · L2 one formula · L3 formula + Pythagoras · L4 coordinates / hidden shape · L5 proof & design. In a kite, which diagonal bisects the other? ::: The axis of symmetry (long diagonal) bisects the other diagonal at $90^{\circ }$; the axis is itself split unequally. When do you subtract under the square root? ::: When the known side is already the hypotenuse and you seek a shorter leg: $\sqrt{{\text{hyp}}^2-{\text{leg}}^2}$.