Quadrilaterals Worksheet

(b) The intersection of its diagonals

1. The diagonals bisect each other at O.
2. A 180° turn about O exchanges both pairs of opposite vertices.

(c) 2:1

1. The centroid lies two thirds of the way from vertex A to midpoint M.
2. Thus AG = 2AM/3 and GM = AM/3.

(b) Two reflection axes and half-turn symmetry

1. Its axes pass through the midpoints of opposite sides.
2. A 180° turn exchanges opposite vertices, but a 90° turn interchanges unequal side lengths.

(b) 9 cm

1. The segment joining two side midpoints is half the third side.
2. PQ = 18/2 = 9 cm.

AG = 10 cm and GM = 5 cm.

1. The centroid divides each median in the ratio 2:1 measured from the vertex.
2. The three ratio-parts together have length 15, so each part is 5 cm.
3. AG contains two parts and is 10 cm; GM contains one and is 5 cm.

The union of AB, BC, CD, DA is a quadrilateral; opposite sides equal imply a parallelogram.

1. Join A to B to C to D to A, with four distinct coplanar vertices and no three collinear. A simple quadrilateral has no crossings other than adjacent sides meeting at vertices.
2. In simple ABCD draw AC. Given AB = CD and BC = DA, AC is common.
3. Triangles ABC and CDA are congruent by SSS.
4. Matching alternate interior angles imply AB ∥ CD and BC ∥ AD, giving a parallelogram.

PQ is parallel to BC and half its length.

1. Extend PQ through Q to S with QS = QP. Since AQ = QC and vertical angles at Q are equal, triangles AQP and CQS are congruent by SAS.
2. Hence AP = CS and AP ∥ CS. With AP = PB, the segment PB is equal and parallel to CS.
3. PBCS is a parallelogram, so PS ∥ BC and PS = BC.
4. P, Q, S are collinear and PS = 2PQ. Therefore PQ ∥ BC and PQ = BC/2.

Each diagonal bisects the other; the bisection condition characterises parallelograms.

1. For parallelogram ABCD with diagonals meeting at O, AB ∥ CD and AB = CD. Alternate angles make triangles AOB and COD congruent by ASA.
2. Thus AO = OC and BO = OD.
3. Conversely assume these two equalities. Vertical angles give triangles AOB and COD congruent by SAS.
4. Corresponding alternate angles imply AB ∥ CD. Applying SAS to AOD and COB gives AD ∥ BC, so ABCD is a parallelogram.

Both conditions imply a parallelogram.

1. For (i), if AB = CD and AB ∥ CD, draw AC. The equal included alternate angles and common AC give SAS congruence of ABC and CDA.
2. The other alternate angles imply BC ∥ AD.
3. For (ii), let ∠A = ∠C and ∠B = ∠D. The 360° sum gives ∠A + ∠B = 180°.
4. Supplementary same-side interior angles imply AD ∥ BC; using ∠B + ∠C = 180° gives AB ∥ CD.

AQ = QC.

1. Let R be the actual midpoint of AC.
2. The midpoint theorem gives PR ∥ BC.
3. Through P there is only one line parallel to BC, so PQ and PR are the same line.
4. Its intersection with AC is unique; Q = R and therefore AQ = QC.

G has position vector (a + b + c)/3 and divides every median 2:1.

1. The midpoint M of BC has vector (b + c)/2.
2. The point two thirds of the way from A to M has vector a + (2/3)((b + c)/2 − a) = (a + b + c)/3.
3. The same calculation starting from B and C produces exactly the same vector.
4. Thus this one point lies on all three medians, proving concurrence.
5. Its vertex-to-point and point-to-midpoint fractions are 2/3 and 1/3, so each ratio is 2:1. Coordinate calculations may be written component-wise instead of using vectors.

PQRS is a parallelogram when AC and BD are not parallel; parallel diagonals make it collapse to a segment.

1. In triangles ABC and ADC, the midpoint theorem gives PQ ∥ AC, SR ∥ AC and PQ = SR = AC/2.
2. Similarly QR and PS are parallel to BD and have length BD/2.
3. Both pairs of opposite sides are parallel, giving the midpoint parallelogram; the reasoning does not require the original quadrilateral to be convex.
4. If AC ∥ BD, all four midpoint sides are collinear, so the figure is a degenerate segment rather than a two-dimensional parallelogram.

D = (3,6); centre (4,4). General parallelogram: half-turn; rectangle and rhombus: two reflection axes; square: four and quarter-turns.

1. Diagonals have the same midpoint, so A + C = B + D coordinate-wise. D = A + C − B = (3,6).
2. The centre is the midpoint of AC: (4,4). A 180° turn about it exchanges opposite vertices.
3. A general parallelogram need not have a reflection axis. A non-square rectangle has two midpoint axes; a non-square rhombus has two diagonal axes.
4. A square has four reflection axes and rotations of 90°, 180°, 270° and 360°. All parallelograms have the half-turn symmetry.

Every non-degenerate simple quadrilateral, convex or concave, admits a tiling by congruent copies.

1. Make congruent cutouts and place neighbours by 180° turns about side midpoints so shared edges match.
2. At the resulting vertex configurations, one copy of each of the four interior angles meets; the sum is 360° (including a reflex interior angle when concave).
3. Continue the matching placements in repeated rows; the paired arrangements give a repeating pattern without gaps or overlaps.
4. The tile is the filled region of a simple quadrilateral. A self-crossing four-edge figure does not automatically define the same kind of filled tile. Use paper manipulation to check the matching.

The inner figure is a rhombus of side 5 cm and area 24 cm².

1. Each inner side is the hypotenuse of a corner triangle with legs 4 and 3; its length is 5 cm.
2. All four sides are therefore equal; the midpoint figure is a parallelogram by Varignon, hence a rhombus.
3. The rectangle area is 48 cm². Four corner triangles each have area (1/2) × 4 × 3 = 6 cm².
4. Subtract 24 cm² of corners from 48 cm² to get inner area 24 cm².