Exercise practice 2: Quadrilaterals

4.1 Figure it Out, page 107

The second exercise block has eleven questions, and the final one has seven subparts. Each question is stated, answered and then worked in steps.

4.1.1 FIO 3.1 Two equilateral triangles

Question. Join two equilateral triangles of side 4 cm. Find all sides and angles of the quadrilateral formed.

Answer. A rhombus with four sides of 4 cm and angles 120°, 60°, 120°, 60°.

Key steps. Each equilateral triangle has three sides of 4 cm and three angles of 60°. Join them along one side. The outer boundary has four sides, each 4 cm, because each comes from a triangle side. At the two vertices where the triangles are joined, the quadrilateral angle is 60° + 60° = 120°. The other two vertices are vertices of the equilateral triangles, so those angles are 60° each. The sum is 120 + 60 + 120 + 60 = 360°.

4.1.2 FIO 3.2 Constructing a kite

Question. Construct a kite whose diagonals are 6 cm and 8 cm.

Answer. A kite with diagonals 6 cm and 8 cm, the longer diagonal bisecting the shorter at 90°.

Key steps. Draw PQ = 6 cm and its perpendicular bisector, meeting PQ at T. On the perpendicular mark R and S with TR = TS = 4 cm, so RS = 8 cm. Join P-R, R-Q, Q-S and S-P. In the quadrilateral PRQS, the diagonal RS bisects PQ at right angles. Triangles PRT and RQT are congruent by SAS (PT = QT, ∠PTR = ∠QTR = 90°, RT common), so PR = RQ. Similarly SP = SQ. Thus PRQS has two pairs of adjacent equal sides and is a kite; its diagonals are 6 cm and 8 cm.

4.1.3 FIO 3.3 Angles in a trapezium

Question. Work out the missing angles in the trapeziums below. (i) Trapezium PQRS with PQ parallel to RS, ∠P = 135° and ∠Q = 105°. (ii) Isosceles trapezium ABCD with AD = BC and ∠D = 100°.

Answer. (i) ∠S = 45°, ∠R = 75°. (ii) ∠A = ∠B = 80°, ∠C = 100°.

Key steps. (i) Along the two parallel sides, each pair of angles lying on the same side of a cutting line adds to 180°. With PQ parallel to RS, ∠P + ∠S = 180°, so ∠S = 180 - 135 = 45°. Also ∠Q + ∠R = 180°, so ∠R = 180 - 105 = 75°. Check: 135 + 105 + 75 + 45 = 360. (ii) In isosceles trapezium ABCD, AD = BC, so the angles at the equal sides match: ∠A = ∠B and ∠C = ∠D. Since AB is parallel to DC, ∠A + ∠D = 180°, so ∠A = 180 - 100 = 80°. Hence ∠B = 80°. Also ∠C = ∠D = 100°. Check: 80 + 80 + 100 + 100 = 360.

4.1.4 FIO 3.4 Venn diagram of quadrilaterals

Question. Make a Venn diagram containing these five types: rhombuses, rectangles, squares, kites and parallelograms. Use it to answer: (i) Which shape belongs to both the kite set and the parallelogram set? (ii) Is there a shape that is a kite and a rectangle at the same time? (iii) Must every kite be a rhombus? State how the two sets are related.

Answer. (i) A rhombus, and also a square. (ii) Only a square can be both; a non-square rectangle is not a kite. (iii) No; every rhombus is a kite, but not every kite is a rhombus.

Key steps.

• A rhombus has all four sides equal, so it satisfies the kite condition (two pairs of adjacent equal sides) and the parallelogram condition (opposite sides equal), so it lies in both sets. A square lies in the same overlap and also inside the rectangle set.
• A kite can be a rectangle only when all four sides are equal, which makes it a square. If a rectangle has unequal adjacent sides, it is not a kite.
• A rhombus is a kite because its four equal sides give two non-overlapping adjacent equal pairs. A kite with pairs 5 cm and 8 cm is not a rhombus. So the rhombus set sits inside the kite set.

4.1.5 FIO 3.5 Two rectangles PAIR and RODS

Question. The figure shows two rectangles, PAIR and RODS. Calculate ∠IOD.

Answer. ∠IOD = 30°.

Key steps. In rectangle PAIR, the diagonal AR makes a 30° angle with the side AI. The two rectangles are drawn with corresponding sides parallel: OI is parallel to the diagonal AR, and OD is parallel to the side AI. The angle between two lines equals the angle between their parallels, so ∠IOD = ∠RAI = 30°. The 5 cm marks show the equal corresponding sides that keep the two rectangles in the same orientation.

4.1.6 FIO 3.6 Square from a diagonal

Question. Construct a square whose diagonal measures 6 cm, using only a compass and a straight edge (no protractor).

Answer. A square with a 6 cm diagonal.

Key steps. Draw AB = 6 cm and mark its midpoint O. Using a compass, construct the perpendicular bisector of AB at O. Mark C and D on this perpendicular with OC = OD = 3 cm, so CD = 6 cm. Join A-C-B-D-A. The diagonals AB and CD are equal, bisect each other at O, and meet at 90°. That combination forces four equal sides and four right angles, so ACBD is the required square.

4.1.7 FIO 3.7 Midpoints of a square

Question. CASE is a square and U, V, W, X are the midpoints of its sides. What quadrilateral is UVWX? Prove it and describe other square-in-square constructions.

Answer. UVWX is a square.

Key steps. Let each side of the outer square be s. Each midpoint is s/2 from the nearest vertex. In the corner triangle CUV, CU = CV = s/2 and ∠C = 90°, so the two base angles are 45° and UV is the hypotenuse. The same calculation at the other three corners gives the same length for UX, XW and WV. At V, the straight angle along the side of the outer square is made of three angles: 45° from the corner triangle at V, the angle of UVWX at V, and 45° from the neighbouring corner triangle. Hence the angle of UVWX at V is 180 - 45 - 45 = 90°. Repeating gives four right angles. Four equal sides and four right angles make UVWX a square. The textbook figure (b) rotates the inner square so its vertices lie on the outer sides; starting the midpoint construction from a different side gives more examples.

4.1.8 FIO 3.8 Four equal sides and one right angle

Question. A four-sided figure has all its sides equal and one of its angles measures 90°. Must it be a square? Give a proof.

Answer. Yes, it is a square.

Key steps. Having four equal sides makes the figure a rhombus. A rhombus has matching opposite angles, and any two neighbouring angles add to 180°. Suppose ∠A = 90°. Then the adjacent ∠B = 180 - 90 = 90°, and the opposite ∠C = ∠A = 90°, so ∠D = 90° as well. All four angles are right angles and all four sides are equal, which is exactly a square.

4.1.9 FIO 3.9 Equal opposite sides

Question. What quadrilateral has equal opposite sides? Justify.

Answer. A parallelogram.

Key steps. In quadrilateral ABCD let AB = CD and BC = AD. Draw diagonal AC. In △ABC and △CDA, AB = CD, BC = DA and AC is common, so the triangles are congruent by SSS. Hence ∠BAC = ∠DCA and ∠BCA = ∠DAC. The first pair are alternate angles for AB and CD with transversal AC, so AB is parallel to CD. The second pair are alternate angles for BC and AD, so BC is parallel to AD. Both pairs of opposite sides are parallel, so ABCD is a parallelogram.

4.1.10 FIO 3.10 The dart-shaped quadrilateral

Question. For a dart-shaped four-sided figure like the one shown, does the interior angle total still come to 360°? Explain your reasoning and check by measuring.

Answer. Yes, the sum is 360°.

Key steps. Choose the diagonal that lies inside the figure, from the reflex vertex to the opposite vertex. It splits the dart into two triangles. The angles of each triangle add to 180°, and the six triangle angles together are exactly the four angles of the quadrilateral. So the total is 180 + 180 = 360°. Construct the figure and measure the four angles: the reflex angle is still counted as one interior angle and the total comes to 360°.

4.1.11 FIO 3.11 True or false

Question. State whether these are true or false, with reasons. (i) If a four-sided figure has two equal diagonals that halve one another, it has to be a square. (ii) Any four-sided figure with three right angles is necessarily a rectangle. (iii) If the diagonals of a four-sided figure cut each other in half, the figure must be a parallelogram. (iv) A four-sided figure with perpendicular diagonals is necessarily a rhombus. (v) If opposite angles of a four-sided figure match, the figure must be a parallelogram. (vi) When all four angles of a quadrilateral are equal, the figure is a rectangle. (vii) Every isosceles trapezium is also a parallelogram.

Answer. (i) False. (ii) True. (iii) True. (iv) False. (v) True. (vi) True. (vii) False.

Key steps. (i) A rectangle provides a counterexample: its diagonals are equal and halve each other, yet its sides can differ. A square needs the extra condition that the diagonals meet at 90°, or that all sides are equal. (ii) The fourth angle is 360 - 3 × 90 = 90°, so all four angles are right angles, which is a rectangle. (iii) When each diagonal halves the other, the four small triangles formed at the crossing are congruent in pairs, which forces the opposite sides to be equal and parallel. That is a parallelogram. (iv) A kite may have diagonals at right angles and still not have four equal sides. The perpendicular condition alone does not force all four sides to be equal. (v) Suppose the opposite angles match: ∠A = ∠C and ∠B = ∠D. then 2∠A + 2∠B = 360°, so ∠A + ∠B = 180°. These are co-interior angles for AD and BC with transversal AB, so AD is parallel to BC. Similarly AB is parallel to DC, so the figure is a parallelogram. (vi) All angles equal in a quadrilateral means each is 360/4 = 90°, which is a rectangle. (vii) An isosceles trapezium has exactly one pair of parallel sides by its usual picture; its equal non-parallel sides are not parallel, so it is not a parallelogram.