Exercise practice 1: Quadrilaterals

3.1 Figure it Out, page 94

These are the five questions from the first exercise block of the chapter. Each answer is derived below with the property used.

3.1.1 FIO 1.1 Angles inside a rectangle

Question. Work out every marked angle in each rectangle below. (i) Rectangle ABCD with ∠ABD = 30°. (ii) Rectangle PQRS whose diagonals meet at O with ∠POS = 110°.

Answer. (i) ∠ADB = 60°, ∠DBC = 60°, ∠BDC = 30°, ∠CAB = 30°, ∠CAD = 60°, ∠ACD = 30°, ∠ACB = 60°. (ii) ∠QOR = 110°, ∠QOP = ∠ROS = 70°, ∠OQR = ∠ORQ = 35°, ∠OQP = ∠OPQ = 55°, ∠ORS = ∠OSR = 55°.

Key steps. (i) △ABD is right angled at A, so ∠ADB = 60°. The diagonals are equal and bisect, so OA = OB; △AOB is isosceles and ∠CAB = ∠ABD = 30°. Then ∠CAD = 90 - 30 = 60°. In right triangle ACD, ∠ACD = 30°. In right triangle ABC, ∠ACB = 60°. The diagonal BD splits the 90° angle at B into 30° and 60°, and at D into 60° and 30°. (ii) Because the two diagonals of a rectangle have the same length and cut each other in half, OQ = OR and OP = OQ. Angles around O: ∠QOR = ∠POS = 110° (vertically opposite) and ∠QOP = ∠ROS = 180 - 110 = 70°. Triangle OQR is isosceles with vertex angle 110°, so its base angles are (180 - 110)/2 = 35°. Triangle OQP is isosceles with vertex angle 70°, so its base angles are 55°. Triangle ORS repeats the same pattern.

3.1.2 FIO 1.2 Quadrilaterals with equal diagonals

Question. Construct a four-sided figure with two 8 cm diagonals of equal length. The diagonals must halve each other. For parts (i) to (iv), make the crossing angle 30°, 40°, 90° and 140° respectively.

Answer. In every case the quadrilateral is a rectangle; in case (iii) it is a square.

Key steps. Draw a segment AB 8 cm long and mark its midpoint O. At O draw a ray making the required angle with OA. Mark C and D on that ray with OC = OD = 4 cm. Join AC, CB, BD and DA. Both diagonals measure 8 cm, O is the midpoint of each one, and their crossing angle is the chosen angle. A four-sided figure whose diagonals match in length and halve one another is a rectangle, so all four constructions give rectangles. When the angle is 90°, the diagonals are equal, bisect each other and are perpendicular, which forces all four sides to be equal, so that figure is a square.

3.1.3 FIO 1.3 Perpendicular diameters

Question. A circle has centre O. PL and AM are two perpendicular diameters. What figure is APML? Reason or experiment.

Answer. APML is a square.

Key steps. OA = OM = OP = OL because all are radii of the same circle. The diagonals AM and PL are diameters, so AM = PL. They bisect each other at O, the centre, and they are perpendicular because the diameters are perpendicular. A quadrilateral whose diagonals are equal, bisect each other and meet at 90° is a square. Also, the four chords AP, PM, ML and LA are equal because each is a chord subtending the same central angle of 90°.

3.1.4 FIO 1.4 Making a right angle with sticks

Question. You have no paper, two sticks that are the same length, and a piece of thread. How can you produce a perfect right angle?

Answer. Use the sticks as the equal diagonals of a quadrilateral, tied so that they bisect each other, and pass the thread around their four endpoints. Any corner of the thread quadrilateral is 90°.

Key steps. Call the two sticks AB and CD. Mark their midpoints and tie them at the common midpoint O so that OA = OB = OC = OD. Tie one end at A, run the thread via C to B, then via D back to A. The thread forms quadrilateral ACBD. Its diagonals AB and CD are equal and bisect each other, so ACBD is a rectangle. Every angle of a rectangle is 90°, so ∠C and the other three corners are right angles. The construction works because the rectangle property has already been proved, not because the sticks happen to look square.

3.1.5 FIO 1.5 Parallel and equal opposite sides

Question. Opposite sides of a rectangle are parallel. Can "opposite sides parallel and equal" be taken as the definition of a rectangle? Is every quadrilateral with opposite sides parallel and equal a rectangle?

Answer. No. Such a quadrilateral is a parallelogram, and a parallelogram need not be a rectangle.

Key steps. Take two pairs of parallel lines that meet at an angle other than 90°. With side lengths 4 cm and 5 cm, the quadrilateral has opposite sides parallel and equal but its angles are 30°, 150°, 30° and 150°. A rectangle needs all four angles to be 90°, and this figure fails that test. The condition "opposite sides parallel and equal" defines a parallelogram; adding "all angles are 90°" narrows it to a rectangle.

3.2 Figure it Out, page 102

3.2.1 FIO 2.1 Remaining angles

Question. Work out the angles still missing in each quadrilateral below. (i) Parallelogram RAPE with ∠A = 40°. (ii) Parallelogram PQRS with ∠P = 110°. (iii) Rhombus UVXW with diagonal VX and ∠XVU = 30°. (iv) Quadrilateral AOIE with diagonal OE bisecting the angles at O and E, and ∠AOE = ∠EOI = ∠OEI = 20°.

Answer. (i) ∠P = 40°, ∠R = ∠E = 140°. (ii) ∠R = 110°, ∠Q = ∠S = 70°. (iii) ∠V = ∠X = 60°, ∠U = ∠W = 120°; the diagonal halves the 60° angles into 30° each. (iv) ∠O = ∠E = 40°, ∠A = ∠I = 140°.

Key steps. (i) In a parallelogram, opposite angles match: ∠P = ∠A = 40°. Neighbouring angles add to 180°, so ∠R = ∠E = 180 - 40 = 140°. (ii) ∠R = ∠P = 110° (opposite). ∠Q = ∠S = 180 - 110 = 70° (adjacent). (iii) Since a rhombus is also a parallelogram, its opposite angles match and neighbouring angles add to 180°. A rhombus diagonal splits the angles at its ends into two equal parts, so ∠V = 2 × 30 = 60°, ∠X = ∠V = 60°, and ∠U = ∠W = 180 - 60 = 120°. (iv) The diagonal OE bisects the angles at O and E, so each half is 20° and ∠O = ∠E = 40°. The remaining angles of the quadrilateral are equal: ∠A = ∠I = (360 - 80)/2 = 140°.

3.2.2 FIO 2.2 Parallelogram from diagonals

Question. Use what you know about diagonals to construct a parallelogram with diagonals measuring 7 cm and 5 cm that cross at 140°.

Answer. A parallelogram whose diagonals are 7 cm and 5 cm, crossing at 140°.

Key steps. Draw AB = 7 cm and mark its midpoint O. At O draw a ray making 140° with OA. Mark C and D on this ray with OC = OD = 2.5 cm, because in a parallelogram each diagonal cuts the other into equal halves. Join AC, CB, BD and DA. In quadrilateral ACBD, the diagonals AB and CD bisect each other, so it is a parallelogram; the diagonals have the required lengths and angle.

3.2.3 FIO 2.3 Rhombus from diagonals

Question. Use diagonal properties to construct a rhombus with diagonals 4 cm and 5 cm.

Answer. A rhombus with diagonals 4 cm and 5 cm.

Key steps. Draw AB = 5 cm and mark midpoint O. Draw the perpendicular to AB at O. Mark C and D on the perpendicular with OC = OD = 2 cm, so CD = 4 cm. Join A-C-B-D-A. The diagonals AB and CD are perpendicular and bisect each other. In triangle AOC, AO = 2.5 cm and OC = 2 cm; the four triangles around O are congruent, so all four sides of ACBD are equal. Hence ACBD is a rhombus with the required diagonals.