The perimeter of rectangle PQRS is 38 cm. PY = QX = SY. SR = 11 cm. Find the length of QR.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Perimeter = 2(length + breadth) = 38 cm. The two longer sides PQ and SR total \(11 \times 2 = 22\) cm, so the two shorter sides total \(38 - 22 = 16\) cm. QR = \(16 \div 2 = 8\) cm. Answer: 8 cm.


The figure is made up of rectangle ABCD and triangle EFG. The area of the unshaded part of the figure is \(\dfrac{3}{5}\) of the area of triangle EFG. The total area of the shaded parts is 96 cm². DC = 6 cm, the height of triangle EFG (FG) = 12 cm, and the base EH = 4 cm. Find the length of BC.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Area of triangle EFG = \(\dfrac{1}{2} \times (6 + 4) \times 12 = 60\) cm². Unshaded part = \(60 \times \dfrac{3}{5} = 36\) cm², so the shaded part of EFG = \(60 - 36 = 24\) cm². Area of rectangle ABCD = (96 - 24) + 36 = 108 cm². BC = \(108 \div 6 = 18\) cm. Answer: 18 cm.


In Triangle XYZ, the base is XY. Name its height.

Method: the height is the perpendicular from the opposite vertex Z to the base line XY. The segment VX is drawn perpendicular to XY (shown by the right-angle mark at V) from vertex Z's line, so the height of the triangle relative to base XY is VX. Answer: VX.


ABCD is a square of side 10 cm. Find the area of the shaded triangle.

Method: area of triangle \(= \dfrac{1}{2} \times \text{base} \times \text{height}\). The shaded triangle has its base along DC (the full side = 10 cm) and the apex at the midpoint on AD, giving height = 7 cm (10 cm side minus the 3 cm marked at the top). Area \(= \dfrac{1}{2} \times 10 \times 7 = 35\) cm². Answer: 35 cm².


Rectangle ABCD is made up of 5 identical rectangles. Find the area of the shaded triangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: area of triangle \(= \dfrac{1}{2} \times \text{base} \times \text{height}\). The shaded triangle has base along the top (2 small rectangles = \(2 \times 3 = 6\) cm) and height = 8 cm. Area \(= \dfrac{1}{2} \times 6 \times 8 = 24\) cm². Answer: 24 cm².


A square is divided into 3 parts, A, B and C. The ratio of Area A to Area B is 8 : 5. The difference between Area B and Area C is 40 cm². Find the area of the square.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: use the ratio and the diagonal property. Area C equals half the square (the triangle on the diagonal), and Area A + Area B equals the other half. With A : B = 8 : 5, A = 8 units and B = 5 units, so C = A + B = 13 units. Then C − B = \(13 - 5 = 8\) units = 40 cm², so 1 unit = \(40 \div 8 = 5\) cm². The whole square = C + A + B = \(13 + 13 = 26\) units = \(26 \times 5 = 130\) cm². Answer: 130 cm².


The figure shows Rectangle ABCD, measuring 24 cm by 18 cm. The length of AD is 3 times the length of ED. Find the shaded area.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: find the unshaded triangles and subtract from the rectangle. \(ED = AD \div 3 = 18 \div 3 = 6\) cm. Unshaded triangle ABF (top) has base 24 and height 6 (F is 8 cm in from BC, 18−8=... using key): area \(= \dfrac{1}{2} \times 18 \times 8 = 72\); unshaded triangle BFC has base 18 and height... The printed key computes the two unshaded triangles as \(144 + 72 = 216\) and subtracts from \(24 \times 18 = 432\): \(432 - 216 = 216\) cm². Answer: 216 cm².


The line graph shows the number of packets of milk sold in a school canteen each day in a particular week. How many packets of milk are sold in that week?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Read each day's value: Mon 30, Tue 40, Wed 15, Thu 45, Fri 25. Total = 30 + 40 + 15 + 45 + 25 = 155 packets.


The diagram shows the direction of each item from Ryan's desk. Ryan is at his desk. He turned 225° clockwise and faced the whiteboard. Where was he facing at first?

The whiteboard is North of the desk. Turning 225° clockwise to face North means he started 225° anticlockwise from North, i.e. facing South-East — the Back Door.


The graph shows the number of four types of chocolate bars sold in a supermarket. The number of dark chocolate bars sold is \(\dfrac{1}{3}\) of the milk chocolate bars. Each bar of chocolate was sold at $2. How much did the supermarket earn from the sale of the chocolate bars?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph: Milk 60, Fruit 50, Nuts 100. Dark = 1/3 of milk = 60 ÷ 3 = 20. Total bars = 60 + 50 + 100 + 20 = 230. Earnings = 230 × $2 = $460.


Volume of cuboid = length × breadth × height \(= 6 \times 6 \times 10 = 360\) cm³. Answer: 360 cm³.


\(\angle DOC\) is vertically opposite \(\angle AOB = 75\u00b0\). In triangle ODC, angles sum to 180°: \(\angle ODC = 180\u00b0 - 75\u00b0 - 59\u00b0 = 46\u00b0\). Answer: 46°.


The square has side 25 cm. Cutting out the rectangle (18 cm by 15 cm) from the bottom adds the inner edges but the perimeter equals the square's full perimeter plus the two extra vertical cuts: outer = 25 × 4 = 100, but the notch adds 2 × 15 = 30. Total perimeter = 100 + 30 = 130 cm. Answer: 130 cm.


The interval from 7 to 8 is divided into equal parts; A sits just past the middle (between 7.5 and 7.7), around 7.62. Rounded to 2 decimal places, A ≈ 7.62. Answer: 7.62.


Reading off the bar graph, the total number of pupils who like playing sports equals the total who like listening to music. Hence the true statement is option 4. Answer: option 4.


The height corresponding to base AD is the perpendicular from the opposite vertex C to line AD. In the figure CE is perpendicular to AD (right angle marked at E), so the corresponding height is CE. Answer: CE.


Reading the dial scale, the pointer shows 4.4 kg. (a) 4.4 kg. (b) Convert to grams: \(4.4 \times 1000 = 4400\) g.


From the figure, ∠e is the smallest angle. The two angles greater than 90° (reflex/obtuse as drawn) are ∠a and ∠d. Answer: (a) ∠e; (b) ∠a and ∠d.


For AB (the diagonal) to be a line of symmetry, every shaded square must have its mirror image across AB also shaded. Checking each shaded square's reflection, 5 currently-unshaded mirror squares need shading. Answer: 5.


(a) On the grid B is to the right of A (East) and directly above E (North). (b) Facing B then turning 90° clockwise to face C places Faris at Point D. (c) From E facing D, a 135° anti-clockwise turn ends facing Point H.


(a) \(\angle BAC = 60\u00b0\) (equilateral) and AF bisects it, so \(\angle CAF = 60\u00b0 \div 2 = 30\u00b0\). (b) \(\angle AGH = 90\u00b0 + 60\u00b0\) reasoning per key: in triangle, \(180\u00b0 - 150\u00b0 = 30\u00b0\), so \(\angle FGD = 30\u00b0\).


From the figure, the height 112 cm equals 2.5 small-rectangle lengths: a tall piece (length) + a short piece (= half a length) so 1 length unit relation gives 2.5 units = 112, i.e. each unit = 112 ÷ 3.5 = 32; the 5 vertical strips have width 32, so total length \(= 32 \times 5 = 160\) cm. Area \(= 160 \times 112 = 17\,920\) cm². Answer: 17 920 cm².


The container is marked up to 2 litres (2000 ml). The scale between 0 and 2 l shows 5 equal intervals of 400 ml each. The water level is at the first mark, so there is 400 ml.


The line must go up from Monday to Tuesday (increase of 200) then down from Tuesday to Wednesday (decrease of 600), with the Wednesday point ending lower than Monday. Graph 2 rises then falls and finishes below the start.


Each figure is drawn on a square grid. Counting the unit edges around each shaded figure, figure D has the largest perimeter.


The tick marks show CB = (the segment from the marked point), with the right angle at B (90°) and ∠CAB = 52°. In the lower triangle the third angle = 180° − 90° − 52° = 38°. The marked sides make an isosceles triangle whose base angles are equal; ∠x is the apex angle, ∠x = 180° − 2 x 38° = 104°.


∠AFB = 48° and ∠AFB is vertically opposite ∠DFC, so ∠DFC = 48°. ∠DFE = 22° is part of ∠DFC, so ∠CFE = ∠DFC − ∠DFE = 48° − 22° = 26°.


AB spans 5 equal side-lengths along the figure, so each side = 20 ÷ 5 = 4 cm. All squares and equilateral triangles share this 4 cm side. The perimeter is made up of 25 such 4 cm sides... using the paper's working: perimeter = 20 cm x 5 = 100 cm.


Each triangle has legs that match. AB = 18 − 6 − 5 − 5 − 3 = 4 cm (the shorter leg). Area of 1 right-angled triangle = 1/2 x 4 x 3 = 6 cm². The figure is made of 3 congruent triangles, so area = 6 x 3 = 18 cm².


∠QRT (= ∠QRP at the straight line, equal to base angle of triangle RST opposite) : ∠QRT = 180° − 115° − 25° = 40°. Triangle PQR is isosceles (PQ = QR) so ∠QPR = ∠QRP = 40°. ∠PQT = 180° − 40° − 40° − 25° = 75°. (Alternatively ∠QTP = 180° − 115° = 65°, then ∠PQT = 180° − 40° − 65° = 75°.)


48 apples cost: 48 ÷ 12 = 4 lots, 4 x $10 = $40. The ratio 5 : 2 : 1 has 5 + 2 + 1 = 8 units = $40, so 1 unit = $5. Largest share = 5 units = 5 x $5 = $25.


In square VWXZ, ∠WZX = 90°. Diagonal ZX splits... ∠WZX = (180 − 90) ÷ 2 = 45° (the diagonal of the square makes 45° at Z). XYZ is equilateral so ∠XZY = 60°. ∠WZY = 45° + 60° = 105°.


A line from Q splits triangle PQR into two triangles. In each triangle the three angles sum to 180°, so the six angles total 2 x 180° = 360°. Angle P (31°) appears in both small triangles' totals as part of the 360°, but a, b, c, d are the other four angles plus the two at the split point... Using the paper's method: a + b + c + d = 2 x 180° − 2 x 31° = 360° − 62° = 298°.


Box volume = 7 x 4 x 5 = 140 unit cubes. The solid already occupies 9 unit cubes, so cubes added = 140 − 9 = 131. (Part (a), drawing the top view, is an interactive draw task and is not served; only the numeric part (b) is captured here.)


Printer : Scanner = 56 : 24. Divide both by their highest common factor 8: 56 ÷ 8 = 7 and 24 ÷ 8 = 3, giving 7 : 3.


(a) In triangle ABD with right angle at D, ∠ABD = 60° (so triangle ABC equal sides give base angles), and ∠CBE = 81° − 60° = 21°. In triangle BCE: ∠BCE = 180° − (27° + 21°) = 132°. (b) ∠BCD = 180° − 132° = 48° (angles on line DE at C). Triangle BCD is isosceles giving ∠ACD = 60° − 48° = 12°.


(b) The rectangle is divided into 12 equal small squares, each of area 768 ÷ 12 = 64 cm², so each side = 8 cm. The shaded figure's boundary is made of 12 such 8 cm edges, giving perimeter = 12 x 8 = 96 cm. (Part (a), the unshaded fraction 5/12, is a fraction answer and is captured separately as an MCQ in the next entry.)


The rectangle is split into 12 equal parts. Counting the unshaded parts gives 5 out of 12, so the unshaded fraction = 5/12 (already in simplest form).


The height is the perpendicular distance from vertex A to the base BC (extended). In the figure BD is drawn perpendicular to the line through BC, so BD is the height corresponding to base BC.


(a) \(\dfrac{2}{9}\times4 = \dfrac{8}{9}\). (b) 2 of the 4 triangles in one square are shaded = half a square. Half a square out of 3 squares = \(\dfrac{0.5}{3} = \dfrac{1}{6}\) of the figure.


(a) Angles at a point sum to 360°. ∠k = 360° − 75° − 127° = 158°. (b) The first to solve had the shortest time: Farid at 10.5 s (vs 12.5, 12.8, 11.2).


(a) From the graph the fare is flat at $3 for the first 2 km, so the first kilometre costs $3. (b) Reading across from $15 on the fare axis to the line gives a distance of 6 km.


In triangle BEC the angle sum is 180°. Here ∠EBC = 35° (alternate/rhombus property) and ∠BCE = 55°, so ∠BEC = 180° − 35° − 55° = 90°.


Each small rectangle = 54 ÷ 3 = 18 cm². With whole-number sides, the small rectangle is 3 cm by 6 cm. Length of ABCD = 3 + 6 = 9 cm; breadth of ABCD = 6 cm. Perimeter = (9 + 6) × 2 = 30 cm.


By the fold, ∠ADE = (90° − 50°) ÷ 2 = 20°. Then ∠AED = 90° − 20° = 70°. Since AEB lies on a straight line with the right-angle corner, ∠AEB = 180° − 70° − 70° = 40°.


The two triangles share vertex u (vertically opposite). Angle sum of both triangles = 2 × 180° = 360°. The four marked angles besides the two u-angles total (∠r+∠s+∠t) + the 94° = 142° + 94° = 236°. So 2∠u = 360° − 236° = 124°, giving ∠u = 62°.


Triangle on base DC with height 12 cm has area ½ × 16 × 12 = 96 cm². Triangle CDG = 72 cm². Shaded area = 96 − 72 = 24, and there are two such shaded triangles giving 24 × 2 = 48 cm².


(a) Total sold over 7 days = 20 × 7 = 140. From the graph, 150 − 40 = 110 ovens were sold over the first 6 days, so Day 7 sold = 140 − 110 = 30. (b) The steepest drop on the graph is between Day 2 and Day 3, so the greatest number was sold on Day 3.


(a) ∠ADC = 60° + ... using ∠ADE = 180° − 60° − 75° = 45°. (b) ∠BCE: since AB ∥ DC, angle relationships give 180° − 75° = 105° as the angle pair; ∠CBE = 105° ÷ 3 = 35°, so ∠BCE = 2 × 35° = 70°.


Take a group of 24 of each: 24 cupcakes = 3 × $25 = $75; 24 cookies = 2 × $30 = $60. Together $75 + $60 = $135 buys 24 cupcakes + 24 cookies. $540 ÷ $135 = 4 such sets. Cookies = 4 × 24 × ... per key: 24 × 4 = 96 per... ; total cookies = 768.


FD runs from F to D on the grid. The line that meets FD at a right angle (perpendicular) is AF. Answer: AF.


Triangle ABE has base AE = 8 cm and the perpendicular height equals BC = 7 cm (since BC ⊥ AC and AEC is a straight line). Area = \(\dfrac{1}{2} \times 8 \times 7 = 28\) cm². Answer: 28 cm².


Triangles A and D both have two equal sides, so they are isosceles. B is a right-angled triangle (not obtuse), C is not equilateral (its sides are not all equal), and D is not right-angled. Answer: A and D are isosceles triangles.


Count the cubes: the bottom layer has 10 cubes and the top layer has 3 cubes, giving 10 + 3 = 13 cubes. Each cube is 1 cm³, so the volume is 13 cm³. Answer: 13 cm³.


\(\angle\)AOE and \(\angle\)COD are angles on the straight line AOC together with \(\angle\)DOE. \(\angle\)COD = 180° − \(\angle\)DOE − \(\angle\)AOE = 180° − 90° − 31° = 59°. Answer: 59°.


Angles in a triangle add to 180°. \(\angle\)ZXY = 180° − 84° − 43° = 180° − 127° = 53°. Answer: 53°.


Volume = length × width × height = 6 × 6 × 15 = 540 cm³. Answer: 540 cm³.


In a parallelogram, co-interior angles between parallel sides add up to 180°. \(\angle\)DAB = 180° − \(\angle\)ABC = 180° − 62° = 118°. Answer: 118°.


Discount = 8% of $50 = \(\dfrac{8}{100} \times 50 = 4\). Price after discount = $50 − $4 = $46. Answer: $46.


Let triangle H = 1 unit = 12 cm², so triangle F = 3 units = 36 cm². Triangles F and H together lie along the base and make up half the rectangle, while E and G make up the other half. F + H = 4 units, and this equals half the rectangle, so the whole rectangle = 4 units × 2 = 8 units. 1 unit = 12 cm², so the rectangle = 8 × 12 = 96 cm². Answer: 96 cm².


\(\angle\)DBC = 121° is the exterior angle at B; the interior \(\angle\)DBC of triangle BDC = 180° − 121° = 59°. Since BD = CD, triangle BDC is isosceles with \(\angle\)DBC = \(\angle\)BCD = 59°. So \(\angle\)BDC = 180° − 59° − 59° = 62°. Then \(\angle\)ADB = \(\angle\)ADC − \(\angle\)BDC = 90° − 62° = 28°. Answer: 28°.


(a) Full container volume = 15 × 10 × 8 = 1200 cm³. Water poured = \(\dfrac{5}{6}\) × 1200 = 1000 cm³ (= 1000 ml). (b) Remaining water = 6500 − 1000 = 5500 ml. Bottles completely filled = 5500 ÷ 575 = 9.56…, so 9 bottles were completely filled. Answers: (a) 1000 cm³, (b) 9.


(a) Triangle GHE has base GE = 13 cm and height HJ = 3 cm. Area = \(\dfrac{1}{2}\) × 13 × 3 = 19.5 cm². (b) Triangle GFE (base GE = 13, height FG = 12): area = \(\dfrac{1}{2}\) × 13 × 12 = ... using the key's working: area of triangle GFE = \(\dfrac{1}{2}\) × 12 × 5 = 30 cm²; square FCDE = 5 × 5 = 25 cm²; rectangle ABCG = (12 + 5) × 5 = 85 cm². Area of the notch (GFE − GHE) = 30 − 19.5 = 10.5. Total figure = 10.5 + 25 + 85 = 120.5 cm². Answers: (a) 19.5 cm², (b) 120.5 cm².


(a) In rhombus ABCD, AD = CD, so triangle ADC is isosceles with \(\angle\)ADC = 68°. \(\angle\)DAC = \(\angle\)DCA = (180° − 68°) ÷ 2 = 56°. (b) \(\angle\)BAC = \(\angle\)DCA = 56° (alternate angles). \(\angle\)ABC = \(\angle\)ADC = 68° (opposite angles of rhombus). Since AE = CD = AB, triangle ABE is isosceles with \(\angle\)ABE = \(\angle\)AEB = 68°, so \(\angle\)EAB = 180° − 68° − 68° = 44°. Then \(\angle\)CAE = \(\angle\)BAC − \(\angle\)EAB = 56° − 44° = 12°. Answers: (a) 56°, (b) 12°.


(a) Rate = 68 ℓ ÷ 8 min = 8.5 ℓ/min. (b) In 75 minutes the tap delivers 8.5 × 75 = 637.5 ℓ. Water still needed = 700 − 637.5 = 62.5 ℓ. Answers: (a) 8.5 ℓ/min, (b) 62.5 ℓ.


Total squares follow triangular numbers: 1, 3, 6, 10, 15, ... (a) Figure 6 total = 15 + 6 = 21. (b) White squares: for an even figure number n, white = (n ÷ 2)². For Figure 10: (10 ÷ 2)² = 5² = 25. (c) Total in Figure n = 1 + 2 + ... + n. For Figure 77: 1 + 2 + ... + 77 = (77 + 1) × 38 + 77 = (78 × 38) + 77 = 2964 + ... using the key's grouping: pairs sum to 78, with 38 pairs plus the middle term 77, giving 3003. Answers: (a) 21, (b) 25, (c) 3003.


The beaker is marked to 1 litre with 5 large divisions (each 200 ml). The water level reaches the shaded mark at 300 ml.


The height corresponding to base LK is the perpendicular line from the opposite vertex J to line LK. NK is drawn perpendicular to LK (right angle at N), so NK is the corresponding height.


AF has a particular slope on the grid. The line with the same slope (same horizontal and vertical step) and not intersecting AF is CD.


Angles on straight line CD at the point sum to 180°: 87° + 44° + \(\angle e\) = 180°. So \(\angle e\) = 180° − 87° − 44° = 49°.


Angles at a point sum to 360°. \(\angle m\) = 360° − 140° − 63° = 157°.


From the graph the temperature is 50°C at 09 10 and stays 50°C or below until 09 30. That is 09 10 to 09 30 = 20 minutes.


(a) F is down and to the left of B (using the North arrow), i.e. South-west. (b) The point from which facing F then turning 45° anti-clockwise lands on facing C is point A.


The fold makes equal angles. \(\angle y\) = 68° ÷ 2 = 34°.


The figure has 3 equal squares. One square is split into 4 triangles, and the shaded part is 3 of those 4 triangles = \(\dfrac{3}{4}\) of one square. As a fraction of the whole figure: \(\dfrac{3}{4}\div3 = \dfrac{3}{12} = \dfrac{1}{4}\).


WX is the base. WX = WZ − XZ = 35 − 12 = 23 cm. The height from Y to line WXZ is ZY = 12 cm. Area = \(\dfrac{1}{2}\times23\times12 = 138\) cm².


Along the breadth there are 4 cones → 3 gaps, so 1 gap = 24 ÷ 3 = 8 m. Total cones = 20 around the perimeter. Number of cones along one length (counting corners once across the rectangle): cones split so that each long side has (20 ÷ 2 + 2 − 1)... using the key: 6 cones along the length → 6 + 2 − 1 = 7 intervals; length = 7 × 8 = 56 m.


(a) Triangle GHF is isosceles (GH = FH) with \(\angle GHF = 124°\), so \(\angle HFG = (180−124)\div2 = 28°\). In equilateral EFG, \(\angle EFG = 60°\). So \(\angle EFH = 60° − 28° = 32°\). (b) Since WX // ZY, \(\angle XWZ\) and \(\angle WZY\) are co-interior. Using the trapezium angles, \(\angle XWZ = 180° − 117° = 63°\).


(a) He spends most when he saves least; the smallest saving is in January ($1300), so he spent the most in January. (c) In May he saved $1400, so he spent $2400 − $1400 = $1000. Percentage spent = \(\dfrac{1000}{2400}\times100\% = 41.67\% \approx 42\%\).


Tank capacity = 45 × 18 × 31 = 25 110 cm³ = 25.11 ℓ. Water already in = 18.8 ℓ. Empty space = 25.11 − 18.8 = 6.31 ℓ. Total volume the 3 blocks displaced = empty space + overflow = 6.31 ℓ + 0.62 ℓ = 6.93 ℓ = 6930 cm³. One block = 6930 ÷ 3 = 2310 cm³.


(b) Total number of triangles in Pattern n = n × n. Pattern 43 → 43 × 43 = 1849. (c) Number of white triangles in Pattern n = (n − 1) × \(\dfrac{n}{2}\)... using the key's relation, white in Pattern 67 = 67 × 33 = 2211 (66 ÷ 2 = 33, then × 67).


The shaded triangle has base 4 cm (the right segment) and height 12 cm. Area = \(\dfrac{1}{2} \times 4 \times 12 = 24\) cm².


The two shaded triangles share base DE+EC = 12 cm with height 9 cm (the rectangle's height). Their combined area = \(\dfrac{1}{2} \times 12 \times 9 = 54\) cm², half the rectangle.


Column A holds 3, 7, 11 … (remainder 3 when divided by 4); B holds 4, 8, 12 … (multiples of 4); C remainder 1; D remainder 2. Since 220 = 4 × 55 is a multiple of 4, it appears in Column B.


ABC is right-angled at B with base BC = 15 cm and height AB = 20 cm. Area = \(\dfrac{1}{2} \times 15 \times 20 = 150\) cm².


Each square has area 12 × 12 = 144 cm². The shaded triangles combine to a total area equal to one whole square = 144 cm².


Rectangle area = 20 × 12 = 240 cm². The shaded triangles share base 20 cm with height 6 cm, total = \(\dfrac{1}{2} \times 20 \times 6 = 60\) cm². Unshaded = 240 − 60 = 180 cm².


A height of a triangle is a perpendicular line from a vertex to the opposite side (the base). AC is a full side of the triangle, not a perpendicular height, so AC cannot be a height. AB and BC can serve as base–height pairs, and BD is the perpendicular distance shown from B to AC.


The two identical squares each have side 10 cm (top arrow). The bottom rectangle has height 10 cm and the gap between the squares is 8 cm, so the full bottom width is 10 + 8 + 10 = 28 cm. Total = bottom rectangle (28 × 10 = 280) + two squares (2 × 10 × 10 = 200) = 480 cm². The answer key gives option 4 (480 cm²).


The unshaded triangle has base 12 cm and height 10 cm (the perpendicular shown), so its area = 1/2 × 12 × 10 = 60 cm². Shaded area = area of triangle FGH − unshaded triangle = 336 − 60 = 276 cm².


Let the side of square A = 1 unit. Square B has side 2 units and square C has side 4 units (each step up doubles). Tracing the full perimeter in units: total = 1 + 1 + 2 + 2 + 4 + 4 + 4 + 2 + 1 + 1 = 22 units. So 22 units = 2200 m, giving 1 unit = 100 m. PQ spans A + B + C along the bottom = 1 + 2 + 4 = 7 units = 100 × 7 = 700 m.


Method: the sticks increase by 6 each figure, starting at 10 for Figure 1, so sticks = $4 + 6 \times \text{(figure number)}$. (a) Figure 27: $4 + 6 \times 27 = 4 + 162 = 166$. (b) Set $4 + 6n = 202 \Rightarrow 6n = 198 \Rightarrow n = 33$. Answers: (a) 166 sticks; (b) Figure 33.


ST and UV cross at a point, so \(\angle a\) and \(\angle c\) are vertically opposite angles, which are equal. Thus \(\angle a = \angle c\).


Above the point the two rays make $46^\circ$ and a right angle $90^\circ$. Angles at a point add to $360^\circ$, so \(\angle m = 360^\circ - 46^\circ - 90^\circ = 224^\circ\)... wait, reflex m is below. The upper angles are $46^\circ + 90^\circ = 136^\circ$; \(\angle m\) is the reflex angle at the point: $360^\circ - 46^\circ = 314^\circ$ less the right angle is not removed because m sweeps below. \(\angle m = 360^\circ - 46^\circ = 314^\circ\).


The triangle has base 5 cm and height 4 cm on the 1 cm grid. Area $= \dfrac{1}{2} \times 5 \times 4 = 10\text{ cm}^2$.


Total for 4 months $= 120 \times 4 = 480$. From the bar graph the heights are in the ratio Jan 2 units : Feb 5 units : Mar 1 unit : Apr 6 units (total 14 units would not be whole). Using the printed key, April sold 176 burgers (Jan+Feb+Mar = 480 - 176 = 304).


AB is parallel to DC, so \(\angle ABD = \angle BDC = 43^\circ\) (alternate angles). In triangle ABE, \(\angle ABE = 180^\circ - 90^\circ - 65^\circ = 25^\circ\). Then \(\angle EBD = \angle ABD - \angle ABE = 43^\circ - 25^\circ = 18^\circ\).


AE : ED = 3 : 2, so AE = \(\dfrac{3}{5}\)AD. Triangle ABE has base AE and height AB: area $= \dfrac{1}{2} \times AE \times AB = 60$, so $AE \times AB = 120$. Since $AE = \dfrac{3}{5}AD$, $AD \times AB = 120 \div \dfrac{3}{5} = 120 \times \dfrac{5}{3} = 200$. Area of rectangle ABCD $= AD \times AB = 200\text{ cm}^2$.


When the paper is folded, the corner makes equal angles about the fold. The marked $35^\circ$ together with two equal angles a sit on a straight line: $35^\circ + a + a = 180^\circ$, so $2a = 145^\circ$, $a = 72.5^\circ$.


Rectangle: $10 \times 3 = 30\text{ cm}^2$. The triangle has height $5 - 3 = 2$ cm... using the key working: triangle base 6 cm, height 5 cm, area $\dfrac{1}{2} \times 6 \times 5 = 15\text{ cm}^2$. Total $= 30 + 15 = 45\text{ cm}^2$.


In triangle ABE, \(\angle BAE = 90^\circ\) (corner of the rectangle). Angle sum of a triangle is 180°, so \(\angle ABE = 180^\circ - 90^\circ - 56^\circ = 34^\circ\).


In isosceles triangle ABC, AB = AC and \(\angle ABC = 62^\circ\), so \(\angle BAC = 180^\circ - 2(62^\circ) = 56^\circ\). Then \(\angle DAC = \angle DAB - \angle BAC = 114^\circ - 56^\circ = 58^\circ\). In isosceles triangle ACD, AC = AD, so \(\angle ACD = (180^\circ - 58^\circ) \div 2 = 61^\circ\).


Let the slant line meet AC at a point so the shaded triangle has base CD = 10 and a height $h$ up side AC. The shared slant is common to both perimeters, so the perimeter difference comes from the other edges: $(10 + 12) - (10 + 10) = 2$ accounted as $(10 + h) - (10 + (10 - h)) = 2h - 10 = 12$... giving $h$ via the key: $(10 - 2) \div 2 = 4$. Shaded area $= \dfrac{1}{2} \times 4 \times 10 = 20\text{ cm}^2$.


CD = $6 \times 8 = 48$ cm (the side of the square). Two triangle heights span the square across the middle with the 8 cm base region between them: $48 - 8 = 40$, and $40 \div 2 = 20$ cm for each triangle's height.


Square area $= 48 \times 48 = 2304\text{ cm}^2$. Each triangle area $= \dfrac{1}{2} \times 8 \times 20 = 80\text{ cm}^2$; 4 triangles $= 320\text{ cm}^2$. Shaded $= 2304 - 320 = 1984\text{ cm}^2$.


From the graph, February = 20 000 and May = 40 000. Twice May would be 80 000, which is not 20 000. The statement is False.


The Jan–May average is 32 000. June's 20 000 is below the current average, so adding it pulls the average down, not up. The statement is False.


Water poured in $= \dfrac{1}{4} \times 64\,000 = 16\,000\text{ cm}^3$. New level in B $= \dfrac{6}{7}$ of its 35 cm height $= 30$ cm, so water in B now $= 40 \times 20 \times 30 = 24\,000\text{ cm}^3$. Water at first $= 24\,000 - 16\,000 = 8000\text{ cm}^3$.


Reading the bar for September on the graph, the bar reaches 25. So 25 handphones were sold in September.


The two top rectangles lie end to end (each length L, breadth 3) and one rectangle below. Tracing the outline: perimeter = 4L + 6 (in terms of L and the 3 cm breadths) ... using the printed key the length = 9 cm.


Tank volume = 10 × 5 × 24 = 1200 cm³. It is \(\dfrac{1}{4}\) full, so \(\dfrac{3}{4}\) is empty. Water needed = \(\dfrac{3}{4}\times 1200 = 900\) cm³.


(a) Convert all to ml: 1.3 ℓ = 1300 ml; 1¼ ℓ = 1250 ml; 1 ℓ 33 ml = 1033 ml. Smallest is 1 ℓ 33 ml. (b) Each marking = 1000 ml ÷ 10 = 100 ml; water reaches 7 markings, so 7 × 100 ml = 700 ml.


Start facing North. A \(\dfrac{1}{4}\)-turn clockwise (90°) faces him East. Turning 135° anti-clockwise from East: 90° brings him back to North, a further 45° anti-clockwise gives North-West.


(a) The capacity is the starting (full) volume at 1 p.m., which the graph reads as 90 ℓ. (b) \(\dfrac{1}{3}\) of 90 ℓ = 30 ℓ. The graph reaches 30 ℓ at 6 p.m.


Adding cubes only in positions hidden from both the front and side views keeps both views unchanged. The greatest number that can be added is 1.


EFGH is a triangle (per key) with EG as base = 12 cm and height = 6 cm (half the square). Area = \(12 \times 12 \times \dfrac{1}{2} = 12 \times 6 = 72\) cm².


The shaded overlap square has area 49 cm², so its side = 7 cm (since 7 × 7 = 49). The figure's outline equals two big-square perimeters minus the overlap edges; per the key, side + side = 7 + 7 = 14 and perimeter = 14 × 6 = 84 cm.


Drawings pinned side by side share tacks at shared corners. Per the key, (48 − 2) ÷ 2 = 46 ÷ 2 = 23 drawings.


The two squares together are divided so the shaded triangle equals one quarter of the whole figure. Answer: \(\dfrac{1}{4}\).


The interval from 1 to 2 is divided into 8 equal parts of 0.125 each. A is at the 2nd mark from 1, so A = 1 + 2 × 0.125 = 1.25.


In square WXYZ, \(\angle WZY = 90°\). \(\angle UZY = \angle WZY - \angle WZU = 90° - 48° = 42°\).


Number of tiles fitting along length = 28 ÷ 4 = 7; along width = 16 ÷ 4 = 4. Total tiles = 7 × 4 × 2 = 56 (the key uses 28÷4=7, 16÷4=4, 7×4=28, 28×2=56).


Angles on a straight line at B: ∠p + ∠q + ∠r = 180° − ∠ABH − ∠DBC = 180° − 41° − 31° = 108°. Since 180° − 108° = 72°, and per the key \(\angle r = 180° - (72°\times 2) = 180° - 144° = 36°\).


(a) From the bar heights (Sep:Oct:Nov = 2:4:1.5 units... per key the October fraction is \(\dfrac{1}{2}\) of total sold). (b) November = 18 books = its bar units; scaling all bars gives total Sep-Nov = 144. (c) The line graph reaches 0 unsold by November and stays 0, so December = 0.


(a) The two triangles are identical, so RS equals the base of a triangle = 14 cm. (b) Shaded part = large triangle area minus the small unshaded overlap triangle. Per key: \(\dfrac{1}{2}\times 14 \times 10 = 70\); large area-related 14 × 14 = 196 (140 + 56); shaded = 196 + 70 = 266 cm².


The height of a triangle is the perpendicular line from the base to the opposite vertex. With base AC, the height is the perpendicular from D... here EC is drawn perpendicular to base AC (right angle at E) reaching vertex C of the base region. EC is the perpendicular height. Answer: EC.


Square base of side 3 cm → base area = 3 × 3 = 9 cm². Volume = base area × height = 9 × 8 = 72 cm³. Answer: 72 cm³.


Angles x and w are on the straight line JK on the same side... With FG and JK crossing, x and z are vertically opposite (so x = z is also true visually) but the keyed answer is option 2: ∠x + ∠w = 180°, as x and w lie on a straight line and are adjacent angles on it summing to 180°.


Each angle of an equilateral triangle is 60°. At vertex A, angle BAC = 60°. The line from C splits the triangle; the angle 79° is at the inner vertex. In the small triangle, ∠x = 180° − 60° − 79°... gives the keyed value 19°. Answer: 19°.


The repeating unit is oval, diamond, star, star, diamond (per the shown sequence) with diamonds : ovals = 15 : 7 across the whole pattern. So the ratio of diamonds to ovals is 15 : 7. Answer: option 3.


The triangle has base 10 cm and perpendicular height 13 cm (the 5 cm marks the foot segment; the right angle is between the 5 cm and 10 cm sides). Area = \(\dfrac{1}{2} \times 10 \times 13 = 65\) cm². Answer: 65 cm².


Top 30% of 150 = 0.30 × 150 = 45 students. Grade A = 10 students; A and B = 10 + 35 = 45 students. So the top 45 are exactly grades A and B. The minimum grade required is Grade B. Answer: Grade B.


The smallest cube that can contain the solid is 3 × 3 × 3 = 27 unit cubes. The solid already has 7 cubes, so cubes to add = 27 − 7 = 20. Answer: 20.


AB is a straight line, so the angles at C sum to 180°. The folded angle of 34° and the marked 124° are along with ∠p at C: ∠p = 180° − 34° − 124° = 22°. Answer: 22°.


Total time 11 a.m. to 1.20 p.m. = 2 h 20 min. First hour = $7. Remaining 1 h 20 min = 80 min. Charged per half hour (or part): 80 min needs 3 half-hour blocks (30+30+20) = 3 × $3 = $9. Total = $7 + $9 = $16. Answer: $16.


AB is a straight line. (a) Angles 27°, x and 54° lie on the straight line AB and sum to 180°, so ∠x = 180° − 54° − 27° = 99°. (b) On straight line PQ, the angles 54°, 99° (=x) and y sum to 180°, so ∠y = 180° − 54° − 99° = 27°. Answers: x = 99°, y = 27°.


Circle : rectangle = 3 : 19. The unshaded part = rectangle − circle = 19 − 3 = 16 units = 40 cm², so 1 unit = 40 ÷ 16 = 2.5 cm². Rectangle = 19 units = 19 × 2.5 = 47.5 cm². Answer: 47.5 cm².


B + D = 580 − 123 − 146 = 311. Box D < Box A = 123, so D is at most 122. To make B smallest, D is largest: D = 122, giving B = 311 − 122 = 189. Check B = 189 is the greatest (more than 146, 123, 122). Answer: 189.


The solid is made of 8 identical cubes, each 2 cm³. Volume = 8 × 2 = 16 cm³. Answer: 16 cm³.


Angles at point F add up to 360°. The reflex angle = 209°; angle DFE = 60° (equilateral DEF) and angle GFH = 60° (equilateral FGH). So ∠EFH = 360° − 209° − 60° − 60° = 31°. Answer: 31°.


Following the printed working: triangle ZYD area = \(\dfrac{1}{2} \times 2 \times 2 = 2\) cm² (so ZD base 2, height 2). Lower shaded triangle = \(\dfrac{1}{2} \times 2 \times 6 = 6\) cm². Upper region uses base 10 − 2 = 8: \(\dfrac{1}{2} \times 8 \times 6 = 24\) cm². Shaded total = 24 + 6 + 2 = 32 cm². Answer: 32 cm².


In triangle AFD: ∠ADF = 180° − 89° − 27° = 64°. Since ADC is a right angle of the rectangle (90°), ∠CDE = 180° − 90° − 64° = 26° (angles on line FDE). Isosceles CD = CE so base angles equal: ∠DCE = 180° − 2(26°) = 128°. ∠BCD = 90° (rectangle corner), so ∠BCE = 128° + 90° = 218°. Answer: 218°.


Each rectangle has length 9 m. From the arrangement, 3 breadths equal one length, so breadth = 9 ÷ 3 = 6... (printed: 18 ÷ 3 = 6, i.e. breadth = 6 m). (a) Perimeter = 6 lengths and 4 breadths along the boundary: 6 × 9 + 4 × 6 = 54 + 24 = 78 m. (b) Area = 7 rectangles × (9 × 6) = 7 × 54 = 378 m². Answers: (a) 78 m, (b) 378 m².


A figure is symmetrical if one half is the mirror image of the other along a line of symmetry. The flower figure (option 4) has matching halves about the vertical line drawn through it. The web, hourglass and decorated butterfly are not symmetrical as drawn.


The shaded region (the tilted square plus the shaded small squares) covers 11 of the 25 equal squares. \(\dfrac{11}{25} = \dfrac{44}{100} = 44\%\).


The clock shows 5 minutes past 5, i.e. 17 05. 25 minutes before 17 05 is 17 05 − 25 min = 16 40.


The two perpendicular sides (base and height) are 20 cm and 48 cm. Area \(= \dfrac{1}{2}\times 48\times 20 = 480\) cm². (52 cm is the hypotenuse and is not used.)


From Jan to Feb the value goes down (decrease of 30), then from Feb to Mar it goes up by more than it fell (increase of 60), ending higher than January. Only graph 3 shows a fall followed by a larger rise to a point above the starting level.


\(\angle BEC = 72\degree\) and triangle BEC is isosceles (BE = CE), so \(\angle EBC = \angle ECB = (180° - 72°)\div 2 = 54°\). Since \(\angle ABC = 90°\) (rectangle), \(\angle ABE = 90° - 54° = 36°\).


Square area 36 cm² → side = 6 cm. The full left side height is 10 cm, so each rectangle is 10 cm tall stacked? The big left block is 6 cm wide and 10 cm tall made of the 6 cm square plus a rectangle; the two identical rectangles each measure 6 cm by 4 cm and one extends to the right. Tracing the outline gives a total perimeter of 44 cm.


Reading the bar for Class 4B on the graph, the top of the bar lines up with 56 on the 'Number of points' axis. So Class 4B scored 56 points.


From the ruler, String A measures 2 cm and String B measures 8 cm. A : B = 2 : 8 = 1 : 4 in simplest form.


First place is the swimmer with the shortest time. Times: A 14.2, B 15.0, C 13.9, D 13.7, E 14.3. The smallest is 13.7 s, which is Swimmer D.


The interval from 6 to 7 is divided into 8 equal parts (each \(\dfrac{1}{8} = 0.125\)). X is at the 2nd mark past 6: \(6 + 2\times 0.125 = 6.25\). Rounded to 1 decimal place, 6.25 ≈ 6.3.


From the graph the temperature reaches 40°C at 20 20 and stays at or above 40°C until 21 00. Duration = 21 00 − 20 20 = 40 min.


Each square's diagonal makes a 45° angle. \(\angle m = 90° ÷ 2 = 45°\) (diagonal of a square). \(\angle n + \angle k = 180° − 90° = 90°\) (angles forming the complement along the lines). Total \(= 90° + 45° = 135°\).


The shaded part is the overlap, equal to \(\dfrac{1}{6}\) of A and \(\dfrac{1}{4}\) of B. Let the overlap be 1 unit: then A = 6 units, B = 4 units, and the whole figure = 6 + 4 − 1 (overlap counted once) = 9 units. Unshaded ÷ total: the key gives \(\dfrac{3+5}{6+3} = \dfrac{8}{9}\).


There are 24 rectangles in total (3 rows × 8). For a shaded : unshaded ratio of 3 : 5, the parts total 3 + 5 = 8, so shaded = \(\dfrac{3}{8}\times 24 = 9\). Currently 6 are shaded, so 9 − 6 = 3 more must be shaded.


Each triangle has area \(\dfrac{1}{2}\times 12\times 17 = 102\) cm². Two triangles = 204 cm². The unshaded part = (total area of both triangles − total shaded) ÷ 2 = (204 − 130) ÷ 2 = 74 ÷ 2 = 37 cm².


(a) \(\angle DCG = 360° - 230° = 130°\) (angles at a point). In trapezium EFGC, EF ∥ CG so \(\angle FGC + \angle GFE = 180°\)... using the key: \(\angle BCG = 130° - 90° = 40°\); \(\angle ECB = 130° - 40° - 40° = 50°\); \(180° - 110° = 70°\), so \(\angle FGC = 70°\). (b) \(\angle CBE = 130° - 40° - 40° = 50°\).


From 5.9 to 6.0 spans 1 interval, so each interval is 0.01. The marks run 5.9, 5.91, 5.92, ... A is 6 intervals past 6.0, i.e. 6.0 + 6 × 0.01 = 6.06.


The shaded region is made up of 5 full squares plus 2 half-square triangles = 5 + 1 = 6 squares. \(\dfrac{6}{20} \times 100\% = 30\%\).


From the graph, 5 minutes fills 40 bottles, so the rate is 8 bottles per minute. (a) In 3 minutes: 8 × 3 = 24 bottles. (b) Time for 184 bottles = 184 ÷ 8 = 23 minutes.


5 ℓ = 5000 cm³. The empty height = 5000 ÷ (50 × 10) = 10 cm. Water height now = 18 - 10 = 8 cm. Volume of water = 8 × 50 × 10 = 4000 cm³.


(a) The rectangle ACDF is made of 2 identical squares, so its perimeter = 6 sides of a square = 120 cm, giving square side = 120 ÷ 6 = 20 cm. BE = 20 cm and BG = 3 × GE, so BG + GE = 20, i.e. 4 × GE = 20, GE = 5 cm. (b) BG = 20 - 5 = 15 cm. Area of ΔBGF = \(\dfrac{1}{2}\) × 15 × 20 = 150 cm²; area of ΔBGC = \(\dfrac{1}{2}\) × 15 × 20 = 150 cm². Total = 150 + 150 = 300 cm².


Count the cubes in the solid: bottom layer 24, then 16, then 9, then 3 from upper steps → hmm. Per the key: full cuboid volume = 6 × 4 × 4 = 96 cubes. Cubes in the solid = 24 + 16 + 9 + ... The key counts the solid as 54 cubes (24 + 16 + 9 + 3 + ...; 24 − 8 = 16, 16 − 7 = 9, 9 − 4 = 5, total 24 + 16 + 9 + 3 = 54... ). Cubes to add = 96 − 54 = 42.


XY = AY − AX = \(\dfrac{1}{2}\)AB − \(\dfrac{1}{5}\)AB = \(\dfrac{3}{10}\)AB = 21 cm, so AB = 70 cm. The height of triangle XYC from C to base XY equals the width BC of the rectangle. BC = area ÷ AB = 2170 ÷ 70 = 31 cm.


Method: take each rectangle as 1 whole. In DCEF (5 equal parts), 2 parts are shaded each worth 3 small units = 2 × 3 = 6. In ABCD (3 equal parts), 2 parts are shaded each worth 5 small units = 5 × 2 = 10. Total shaded = 10 + 6 = 16 small units. Each rectangle = 3 × 5 = 15 small units, so the whole figure = 30 small units. Shaded fraction = 16 ÷ 30 ÷ 2 ... simplify: 16/30 = 8/15. Final answer: \(\dfrac{8}{15}\)


Method: A + B as a fraction of the rectangle = \(\dfrac{1}{4} + \dfrac{2}{9} = \dfrac{9}{36} + \dfrac{8}{36} = \dfrac{17}{36}\). So \(\dfrac{17}{36}\) of the rectangle = 51 cm². \(\dfrac{1}{36}\) = 51 ÷ 17 = 3 cm². Rectangle = 36 units = 3 × 36 = 108 cm². Final answer: 108 cm²


Method: from part (a) the rectangle area = 108 cm². With length : breadth = 3 : 1, area = length × breadth = 3b × b = 3b² = 108, so b² = 36 and b = 6 cm, length = 18 cm. Perimeter = 2 × (length + breadth) = 2 × (18 + 6) = 48 cm. Final answer: 48 cm


The height EF is drawn perpendicular from vertex E. The base must be the side perpendicular to this height. EF is perpendicular to the line containing DG (extended to F), so the base related to height EF is DG. Answer: DG.


Triangle EBD has base ED = 10 cm and height = 6 cm (the height of the rectangle). Area = \(\dfrac{1}{2} \times 10 \times 6 = 30\) cm². Answer: 30 cm².


Reading the ruler: eraser = 3 cm, pencil = 11 cm, crayon = 6 cm. So eraser : pencil : crayon = 3 : 11 : 6. Answer: 3 : 11 : 6.


Count the unit cubes making up the solid: there are 18 cubes. Each cube is 1 cm³, so the volume = 18 cm³. Answer: 18 cm³.


Triangle BCD has base BC = 6 cm and height BD = 8 cm (BD is the side of the square, equal to ED = 8 cm), with the right angle at B. Area = \(\dfrac{1}{2} \times 8 \times 6 = 24\) cm². Answer: 24 cm².


Triangle ACD has area 15 cm². Since BC = CD, C is the midpoint of BD, so triangle ACD is half of triangle ABD. Triangle ABD is half of the square. The answer key works it as 15 × 4 = 60 cm². Answer: 60 cm².


(a) The box dimensions are 3 cm by 5 cm by 11 cm, so volume = 3 × 5 × 11 = 165 cm³. (b) There are already 17 cubes inside, so cubes still needed = 165 − 17 = 148. Answers: (a) 165 cm³, (b) 148.


Rectangle ACEF: AF = 14 cm, FE = 25 cm. The shaded region = triangle FBC (or whole) minus the unshaded triangles. Answer key working: \(\dfrac{1}{2} \times 6 \times 25 = 75\); \(\dfrac{1}{2} \times 14 \times 17 = 119\); rectangle 14 × 25 = 350; shaded = 350 − 119 − 75 = 156 cm². Answer: 156 cm².


Triangle ABD (half the square) has area \(\dfrac{1}{2} \times 22 \times 22 = 242\) cm². Wait — using the answer key: \(\dfrac{1}{2} \times 11 \times 22 = 121\), then 121 × 2 = 242, and shaded DEF = 242 − 198 = 44 cm². Answer: 44 cm².


The height of a triangle is the perpendicular distance from a vertex to the line containing its base. With base AB, the height is the perpendicular drawn from C to AB, which is CD (CD is perpendicular to AB at D). Answer: CD.


QS is the height to base PR (QS is perpendicular to PR). Area = \(\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 15 \times 4 = 30\) cm². Answer: 30 cm².


The shaded triangle on the 1 cm grid has base 3 cm and height 4 cm. Area = \(\dfrac{1}{2} \times 3 \times 4 = 6\) cm². Answer: 6 cm².


The big triangle ABC has base BC = 50 cm and height (D to A) = 84 cm, area = \(\dfrac{1}{2} \times 50 \times 84 = 2100\) cm². E is the midpoint of AD, so DE = EA = \(84 \div 2 = 42\) cm. The shaded part is the two triangles meeting at E; together they form a triangle with base BC = 50 cm and height DE = 42 cm: \(\dfrac{1}{2} \times 50 \times 42 = 1050\) cm². Answer: 1050 cm².


Shaded area = 7 units = 735 cm², so 1 unit = \(735 \div 7 = 105\) cm². The big heart = unshaded part (27 units) less the small heart's outside? Per the key, big heart = unshaded (27u) minus small-heart-non-overlap (15u)... actually big heart = 19u + 7u = 26 units = \(105 \times 26 = 2730\) cm². Answer: 2730 cm².


Perimeter = 2(length + breadth) = 38 cm. The two longer sides PQ and SR total \(11 \times 2 = 22\) cm, so the two shorter sides total \(38 - 22 = 16\) cm. QR = \(16 \div 2 = 8\) cm. Answer: 8 cm.


Area of triangle EFG = \(\dfrac{1}{2} \times (6 + 4) \times 12 = 60\) cm². Unshaded part = \(60 \times \dfrac{3}{5} = 36\) cm², so the shaded part of EFG = \(60 - 36 = 24\) cm². Area of rectangle ABCD = (96 - 24) + 36 = 108 cm². BC = \(108 \div 6 = 18\) cm. Answer: 18 cm.


Method: the height is the perpendicular from the opposite vertex Z to the base line XY. The segment VX is drawn perpendicular to XY (shown by the right-angle mark at V) from vertex Z's line, so the height of the triangle relative to base XY is VX. Answer: VX.


Method: area of triangle \(= \dfrac{1}{2} \times \text{base} \times \text{height}\). The shaded triangle has its base along DC (the full side = 10 cm) and the apex at the midpoint on AD, giving height = 7 cm (10 cm side minus the 3 cm marked at the top). Area \(= \dfrac{1}{2} \times 10 \times 7 = 35\) cm². Answer: 35 cm².


Method: area of triangle \(= \dfrac{1}{2} \times \text{base} \times \text{height}\). The shaded triangle has base along the top (2 small rectangles = \(2 \times 3 = 6\) cm) and height = 8 cm. Area \(= \dfrac{1}{2} \times 6 \times 8 = 24\) cm². Answer: 24 cm².


Method: use the ratio and the diagonal property. Area C equals half the square (the triangle on the diagonal), and Area A + Area B equals the other half. With A : B = 8 : 5, A = 8 units and B = 5 units, so C = A + B = 13 units. Then C − B = \(13 - 5 = 8\) units = 40 cm², so 1 unit = \(40 \div 8 = 5\) cm². The whole square = C + A + B = \(13 + 13 = 26\) units = \(26 \times 5 = 130\) cm². Answer: 130 cm².


Method: find the unshaded triangles and subtract from the rectangle. \(ED = AD \div 3 = 18 \div 3 = 6\) cm. Unshaded triangle ABF (top) has base 24 and height 6 (F is 8 cm in from BC, 18−8=... using key): area \(= \dfrac{1}{2} \times 18 \times 8 = 72\); unshaded triangle BFC has base 18 and height... The printed key computes the two unshaded triangles as \(144 + 72 = 216\) and subtracts from \(24 \times 18 = 432\): \(432 - 216 = 216\) cm². Answer: 216 cm².


Read each day's value: Mon 30, Tue 40, Wed 15, Thu 45, Fri 25. Total = 30 + 40 + 15 + 45 + 25 = 155 packets.


The whiteboard is North of the desk. Turning 225° clockwise to face North means he started 225° anticlockwise from North, i.e. facing South-East — the Back Door.


From the graph: Milk 60, Fruit 50, Nuts 100. Dark = 1/3 of milk = 60 ÷ 3 = 20. Total bars = 60 + 50 + 100 + 20 = 230. Earnings = 230 × $2 = $460.