Work through the levels in order. Type your answer and press Check — open the worked solution if you get stuck. Answers like -6, 3/4 or 0.25 are typed exactly as shown. A ⭐ Bonus Challenge at the end is extra credit — skip it if the rest was hard work.
Level 1 · Warm-up
1 ABCD is a rectangle with AB = 11 cm and BC = 8 cm. P is a point on DC. What is the total area of the two shaded triangles APD and BPC?
Show worked solution
Triangle APB has base AB = 11 and height BC = 8, because P is on the opposite side, so its area is ½ × 11 × 8 = 44 cm2, half of the rectangle. The shaded triangles are the other half: 88 − 44 = 44 cm2, wherever P is. 88 cm2 is the whole rectangle; 22 cm2 halves twice; "It depends" misses that the height of APB never changes.
2 A tilted square sits inside a square of side 8 cm. Its corners cut each side of the big square into 2 cm and 6 cm. Find the area of one corner triangle and the area of the tilted square.
Show worked solution
Each corner triangle is ½ × 2 × 6 = 6 cm2. Tilted square = 8 × 8 − 4 × 6 = 64 − 24 = 40 cm2. Check: 22 + 62 = 4 + 36 = 40.
3 A rectangular garden is 14 m by 10 m. Two straight paths, each 2 m wide, cross it parallel to its sides. Find the area of lawn (the green part), in m2.
Show worked solution
Slide the lawn pieces together and the paths vanish: the lawn becomes one rectangle (14 − 2) by (10 − 2), that is 12 by 8, so its area is 96 m2. Taking both paths from 140 as 140 − 28 − 20 = 92 is wrong, because the square where the paths cross is taken away twice.
Level 2 · Core
4 ABCD is a square. M is the midpoint of BC and N is the midpoint of CD. What fraction of the square is the shaded triangle AMN?
Show worked solution
Take the side as 4, so the square is 16. Corner triangles: ABM = ½ × 4 × 2 = 4, MCN = ½ × 2 × 2 = 2, NDA = ½ × 2 × 4 = 4, total 10. AMN = 16 − 10 = 6, which is 6/16 = \(\dfrac{3}{8}\) of the square. \(\dfrac{1}{4}\) forgets the ½ on triangle MCN; \(\dfrac{1}{2}\) uses the half-of-a-rectangle rule, but MN is not a side of the square; \(\dfrac{5}{8}\) is the unshaded part.
5 Triangle ABC has area 72 cm2. D is on AB with AD : DB = 1 : 2, and E is the midpoint of AC. Find the area of the shaded triangle ADE and of the quadrilateral DBCE.
Show worked solution
Triangles ADC and ABC share apex C with bases on AB, and AD is 1/3 of AB, so ADC = 72 ÷ 3 = 24 cm2. Triangles ADE and ADC share apex D, and AE is half of AC, so ADE = ½ × 24 = 12 cm2. DBCE = 72 − 12 = 60 cm2.
6 The grid is made of 1 cm squares. Find the area of the shaded triangle, in cm2.
Show worked solution
The surrounding rectangle is 7 by 5 = 35 cm2. The three corner triangles are ½ × 7 × 1 = 3.5, ½ × 4 × 5 = 10 and ½ × 3 × 4 = 6, total 19.5. The shaded triangle is 35 − 19.5 = 15.5 cm2.
Level 3 · Challenge
7 Three statements:
1 Joining the midpoints of the sides of any rectangle makes a shape with half the rectangle's area.
2 Two triangles with equal bases and equal areas must be congruent.
3 If P is any point inside a parallelogram, the two triangles joining P to a pair of opposite sides together make half the parallelogram.
Which statements are correct?
1 Joining the midpoints of the sides of any rectangle makes a shape with half the rectangle's area.
2 Two triangles with equal bases and equal areas must be congruent.
3 If P is any point inside a parallelogram, the two triangles joining P to a pair of opposite sides together make half the parallelogram.
Which statements are correct?
Show worked solution
For a rectangle a by b, each corner triangle is ½ × a/2 × b/2 = ab/8, and four of them make ab/2, so the inside shape is the other half (1 ✓). Equal base and equal height fix the area but not the shape: slide the apex along a parallel line (2 ✗). The two triangles share the base length and their heights add to the parallelogram's height, so together they make ½ × base × height (3 ✓). 1 and 3.
8 ABCD is a square of side 8 cm. Quarter circles of radius 8 cm are drawn inside it with centres A and C. Find the exact area of the shaded region where they overlap.
Show worked solution
BD cuts the overlap into two equal parts. Each part is a quarter circle minus triangle ABD: ¼ × π × 82 − ½ × 8 × 8 = 16π − 32. Two parts: 32π − 64 cm2. Sense check: π is a little over 3, so this is a little over 32, and less than the square's 64.
9 ABCD is a square of side 18 cm. E is the midpoint of AB, and CE meets the diagonal BD at F. Find the area of the shaded quadrilateral AEFD, in cm2.
Show worked solution
BE ∥ DC and BE = ½ DC, so triangles FBE and FDC are similar with ratio 1 : 2, and EF is one third of EC. Triangle BEC = ½ × 9 × 18 = 81, so triangle BEF = ⅓ × 81 = 27 (same apex B, base one third). AEFD = triangle ABD − triangle BEF = 162 − 27 = 135 cm2.