Junior geometry problems (ages 11 to 13)

Answer: D, 81 cm²

1. Let the square have side s cm. Cutting it into three identical strips side by side makes each strip s by s/3.
2. Perimeter of a strip: 2(s + s/3) = 2 × 4s/3 = 8s/3.
3. Set 8s/3 = 24, so 8s = 72 and s = 9.
4. Area of the square: 9 × 9 = 81 cm² (D).

Why this works: Naming one length (the side of the square) and writing every other length in terms of it turns a picture into a one-line equation.

Where it leads: Cutting into n identical strips gives perimeter 2s(1 + 1/n): as n grows the strips get thinner and the perimeter approaches 2s.

Answer: D, 105°

1. The ratio 2 : 3 : 7 means the angles are 2x, 3x and 7x for some x.
2. Angles in a triangle add to 180°: 2x + 3x + 7x = 12x = 180°.
3. So x = 15°.
4. The angles are 30°, 45° and 105°; check: 30 + 45 + 105 = 180. ✓
5. The largest angle is 105° (D).

Why this works: A ratio tells you the angles are multiples of one unknown part; the angle sum fixes the size of the part.

Where it leads: The largest angle is more than 90°, so this triangle is obtuse. In general a : b : c is obtuse exactly when one part is more than the sum of the other two.

Answer: 54

1. Interior and exterior angles at a vertex add to 180°, so each exterior angle is 180° − 150° = 30°.
2. The exterior angles of a convex polygon add to 360°, so there are 360 ÷ 30 = 12 sides.
3. From one vertex, diagonals go to every other vertex except itself and its two neighbours: 12 − 3 = 9 diagonals.
4. Doing this from all 12 vertices counts 12 × 9 = 108, but each diagonal is counted from both ends.
5. Number of diagonals: 108 ÷ 2 = 54.

Why this works: Exterior angles of any convex polygon add to 360°, which is the quickest way from an angle to the number of sides. The ‘count from each end, then halve’ trick avoids double counting.

Where it leads: An n-gon has n(n − 3)/2 diagonals: from each corner you can reach all but itself and its two neighbours.

Answer: 14

1. Two dots differ by a steps across and b steps up (0 to 4 each). By Pythagoras the distance is √(a² + b²).
2. Order doesn’t matter, so take 0 ≤ a ≤ b ≤ 4, not both 0.
3. b = 1: a = 0, 1 give 1, 2. b = 2: a = 0, 1, 2 give 4, 5, 8. b = 3: 9, 10, 13, 18. b = 4: 16, 17, 20, 25, 32.
4. That is 2 + 3 + 4 + 5 = 14 values of a² + b², and checking the list 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 32 shows no repeats.
5. Different squared distances mean different distances, so there are 14.

Why this works: Pythagoras turns each distance into a sum of two squares, so counting distances means counting different values of a² + b². Watch for different pairs giving the same sum.

Where it leads: Different pairs can give the same distance when a number is a sum of two squares in two ways, like 25 = 0 + 25 = 9 + 16. That first happens with a 5 by 5 gap.

Answer: 8

1. Take the cuboid as 3 wide, 4 deep and 5 high, built from 3 × 4 × 5 = 60 unit cubes.
2. A cube has no paint if it is not on any painted face.
3. The two side faces across the width are painted, so the outer layer on each side is removed: 3 − 2 = 1 cube wide remains.
4. The two faces across the depth are painted: 4 − 2 = 2 remain.
5. Only the top is painted vertically (the bottom is not): 5 − 1 = 4 layers remain.
6. Unpainted cubes: 1 × 2 × 4 = 8.

Why this works: The unpainted cubes always form a smaller cuboid; each painted face removes one layer in its own direction, so count how many layers each dimension loses.

Where it leads: For a cube painted on all faces, the unpainted inner block is (n − 2)³; the numbers painted on exactly one, two or three faces make a nice table.

Answer: C, 108°

1. Measure both hands clockwise from 12 o’clock.
2. The minute hand turns 360° in 60 minutes, 6° per minute: at 24 minutes it is at 24 × 6° = 144°.
3. The hour hand turns 30° per hour, 0.5° per minute: at 8:24 it is at 8 × 30° + 24 × 0.5° = 240° + 12° = 252°.
4. Difference: 252° − 144° = 108°.
5. 108° is less than 180°, so it is already the smaller angle: 108° (C).

Why this works: Turning each hand into an angle from 12 o’clock makes the problem a subtraction. The common slip is forgetting that the hour hand moves between the hour marks.

Where it leads: The hands overlap 11 times in 12 hours, because the minute hand gains 330° per hour on the hour hand.

Answer: 17

1. The slanted side joins (8, 0) and (0, 6): its equation is x/8 + y/6 = 1, i.e. 6x + 8y = 48.
2. Strictly inside means x ≥ 1, y ≥ 1 and 6x + 8y < 48.
3. x = 1: 8y < 42, so y = 1 to 5 (5 points). x = 2: 8y < 36, y = 1 to 4 (4).
4. x = 3: 8y < 30, y = 1 to 3 (3). x = 4: 8y < 24, y = 1, 2 (2; (4, 3) is on the edge).
5. x = 5: 8y < 18, y = 1, 2 (2). x = 6: 8y < 12, y = 1 (1). x = 7: 8y < 6, none.
6. Total: 5 + 4 + 3 + 2 + 2 + 1 = 17. (Check with Pick’s theorem: area 24 = I + 16/2 − 1 gives I = 17.)

Why this works: Counting column by column is a reliable way to count lattice points. (Pick’s theorem, Area = I + B/2 − 1, gives the same 17 as a check.)

Where it leads: Pick’s theorem gives the same count from the area and the number of boundary points: Area = I + B/2 − 1.

Answer: 70°

1. Angles in a triangle add to 180°: x + x + 20 + 2x − 20 = 4x = 180.
2. So x = 45, and the angles are 45°, 65° and 70°.
3. The largest is 70°. (Check: 45 + 65 + 70 = 180.)

Why this works: Writing every angle in terms of one unknown turns the angle sum into a single equation; always finish by working out all the angles, as the largest may not be the one with the biggest coefficient for every x.

Where it leads: For which values of x would 2x − 20 be the largest angle, and for which would x + 20 be? Comparing expressions like this is the start of inequalities.

Answer: C, 10

1. At each corner, interior + exterior = 180°. With interior = 4 × exterior: 5e = 180, so e = 36°.
2. The exterior angles of any polygon add up to 360°.
3. Number of sides = 360 ÷ 36 = 10 (C).

Why this works: Exterior angles always total 360°, whatever the number of sides, so finding one exterior angle gives the number of sides at once.

Where it leads: If the interior angle is k times the exterior angle, the polygon has 2(k + 1) sides. So the number of sides is always even when k is a whole number.

Answer: 60 cm

1. The square’s side is 36 ÷ 4 = 9 cm, so its area is 81 cm².
2. The rectangle is 3 cm wide with area 81 cm², so it is 27 cm long.
3. Its perimeter is 2 × (27 + 3) = 60 cm.

Why this works: Shapes with the same area can have very different perimeters: long thin shapes have larger perimeters.

Where it leads: Of all rectangles with a given area, the square has the smallest perimeter. Of all shapes with a given area, the circle does (the isoperimetric inequality).

Answer: 168 cm²

1. The diagonal is the hypotenuse of a right-angled triangle with sides 7 and x.
2. x² = 25² − 7² = 625 − 49 = 576, so x = 24.
3. Area = 7 × 24 = 168 cm².

Why this works: A rectangle’s diagonal, two sides and right angle form a right-angled triangle, so Pythagoras links them.

Where it leads: 7, 24, 25 is a Pythagorean triple. All triples come from m² − n², 2mn, m² + n² (here m = 4, n = 3) and their multiples.

Answer: C, 125

1. Each of the 6 faces has area 150 ÷ 6 = 25 cm².
2. So each edge is √25 = 5 cm.
3. Volume = 5³ = 125 cm³ (C).

Why this works: Working back from surface area to edge length to volume is a chain of simple steps.

Where it leads: Doubling every length of a solid multiplies its surface area by 4 and its volume by 8. That is why large animals have trouble losing heat and small ones losing it too fast.

Answer: 60 cm³

1. Let the edges be a, b, c, with ab = 12, bc = 15, ca = 20.
2. Multiplying: (ab)(bc)(ca) = (abc)² = 12 × 15 × 20 = 3600.
3. So the volume abc = √3600 = 60 cm³. (The edges are 4, 3 and 5.)

Why this works: Multiplying the three areas uses each edge twice, which gives the square of the volume directly, without finding the edges first.

Where it leads: The same trick solves xy = p, yz = q, zx = r for any numbers: xyz = √(pqr), then divide by each product to get x, y and z.

Answer: D, 100°

1. Case 1: 40° is the angle between the equal sides. The other two are equal: (180 − 40) ÷ 2 = 70° each.
2. Case 2: 40° is one of the two equal angles. Then the angles are 40°, 40° and 100°.
3. The largest angle possible is 100° (D).

Why this works: ‘Isosceles’ does not say which angle is the odd one out, so both cases must be checked.

Where it leads: Ambiguous cases like this also appear with the sine rule, where two different triangles can fit the same data.

Answer: 7

1. The direction (12, 18) simplifies to (2, 3), because HCF(12, 18) = 6.
2. So the whole-number points are (0, 0), (2, 3), (4, 6), …, (12, 18): steps of (2, 3).
3. From 0 to 6 steps gives 7 points.

Why this works: A step (a, b) along the line hits whole-number points only when the step is a whole multiple of (a/h, b/h), where h is the HCF.

Where it leads: In general the segment from (0, 0) to (m, n) contains HCF(m, n) + 1 lattice points. This is used in Pick’s theorem to count boundary points.

Answer: 36 cm

1. The original rectangle has perimeter 2 × (10 + 8) = 36 cm.
2. Cutting the corner removes 3 cm from one side and 2 cm from the other.
3. It adds two new edges of 2 cm and 3 cm (the inner sides of the notch).
4. So the perimeter is unchanged: 36 cm.

Why this works: Sliding the inner edges of the notch outwards rebuilds the original rectangle’s outline exactly, so the perimeter does not change.

Where it leads: Any ‘staircase’ shape whose edges only go right and up (then left and down) has the same perimeter as its bounding rectangle.

Answer: 15

1. (1, 1) and (7, 1) lie on the line y = 1, so the base is 7 − 1 = 6.
2. The third corner (4, 6) is 6 − 1 = 5 above that line, so the height is 5.
3. Area = ½ × 6 × 5 = 15.

Why this works: Choosing a side parallel to an axis makes the base and height easy to read off.

Where it leads: For any three points the shoelace formula gives the area directly: ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Answer: A, 100 − 25π

1. The circle touches opposite sides, so the side of the square equals the diameter: 10 cm. Square area = 100 cm².
2. Circle area = π × 5² = 25π cm².
3. Area outside the circle = 100 − 25π cm² (A), about 21.5 cm².

Why this works: ‘Shaded area = big shape minus small shape’ avoids working with the awkward curved corners directly.

Where it leads: The circle fills π/4 ≈ 78.5% of its square. In three dimensions a sphere fills π/6 ≈ 52% of its cube, and in high dimensions almost none of it.

Answer: 6

1. Joining the centre to each corner splits the hexagon into 6 equilateral triangles with side 2 cm.
2. An equilateral triangle of side 2 has height √(2² − 1²) = √3, so its area is ½ × 2 × √3 = √3.
3. Hexagon area = 6√3, so k = 6.

Why this works: Cutting a regular polygon into congruent triangles from the centre turns an awkward area into a simple one, times a count.

Where it leads: A regular n-gon with side s has area n s² / (4 tan(180°/n)). As n grows, regular polygons approach circles: Archimedes used 96 sides to pin down π.

Answer: 24

1. Angles around a point add up to 360°.
2. x + 2x + 3x + 4x + 5x = 15x = 360.
3. So x = 24.

Why this works: Turning the angle fact into one equation is all it takes once the unknowns share a single letter.

Where it leads: 1 + 2 + … + n = n(n + 1)/2. For which n can angles x, 2x, …, nx fit round a point with x a whole number of degrees?

Answer: C, 20

1. The horizontal edges: the base has length 5, and the steps on top also add up to 5 across. Total 10.
2. The vertical edges: on the left, the first column’s side and the four risers of the steps climb 1 each, 5 in total; the right-hand side has height 5. Total 10.
3. Perimeter = 10 + 10 = 20 (C), the same as a 5 by 5 square.

Why this works: For a staircase, the steps can be slid out to the edges of the bounding square without changing their total length.

Where it leads: This is the same invariant as the cut-out corner problem. It explains the famous ‘π = 4’ paradox: a staircase hugging a circle always has the square’s perimeter, so lengths do not behave like areas under limits.

Answer: 7 m

1. At the start: height = √(13² − 5²) = √144 = 12 m.
2. After sliding down 7 m the top is 5 m up, so the foot is √(13² − 5²) = 12 m from the wall.
3. The foot moves from 5 m to 12 m: 7 m.

Why this works: The ladder’s length stays fixed, so each position is a right-angled triangle with hypotenuse 13; here the two positions are mirror images.

Where it leads: The midpoint of a sliding ladder traces a quarter circle centred at the foot of the wall. Why? (It is always 6.5 m from the corner.)

Answer: 13

1. The triangle inequality: each side is less than the sum of the other two.
2. x < 7 + 10 = 17 and 10 < 7 + x, so x > 3.
3. x can be 4, 5, …, 16: that is 16 − 4 + 1 = 13 values.

Why this works: If one side were as long as the other two together, the triangle would collapse flat. The triangle inequality captures exactly when a triangle exists.

Where it leads: The triangle inequality is the basis of ‘distance’ in all of maths: any sensible way of measuring distance must satisfy it.

Answer: C, 135°

1. A parallelogram has two pairs of equal opposite angles, and neighbouring angles add to 180°.
2. If one angle is three times another they cannot be opposite (those are equal), so they are neighbours: x + 3x = 180.
3. x = 45, so the angles are 45°, 135°, 45°, 135°. The largest is 135° (C).

Why this works: Neighbouring angles in a parallelogram are co-interior angles between parallel lines, so they sum to 180°.

Where it leads: In any quadrilateral with a pair of parallel sides (a trapezium), the angles along each slanted side add to 180° for the same reason.

Answer: 18 cm²

1. Take CD (length 6) as the base.
2. E lies on AB, which is parallel to CD and 6 cm away, so the height is 6.
3. Area = ½ × 6 × 6 = 18 cm², exactly half the square.

Why this works: Moving E anywhere along AB keeps the same base and height, so the area does not depend on where E is.

Where it leads: Triangles on the same base between the same parallels have equal area. Euclid used this to prove Pythagoras’ theorem by shearing squares into rectangles.

Answer: A, It decreases by 4%

1. New length = 1.2 × old length; new width = 0.8 × old width.
2. New area = 1.2 × 0.8 × old area = 0.96 × old area.
3. So the area decreases by 4% (A).

Why this works: Percentage changes multiply, they do not add: +20% then −20% is ×1.2 ×0.8, not ×1.

Where it leads: (1 + p)(1 − p) = 1 − p²: a rise and fall by the same percentage always leaves you slightly worse off. This is why investment losses hurt more than equal gains help.

Answer: 10

1. Every triangle in the figure has the top corner as a vertex and part of the bottom side as its base.
2. Such a triangle is fixed by choosing its two slanted sides from the 5 lines that leave the top corner (2 original sides + 3 new lines).
3. Number of choices: 5 × 4 ÷ 2 = 10.

Why this works: Matching each triangle to a pair of lines turns a picture count into ‘5 choose 2’.

Where it leads: With n lines from the top corner you get C(n + 2, 2) triangles. Add a line across the triangle and the count changes in a pattern you can predict.

Answer: A, 100 − 9π

1. The four quarter circles all have radius 3, so together they have the area of one full circle of radius 3: 9π.
2. Square area = 100.
3. Area left = 100 − 9π cm² (A), about 71.7 cm².

Why this works: Rearranging pieces in your head (four quarters make a whole) is often quicker than computing each piece.

Where it leads: The four corners of any quadrilateral have angles summing to 360°, so equal-radius sectors at the corners of any quadrilateral also make one full circle.