Geometry puzzles with algebra

The Maths Habit Handout to print

How it works

  1. Work in your team. Each puzzle has its own time.
  2. Each puzzle is worth 5 marks.
  3. Stuck? Ask for a clue. Each clue costs 1 mark.
  4. The clues come in order: clue 1 first.
  5. When the time is up, every clue goes up for one minute.
  6. Solve it in that minute for 1 mark.
  7. Then we go through the solution together.

Teams

teams

Puzzle 14 minutes · 5 marks

The rectangle and the square have the same area. Find the value of x.

(x + 9) cm(x − 3) cm(x + 1) cm(x + 1) cm6464
  1. Clue 1Write the area of the rectangle in terms of x.
  2. Clue 2Write the area of the square in terms of x: (x + 1)².
  3. Clue 3The areas are equal: (x + 9)(x − 3) = (x + 1)².
  4. Clue 4Expand both sides. The x² terms cancel and leave 6x − 27 = 2x + 1.
  1. Write each area in terms of x.(x + 9)(x − 3) and (x + 1)²
  2. The areas are equal.(x + 9)(x − 3) = (x + 1)²
  3. Expand both sides. The x² cancels.x² + 6x − 27 = x² + 2x + 1
  4. Collect the x terms.4x = 28, so x = 7
Clues
4:004 minutes

Puzzle 1: how many clues did each team take?

Puzzle 24 minutes · 5 marks

Find the perimeter of the triangle.

x cm(x + 7) cm(x + 8) cm5, 12, 13
  1. Clue 1The longest side is opposite the right angle: x + 8.
  2. Clue 2Use Pythagoras: x² + (x + 7)² = (x + 8)².
  3. Clue 3Expand and collect on one side: x² − 2x − 15 = 0.
  4. Clue 4Factorise: (x − 5)(x + 3) = 0. A length cannot be negative. Now add the three sides.
  1. The longest side, x + 8, is opposite the right angle.sides x, x + 7, x + 8
  2. Pythagoras.x² + (x + 7)² = (x + 8)²
  3. Expand and collect on one side.x² − 2x − 15 = 0
  4. Factorise. A length cannot be −3.(x − 5)(x + 3) = 0, so x = 5
  5. The sides are 5, 12 and 13.Perimeter = 30 cm
Clues
4:004 minutes

Puzzle 2: how many clues did each team take?

Puzzle 35 minutes · 5 marks

Triangle ABC has a right angle at B. AB : BC = 3 : 4. The area of the triangle is 96 cm². Find its perimeter.

ABC3k4k5k
  1. Clue 1Use the ratio to give the sides a letter: AB = 3k and BC = 4k.
  2. Clue 2The area is ½ × AB × BC, so ½ × 3k × 4k = 96.
  3. Clue 36k² = 96, so k = 4: AB = 12 and BC = 16.
  4. Clue 4Find AC with Pythagoras, then add the three sides.
  1. Give the sides a letter from the ratio.AB = 3k, BC = 4k
  2. Area of a triangle: half the base times the height.½ × 3k × 4k = 96, so 6k² = 96
  3. Divide by 6. A length is positive.k² = 16, so k = 4: AB = 12, BC = 16
  4. Pythagoras for AC, then add the sides.AC = √(12² + 16²) = 20, perimeter = 48 cm
Clues
5:005 minutes

Puzzle 3: how many clues did each team take?

Puzzle 45 minutes · 5 marks

A square of side x cm is cut from the corner of a square of side (x + 4) cm. The shaded shape has an area of 56 cm². Find its perimeter.

(x + 4) cmx cmx cm
  1. Clue 1The shaded area is the big square take away the small square.
  2. Clue 2(x + 4)² − x² = 56. The x² terms cancel.
  3. Clue 38x + 16 = 56, so x = 5 and the big square has sides of 9.
  4. Clue 4Slide the two inside edges out to the corner: the perimeter is the same as the big square's.
  1. Write the shaded area in terms of x.(x + 4)² − x²
  2. The area is 56.(x + 4)² − x² = 56
  3. Expand. The x² cancels.8x + 16 = 56, so x = 5
  4. The two inside edges slide out to the corner, so the perimeter is the big square's.4 × 9 = 36 cm
Clues
5:005 minutes

Puzzle 4: how many clues did each team take?

Puzzle 55 minutes · 5 marks

The cylinder has radius 2x and height 3x. The cone has radius 3x. They have the same volume. Find the height of the cone in terms of x. The volume of a cone is 13πr²h.

2x3x3x4x
  1. Clue 1Volume of a cylinder = πr²h. Square all of the radius: (2x)² = 4x².
  2. Clue 2The cylinder's volume is 12πx³.
  3. Clue 3Call the cone's height h. Its volume is 13π(3x)²h = 3πx²h.
  4. Clue 4Set 3πx²h = 12πx³ and divide both sides by 3πx².
  1. Write both volumes. Call the cone's height h. Square all of 2x and 3x.cylinder 12πx³, cone 3πx²h
  2. The volumes are equal.3πx²h = 12πx³
  3. Divide both sides by 3πx².h = 4x
Clues
5:005 minutes

Puzzle 5: how many clues did each team take?

Puzzle 65 minutes · 5 marks

The rectangle has an area of 40 cm². A semicircle is cut from its bottom edge. Find the shaded area in terms of π.

2x cm(x + 1) cm4
  1. Clue 1Write the area of the rectangle in terms of x: 2x(x + 1).
  2. Clue 2The area is 40: 2x(x + 1) = 40, so x² + x − 20 = 0.
  3. Clue 3Factorise: (x + 5)(x − 4) = 0, so x = 4. The semicircle's radius is half of 8.
  4. Clue 4Shaded area = the rectangle take away the semicircle: 40 − ½ × π × 4².
  1. Write the rectangle's area in terms of x.2x(x + 1)
  2. The area is 40.2x(x + 1) = 40, so x² + x − 20 = 0
  3. Factorise. A length is positive.(x + 5)(x − 4) = 0, so x = 4
  4. The semicircle has radius 4. Take its area from the rectangle's.40 − ½ × π × 4² = 40 − 8π cm²
Clues
5:005 minutes

Puzzle 6: how many clues did each team take?

Puzzle 76 minutes · 5 marks

ABCD is an isosceles trapezium with AB : BC : CD = 1 : 2 : 3 and a perimeter of 48 cm. Find its area as a surd.

DCBAk3k2k2k
  1. Clue 1Use the ratio: AB = k, BC = 2k, CD = 3k. It is isosceles, so AD = 2k too.
  2. Clue 2The perimeter is 8k = 48, so k = 6.
  3. Clue 3Drop a height from A to CD. It lands 6 cm from D: half of 18 − 6.
  4. Clue 4Height² = 12² − 6². Then the area is ½(a + b)h.
  1. The ratio gives the letters. Isosceles, so AD = BC.AB = k, BC = AD = 2k, CD = 3k
  2. The perimeter is 48.8k = 48, so k = 6
  3. Each end sticks out by 6. Pythagoras for the height.h² = 12² − 6² = 108, so h = 6√3
  4. Area of a trapezium.½ × (6 + 18) × 6√3 = 72√3 cm²
Clues
6:006 minutes

Puzzle 7: how many clues did each team take?

Puzzle 86 minutes · 5 marks

Find the length of BC. Give your answer as a surd.

60°ABC(x + 1) cm(x² − 1) cm
  1. Clue 1AB is next to the 60° angle and AC is the hypotenuse: use cos 60° = ½.
  2. Clue 2x + 1x² − 1 = ½, so x² − 1 = 2(x + 1).
  3. Clue 3x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0. x = −1 makes the sides 0.
  4. Clue 4AB = 4 and AC = 8. Use Pythagoras for BC and simplify the surd.
  1. AB is next to the 60°, AC is the hypotenuse.cos 60° = x + 1x² − 1
  2. cos 60° = ½.x + 1x² − 1 = ½, so x² − 1 = 2x + 2
  3. Collect and factorise. x = −1 makes the sides 0.x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0, x = 3
  4. AB = 4 and AC = 8. Pythagoras, then simplify the surd.BC = √(8² − 4²) = √48 = 4√3 cm
Clues
6:006 minutes

Puzzle 8: how many clues did each team take?

Puzzle 96 minutes · 5 marks

AB is the diameter of the semicircle and C is on the arc. Find the shaded area. Give your answer in terms of π.

ABCx cm(x + 2) cm10 cm
  1. Clue 1C is on the arc, so the angle at C is an angle in a semicircle: 90°.
  2. Clue 2Use Pythagoras: x² + (x + 2)² = 10², so x² + 2x − 48 = 0.
  3. Clue 3Factorise: (x + 8)(x − 6) = 0, so x = 6. The triangle's sides are 6 and 8.
  4. Clue 4Shaded area = the semicircle take away the triangle: ½ × π × 5² − ½ × 6 × 8.
  1. C is on the arc, so angle ACB is an angle in a semicircle: 90°.sides x, x + 2, 10
  2. Pythagoras.x² + (x + 2)² = 10², so x² + 2x − 48 = 0
  3. Factorise. A length is positive.(x + 8)(x − 6) = 0, so x = 6: the sides are 6 and 8
  4. The semicircle has radius 5. Take the triangle from it.½ × π × 5² − ½ × 6 × 8 = 252π − 24 cm²
Clues
6:006 minutes

Puzzle 9: how many clues did each team take?

Puzzle 107 minutes · 5 marks

Find the perimeter of the triangle. Give your answer in the form a + b√3.

x cm(x + 2) cm2x cm
  1. Clue 1The longest side is 2x. Use Pythagoras.
  2. Clue 2x² + (x + 2)² = (2x)² gives x² − 2x − 2 = 0.
  3. Clue 3It does not factorise. Complete the square: (x − 1)² = 3.
  4. Clue 4x = 1 + √3, as a length is positive. The perimeter is x + (x + 2) + 2x = 4x + 2.
  1. The longest side is 2x.sides x, x + 2, 2x
  2. Pythagoras.x² + (x + 2)² = (2x)²
  3. Expand and collect: 2x² + 4x + 4 = 4x².x² − 2x − 2 = 0
  4. Complete the square: (x − 1)² = 3. A length is positive.x = 1 + √3
  5. Add the sides: 4x + 2.Perimeter = 6 + 4√3 cm
Clues
7:007 minutes

Puzzle 10: how many clues did each team take?

Final scores

More puzzlesAnswer sheet

  1. The two areas are equal. Find x.

    x = 10
    (x + 8) cm(x − 2) cm(x + 2) cm

  2. Find the value of x.

    x = 3
    x cm(x + 1) cm(x + 2) cm

  3. The area of the rectangle is 24 cm². Find its perimeter.

    Perimeter = 22 cm
    (x + 3) cm(x − 2) cm24 cm²
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