Geometry Worksheets with Answers: PDFs & Worked Examples

239 activities • 12 worksheet topics • original guided practice

Practise selected geometry skills with six worked examples, ten original questions and step-by-step answers. Start with the topic you need, then try the questions before checking your work. The examples cover angle reasoning, areas of triangles and quadrilaterals, circles, Pythagoras, similar shapes and coordinate transformations.

This is a focused KS3 and GCSE revision selection, not a complete course or an exam-board checklist. The area-scale-factor parts are labelled as extensions. Other topics, including constructions, bearings, trigonometry, circle theorems and 3D measures, need separate practice.

How to use this lesson

1. Choose a topic and read its worked example.
2. Cover the answers and attempt the matching questions. Write a reason for each angle calculation.
3. Compare your method with the worked answers, not just the final number.
4. Correct the first wrong step, then try a related worksheet from the directory.

Use a pencil and squared paper. A calculator is useful for circle values and square roots. Keep π or a square root in the calculation until the final step. Round only when asked. All lengths within one formula must use the same unit. Unless a task explicitly asks you to measure, use the stated information rather than measuring the picture.

Topic route: angles → questions 1–2; area → 3–5; circles → 6; Pythagoras → 7–8; similarity → 9; transformations → 10.

Key facts to choose from

• Angles on a straight line total 180°. Angles around a point total 360°. Vertically opposite angles are equal.
• A triangle’s interior angles total 180°. An isosceles triangle’s angles opposite its equal sides are equal.
• When parallel lines are crossed by a transversal, corresponding angles are equal and alternate angles are equal. Check that the lines are parallel before using these facts.
• Triangle area = ½ × base × perpendicular height.
• Parallelogram area = base × perpendicular height.
• Trapezium area = ½ × (sum of the two parallel sides) × perpendicular height. Here “trapezium” uses UK terminology.
• Circle diameter d = 2r, circumference C = πd = 2πr, and area A = πr².
• In a right-angled triangle, a² + b² = c², where c is the hypotenuse opposite the right angle.
• Similar shapes have equal corresponding angles and proportional corresponding lengths. A length scale factor k also multiplies the perimeter by k. Extension: the area is multiplied by k².
• Translation: move every point the same distance in the same direction. Reflection: use a specified mirror line. Rotation: state the centre, angle and direction. Enlargement: state the centre and scale factor.

Six worked examples

Example 1: give a reason for each angle

Triangle ABC has AB = AC and angle BAC = 44°. The line BC is extended through B to D, so D, B and C lie on one straight line in that order. Find angle ABC and the exterior angle ABD.

The two base angles are equal because AB = AC. Their total is 180° − 44° = 136°, so angle ABC = 136° ÷ 2 = 68°.

Angles ABC and ABD lie on a straight line. Therefore angle ABD = 180° − 68° = 112°.

Check: 44° + 68° + 68° = 180°. The exterior angle is also 44° + 68° = 112°.

Watch for: subtracting 44° from 180° finds the total of two equal angles, not one angle.

Example 2: use the perpendicular height

1. A triangle has a base of 12 cm and a perpendicular height of 5 cm.

Area = ½ × 12 × 5 = 30 cm².

2. A trapezium has parallel sides of 9 cm and 15 cm. Their perpendicular separation is 6 cm.

Area = ½ × (9 + 15) × 6 = ½ × 24 × 6 = 72 cm².

Why the formulas make sense: two copies of the triangle can form a parallelogram with the same base and height. Two copies of the trapezium can form a parallelogram with base 9 + 15 = 24 cm and height 6 cm. In each case, one original shape has half the new area.

Watch for: a sloping side is not necessarily the perpendicular height. Area needs square units.

Example 3: distinguish circumference from area

A circular sign has diameter 12 cm. Find its circumference and area, first exactly and then to one decimal place.

Circumference measures the boundary: C = πd = 12π cm ≈ 37.7 cm.

For area, first find r = 12 ÷ 2 = 6 cm. Then A = π × 6² = 36π cm² ≈ 113.1 cm².

Watch for: using 12 as the radius would give four times the correct area. Circumference is a length; area measures the surface enclosed.

Example 4: find a shorter side with Pythagoras

A right-angled triangle has a hypotenuse of 17 cm and one shorter side of 8 cm. Find the remaining side x.

The hypotenuse is opposite the right angle, so x² + 8² = 17².

x² = 289 − 64 = 225, giving x = 15 cm. Take the positive square root because a length is positive.

Check: 8² + 15² = 64 + 225 = 289 = 17². The answer is shorter than the hypotenuse.

Watch for: subtract the squared lengths, not the lengths. Pythagoras needs a right-angled triangle.

Example 5: match corresponding sides in similar shapes

A rectangle is 8 cm long and 5 cm wide. A similar rectangle has a corresponding length of 12 cm. Find its width and perimeter.

The length scale factor, small to large, is k = 12 ÷ 8 = 1.5.

Width = 5 × 1.5 = 7.5 cm.

Perimeter = 2 × (12 + 7.5) = 39 cm.

Check: the smaller perimeter is 2 × (8 + 5) = 26 cm, and 26 × 1.5 = 39 cm.

Extension: the smaller area is 8 × 5 = 40 cm². The larger area is 12 × 7.5 = 90 cm², which is 40 × 1.5². Area scales by 2.25 here, not 1.5.

Watch for: adding 4 cm to both dimensions would produce 12 cm by 9 cm, which is not similar to the original rectangle.

Example 6: translate every vertex

Triangle ABC has vertices A(1, 1), B(4, 1) and C(1, 3). Translate it 2 units left and 3 units up.

Apply the rule (x, y) → (x − 2, y + 3) to each point:

• A′ = (1 − 2, 1 + 3) = (−1, 4)
• B′ = (4 − 2, 1 + 3) = (2, 4)
• C′ = (1 − 2, 3 + 3) = (−1, 6)

The image has the same side lengths, angles and orientation as the original. The translation vector is −2 horizontally and +3 vertically.

Useful rules for question 10: reflection in the y-axis changes (x, y) to (−x, y); a 90° clockwise rotation about the origin changes (x, y) to (y, −x); enlargement by scale factor 2 about the origin changes (x, y) to (2x, 2y). These rotation and enlargement rules depend on the centre being the origin.

Watch for: x is the horizontal coordinate. Moving left makes x smaller even when it crosses zero.

Ten practice questions

Show your working. Give angles in degrees, lengths in the stated length unit and areas in square units. For questions 6 and 7, give both an exact answer and an answer to one decimal place. The parts of question 10 are separate transformations of the original triangle, not a sequence.

1. Triangle ABC has AB = AC and angle BAC = 52°. BC is extended through B to D, with D, B and C collinear in that order. Find angles ABC and ABD. Give a reason for each step.
   
   Answer 1: show the method

   AB = AC, so the base angles are equal. Angle ABC = (180° − 52°) ÷ 2 = 64°. Angle ABD = 180° − 64° = 116°, using angles on a straight line. Check: 52 + 64 + 64 = 180 and 52 + 64 = 116.

2. Two parallel lines are crossed by a transversal. One angle is 67°. At the other intersection, x is the corresponding angle. Angle y is adjacent to x on a straight line. Find x and y, explaining the angle facts used.
   
   Answer 2: show the method

   x = 67°, because corresponding angles are equal when the lines are parallel. y = 113°, because x + y = 180°, so y = 180° − 67°.

3. A triangle has a base of 9.6 cm and a perpendicular height of 4.5 cm. Find its area.
   
   Answer 3: show the method

   Area = ½ × 9.6 × 4.5 = 21.6 cm².

4. A parallelogram has a base of 8 cm, perpendicular height 3.5 cm and sloping side 4 cm. Find its area. Explain why one of the given lengths is not needed.
   
   Answer 4: show the method

   Area = 8 × 3.5 = 28 cm². The 4 cm sloping side is not the perpendicular height, so it is not used in this area calculation.

5. A trapezium has parallel sides of 11 cm and 17 cm and perpendicular height 4.5 cm. Find its area.
   
   Answer 5: show the method

   Area = ½ × (11 + 17) × 4.5 = 14 × 4.5 = 63 cm².

6. A circular badge has diameter 9.2 cm. Find its circumference and area, exactly in terms of π and to one decimal place.
   
   Answer 6: show the method

   r = 9.2 ÷ 2 = 4.6 cm. Circumference = 9.2π cm ≈ 28.9 cm. Area = π × 4.6² = 21.16π cm² ≈ 66.5 cm². Use the calculator’s π value until the final rounding step.

7. The two sides meeting at the right angle of a triangle are 7 cm and 13 cm. Find the hypotenuse, exactly and to one decimal place.
   
   Answer 7: show the method

   c² = 7² + 13² = 49 + 169 = 218. Therefore c = √218 cm ≈ 14.8 cm. The hypotenuse is longer than either leg.

8. A rectangle has diagonal 26 cm and width 10 cm. Find its length and area. Explain why Pythagoras can be used.
   
   Answer 8: show the method

   A diagonal divides a rectangle into two right-angled triangles. Let the length be l. Then l² + 10² = 26², so l² = 676 − 100 = 576. Thus l = 24 cm and area = 24 × 10 = 240 cm².

9. Two triangles are similar. The smaller has side lengths 7.5 cm, 10 cm and 12.5 cm. The side corresponding to 7.5 cm is 12 cm in the larger triangle. Find the other two larger side lengths and the larger perimeter. Extension: the smaller triangle has area 37.5 cm²; find the larger area.
   
   Answer 9: show the method

   k = 12 ÷ 7.5 = 1.6. The remaining sides are 10 × 1.6 = 16 cm and 12.5 × 1.6 = 20 cm. Perimeter = 12 + 16 + 20 = 48 cm. Extension: area factor = 1.6² = 2.56, so area = 37.5 × 2.56 = 96 cm². The original sides form a right-angled triangle, providing an independent check: ½ × 12 × 16 = 96 cm².

10. Triangle PQR has P(2, 1), Q(5, 1) and R(2, 3). Find the image coordinates for each separate transformation: (a) reflection in the y-axis; (b) 90° clockwise rotation about the origin; (c) translation 3 units left and 2 units up; (d) enlargement by scale factor 2, centre the origin.
    
    Answer 10: show the method

    Apply each rule to all three original points:

    • 1. (x, y) → (−x, y): P′(−2, 1), Q′(−5, 1), R′(−2, 3)
    • 2. (x, y) → (y, −x): P′(1, −2), Q′(1, −5), R′(3, −2)
    • 3. (x, y) → (x − 3, y + 2): P′(−1, 3), Q′(2, 3), R′(−1, 5)
    • 4. (x, y) → (2x, 2y): P′(4, 2), Q′(10, 2), R′(4, 6)

Before you move on

Check whether you can explain each of these mistakes:

• Using a sloping edge as the height in an area formula
• Using the diameter where a circle’s radius is needed
• Adding squares when finding a shorter side in a right triangle
• Multiplying an area by the length scale factor instead of its square
• Giving a rotation without its centre or direction

If one is unclear, revisit that example and try a matching topic worksheet below.

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