This set has twelve original questions, ordered from easier to harder, covering all five lessons in constructions and loci. Questions 1 to 4 are warm-ups, 5 to 8 build locus thinking, and 9 to 12 combine conditions and checking.
Attempt each question on paper with a ruler and compasses where a drawing helps, then open the answer. Write the reason for each step. Mark the ones you got wrong and use the routing list at the end.
Questions
1. AB = 12 cm. You construct its perpendicular bisector. How far is the midpoint from A, and what is the smallest whole-number compass radius (in cm) you can use?
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The midpoint is 12 ÷ 2 = 6 cm from A. The radius must be more than 6 cm, so the smallest whole number is 7 cm.
2. Angle ABC = 108° is bisected. Find each of the two equal angles.
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108° ÷ 2 = 54° each.
3. Point X lies on the bisector of angle PQR and is 7.5 cm from arm QP, measured at right angles. How far is X from arm QR?
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Every point on an angle bisector is equally far from both arms, so X is 7.5 cm from QR.
4. Describe the locus of points 6 cm from a fixed point O. Find its length and the area it encloses, each to 1 decimal place.
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It is a circle of radius 6 cm. Circumference = 2 × π × 6 = 12π ≈ 37.7 cm. Area = π × 6² = 36π ≈ 113.1 cm².
5. P is on the perpendicular bisector of a segment of length 16 cm, and P is 8 cm from the midpoint. Find the distance from P to one end of the segment, to 1 decimal place.
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Half the segment is 8 cm. Distance² = 8² + 8² = 128, so the distance is √128 ≈ 11.3 cm. Check: 11.3² ≈ 127.7.
6. Describe the locus of points exactly 3 cm from a line segment of length 10 cm, then find its total length and the area of the region within 3 cm of the segment. Give each to 1 decimal place.
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The locus is two parallel segments of 10 cm joined by semicircles of radius 3 cm at each end.
Length = 10 + 10 + 2 × π × 3 = 20 + 6π ≈ 20 + 18.85 = 38.8 cm.
Area = rectangle 10 × 6 = 60, plus a circle of radius 3: 9π ≈ 28.27. Total ≈ 88.3 cm².
Check: 6π = 18.8496 and 20 + 18.8496 = 38.8496, which rounds to 38.8.
7. In rectangle ABCD, AB = 16 cm and AD = 9 cm. Point P is equidistant from A and B and exactly 10 cm from A. Find how far P is from AB and say whether it is inside the rectangle.
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P lies on the perpendicular bisector of AB, which meets AB at M with AM = 8 cm. MP² = 10² − 8² = 100 − 64 = 36, so MP = 6 cm. Since 6 < 9, P is inside the rectangle, 6 cm from AB.
Check: 8² + 6² = 64 + 36 = 100.
8. Lines BA and BC meet at 60°. Point P is equidistant from both lines and 4 cm from B. How far is P from line BA?
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P lies on the angle bisector, so the angle between BP and BA is 30°. The perpendicular distance from P to BA is BP × sin 30° = 4 × 0.5 = 2 cm.
Check: a 30° right-angled triangle has its shortest side equal to half its hypotenuse, so 2 is half of 4.
9. Towers A and B are 10 km apart. A signal reaches points within 6 km of each tower. Find the length along AB of the points that receive both signals, and the greatest width of the overlap at right angles to AB, to 1 decimal place.
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Tower A reaches to 6 km along AB and tower B reaches back to 10 − 6 = 4 km from A. The overlap runs from 4 km to 6 km, so it is 2 km long along AB.
At the midpoint, 5 km from A, the height satisfies h² = 6² − 5² = 36 − 25 = 11, so h = √11 ≈ 3.317. The overlap extends above and below, so the width is 2 × 3.317 ≈ 6.6 km.
10. Is there a point that is exactly 2 cm from A and exactly 2 cm from B when AB = 5 cm? Explain.
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No. Circles of radius 2 cm about A and B reach at most 2 + 2 = 4 cm along AB, but the centres are 5 cm apart, so the circles never meet.
11. A student constructs the bisector of an 80° angle and measures 38° and 42°. What went wrong, and what should the two angles measure?
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The bisector should give 80° ÷ 2 = 40° on each side. A gap of 2° means the two equal arcs were probably drawn with different radii, or a compass point slipped. Redraw both arcs with one unchanged radius.
12. A rectangular garden ABCD has AB = 12 m and AD = 8 m. A sprinkler at P must be equidistant from A and B and exactly 10 m from A. Find P, then say whether it lies inside the garden.
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P is on the perpendicular bisector of AB, which meets AB at M with AM = 6 m. MP² = 10² − 6² = 100 − 36 = 64, so MP = 8 m. The garden is 8 m wide, so P is on side CD, at its midpoint, 8 m from AB.
Check: PA = √(6² + 8²) = √100 = 10, and PB = 10 by symmetry.
If you got these wrong
- Questions 1, 5, 7, 12: the perpendicular bisector or its Pythagoras step went wrong. Return to constructing a perpendicular bisector.
- Questions 2, 3, 8, 11: the angle bisector was used incorrectly. Return to constructing an angle bisector.
- Questions 4, 6: the locus shape or its length and area were wrong. Return to describing a locus using distance conditions.
- Questions 9, 10: two conditions were not combined properly. Return to combining two loci.
- Questions 7, 11, 12: the answer was not tested against every condition. Return to checking a construction.
Record each slip in the mistake log and retest queue and retry a fresh question after a few days. The non-calculator working trainer can check the square roots and π steps.
If the same mistake keeps appearing, online one-to-one Mathematics tuition lets an experienced teacher see your drawing and working directly.