Geometry and proof · Ages 14–15

The perpendicular bisector's property worksheet

Part of: Centres of a triangle

In the print window, choose “Save as PDF” to download it. The answers print on the last page.

Geometry and proof · Ages 14–15

Numsprite
numsprite.com

Name:Date:Score:／48

Write each answer in the space. For word problems, show your working and write a sentence answer.

1.P lies on the perpendicular bisector of the segment AB, and PA = 16 cm. PB =
Answer:
2.In the picture, PM is the perpendicular bisector of AB. x =
Answer:
3.P lies on the perpendicular bisector of the segment AB, and PA = 9 cm. PB =
Answer:
4.P lies on the perpendicular bisector of the segment AB, and PA = 13 cm. PB =
Answer:
5.PM is the perpendicular bisector of the segment AB, M on AB, with AM = 16 cm and PM = 12 cm. PA = cm, PB = cm
Fill in each blank
Answer:
6.In the picture, PM is the perpendicular bisector of AB. x =
Answer:
7.PM is the perpendicular bisector of the segment AB, M on AB, with AM = 21 cm and PM = 72 cm. PA = cm, PB = cm
Fill in each blank
Answer:
8.P is on the perpendicular bisector of the segment AB, with PA = (2x + 1) cm and PB = (4x − 11) cm. x =
Answer:
9.P is on the perpendicular bisector of the segment AB, with PA = (2x + 3) cm and PB = (4x − 1) cm. x =
Answer:

10.In the picture, PM is the perpendicular bisector of AB. So PA = PB ( ? ). Which reason goes in the gap?
- A. converse of ⊥ bisector theorem
- B. ∠ bisector theorem
- C. prop. of isos. △
- D. ⊥ bisector theorem
Choose the right answer
Answer:
11.A point P is PA = 7 cm and PB = 9 cm from the ends of the segment AB. P lies on the perpendicular bisector of AB.
True or false?
Answer:
12.A point P is PA = 3 cm and PB = 5 cm from the ends of the segment AB. P lies on the perpendicular bisector of AB.
True or false?
Answer:

13.A bus stop P is on the perpendicular bisector of the segment AB between two towers A and B. ∠PAB = 44°. What is ∠APB?
Working:
Answer:
14.Two basketball hoops A and B stand 18 m apart. A child P stands on the perpendicular bisector of AB, 15 m from each hoop. AB's middle M is m from A, and P is m from M.
Fill in each blank
Working:
Answer:
15.A postbox P is equally far from two houses A and B, with PA = (2x + 16) m and PB = (4x − 6) m. What is x?
Working:
Answer:
16.Two villages A and B are 258 m apart. A well P is dug on the perpendicular bisector of AB, 172 m from AB's middle M. AM = m, and the well is m from each village.
Fill in each blank
Working:
Answer:
17.Two basketball hoops A and B stand 42 m apart. A child P stands on the perpendicular bisector of AB, 75 m from each hoop. AB's middle M is m from A, and P is m from M.
Fill in each blank
Working:
Answer:
18.A postbox P is equally far from two houses A and B, with PA = (2x + 11) m and PB = (3x − 7) m. What is x?
Working:
Answer:

Show the answers and working

The perpendicular bisector's property worksheet: Answers and working

1. 16 cm
A point on the perpendicular bisector is equally far from both ends: PB = PA = 16 cm (⊥ bisector theorem).
2. 13 cm
P is on AB's perpendicular bisector, so it is equally far from A and B: x = PA = 13 (⊥ bisector theorem).
3. 9 cm
A point on the perpendicular bisector is equally far from both ends: PB = PA = 9 cm (⊥ bisector theorem).
4. 13 cm
A point on the perpendicular bisector is equally far from both ends: PB = PA = 13 cm (⊥ bisector theorem).
5. 20, 20
PA = 16² + 12² = 20 cm (Pyth. thm.); PB = PA (⊥ bisector theorem).
6. 35°
PA = PB (⊥ bisector theorem), so △PAB is isosceles: x = ∠PAB = 35° (base ∠s, isos. △).
7. 75, 75
PA = 21² + 72² = 75 cm (Pyth. thm.); PB = PA (⊥ bisector theorem).
8. 6
PA = PB (⊥ bisector theorem): 2x + 1 = 4x − 11, so x = 6. Check: PA = PB = 13 cm.
9. 2
PA = PB (⊥ bisector theorem): 2x + 3 = 4x − 1, so x = 2. Check: PA = PB = 7 cm.

10. D. ⊥ bisector theorem
A point on the perpendicular bisector is equally far from both ends: ⊥ bisector theorem.
11. False
Only a point equally far from both ends lies on the perpendicular bisector (converse of ⊥ bisector theorem).
12. False
Only a point equally far from both ends lies on the perpendicular bisector (converse of ⊥ bisector theorem).

Marking: 2 for the working, 1 for the answer, 1 for the sentence.

13. 92°
PA = PB (⊥ bisector theorem), so ∠PBA = 44° (base ∠s, isos. △); ∠APB = 180° − 2 × 44° = 92°.
14. 9, 12
AM = 18 ÷ 2 = 9 m; PM ⊥ AB, so PM = 15² − 9² = 12 m (Pyth. thm.).
15. 11
PA = PB: 2x + 16 = 4x − 6, x = 11, each 38 m; P is on AB's perpendicular bisector (converse of ⊥ bisector theorem).
16. 129, 215
AM = 258 ÷ 2 = 129 m; PA = 129² + 172² = 215 m (Pyth. thm.), and PB = PA (⊥ bisector theorem).
17. 21, 72
AM = 42 ÷ 2 = 21 m; PM ⊥ AB, so PM = 75² − 21² = 72 m (Pyth. thm.).
18. 18
PA = PB: 2x + 11 = 3x − 7, x = 18, each 47 m; P is on AB's perpendicular bisector (converse of ⊥ bisector theorem).

Numsprite · numsprite.com/en/worksheets/centres-of-a-triangle/perpendicular-bisector-theorem