Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.
Step-by-step solution
To find: The circumcircle and the lengths OA, OB, OC
Idea: O is the meeting point of the perpendicular bisectors of the sides, so it is equidistant from A, B and C: OA = OB = OC = the circumradius. Since CA = CB, the triangle is isosceles and O lies on its axis of symmetry, the perpendicular bisector of AB through C.
- Draw the triangle: draw AB = 6 cm. With centre A and radius 7 cm draw an arc; with centre B and radius 7 cm draw another arc cutting it at C. Join CA and CB.1 mark
- Perpendicular bisectors: draw the perpendicular bisectors of AB and BC; they meet at O. (The bisector of AB passes through C, since CA = CB.)½ mark
- Circumcircle: with centre O and radius OA, draw the circle; it passes through A, B and C.½ mark
- Measure: OA ≈ 3.9 cm, OB ≈ 3.9 cm, OC ≈ 3.9 cm. They are equal, because O is equidistant from all three vertices.1 mark
Check: Let M be the midpoint of AB. Then CM = √(72 − 32) = √40 ≈ 6.32 cm. With OM = x: OA2 = 9 + x2 and OC = √40 − x. Setting OA = OC gives 40 − 2√40 x = 9, so x ≈ 2.45 cm and OC ≈ 6.32 − 2.45 = 3.87 cm ✓.
Answer to write in the exam
Steps of construction:
1. Draw AB = 6 cm.
2. With centres A and B and radius 7 cm, draw arcs meeting at C. Join CA, CB.
3. Draw the perpendicular bisectors of AB and BC; they meet at O.
4. With centre O and radius OA, draw a circle through A, B and C.
On measuring: OA = 3.9 cm, OB = 3.9 cm, OC = 3.9 cm
∴ OA = OB = OC ≈ 3.9 cm (radius of the circumcircle)
Common mistakes that cost marks
- Expecting OA, OB, OC to be different because the sides are different. The circumcentre is equidistant from the vertices, so the three lengths must be equal; small differences are only drawing errors.
- Placing O at the centroid (where the medians meet). For a non-equilateral triangle the centroid is a different point.
- Drawing the arcs for C with radius 6 cm from one end. Both CA and CB are 7 cm.
How this can come in the exam
The sides of a triangle are 5 cm, 12 cm and 13 cm. Its circumradius is
- 5 cm
- 6 cm
- 6.5 cm
- 13 cm
Show answer
(C) 6.5 cm
52 + 122 = 132, so the triangle is right-angled and the circumcentre is the midpoint of the hypotenuse: radius = 13 ÷ 2 = 6.5 cm.
O is the circumcentre of ΔABC and OA = 4.2 cm. Find OB + OC and give a reason.
Show answer
O lies on the perpendicular bisectors of the sides, so OA = OB = OC (1 mark). Hence OB + OC = 4.2 + 4.2 = 8.4 cm (1 mark).Try one yourself
Draw an equilateral triangle of side 6 cm and its circumcircle. Measure the radius.
Show answer
The radius is about 3.5 cm (exactly 6 ÷ √3 = 2√3 ≈ 3.46 cm). The circumcentre is inside the triangle and is also the point where the medians meet.
More questions like this
- What is the least possible radius of a circle through two points A and B?
- 1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle? - Equal chords of a circle subtend equal angles at the centre of the circle.
- Chords of a circle that subtend equal angles at the centre are equal.
- Show that the triangle formed by a chord and the centre of the circle is isosceles.