Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Step-by-step solution
To find: The circumcircle of ΔABC, and whether its centre is inside or outside the triangle
Idea: Construct the triangle (two sides and the included angle), then find the circumcentre as the meeting point of two perpendicular bisectors. A triangle with an obtuse angle has its circumcentre outside, on the far side of the longest side.
- Draw the triangle: draw AB = 5 cm. At A construct ∠BAX = 100°. On AX mark C with AC = 4 cm. Join BC (it measures about 6.9 cm).1 mark
- Perpendicular bisectors: draw the perpendicular bisectors of AB and BC. They meet at O.½ mark
- Circumcircle: with centre O and radius OA, draw the circle through A, B and C (OA = OB = OC ≈ 3.5 cm).½ mark
- ∠A = 100° > 90°, so ΔABC is obtuse-angled.½ mark
- In the figure O lies on the other side of BC from A, that is, outside the triangle.½ mark
Check: BC2 = 52 + 42 − 2 × 5 × 4 × cos 100° ≈ 47.9, so BC ≈ 6.92 cm, and the radius = BC ÷ (2 sin 100°) ≈ 3.52 cm. Measured OA should be close to 3.5 cm.
Answer to write in the exam
Steps of construction:
1. Draw AB = 5 cm.
2. At A, draw ∠BAX = 100°. Cut AC = 4 cm on AX. Join BC.
3. Draw the perpendicular bisectors of AB and BC; they meet at O.
4. With centre O and radius OA, draw a circle. It passes through A, B and C.
∠A = 100° > 90° ⇒ ΔABC is obtuse-angled
∴ The centre O lies outside the triangle (beyond side BC). (OA = OB = OC ≈ 3.5 cm)
Common mistakes that cost marks
- Measuring 100° on the inner scale of the protractor when the other scale is needed, giving 80° (an acute angle) by mistake.
- Stopping the perpendicular bisectors at the sides of the triangle. Here they meet outside the triangle, so the lines must be extended.
- Concluding “inside” because the circle surrounds the triangle. The question is about the centre O, not the circle.
How this can come in the exam
The angles of a triangle are 30°, 40° and 110°. Its circumcentre lies
- inside the triangle
- outside the triangle
- on the longest side
- at a vertex
Show answer
(B) outside the triangle
The triangle has an obtuse angle (110°), so its circumcentre lies outside it.
Assertion (A): The circumcentre of an obtuse-angled triangle lies outside the triangle.
Reason (R): The perpendicular bisectors of the sides of an obtuse-angled triangle do not meet.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.
Show answer
(C) A is true, but R is false.
A is true. R is false: the perpendicular bisectors of any triangle meet at one point (the circumcentre); for an obtuse triangle this point is simply outside.
Try one yourself
Draw ΔXYZ with XY = 4.5 cm, ∠X = 120° and XZ = 3.5 cm, and its circumcircle. Where is the centre?
Show answer
∠X = 120° is obtuse, so the centre lies outside the triangle, beyond YZ. (YZ ≈ 6.9 cm, radius ≈ 4.0 cm.)
More questions like this
- 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.
- 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.