Centres of a Triangle: Centroid, Incentre, Circumcentre, Orthocentre
Centres of a Triangle: Centroid, Incentre, Circumcentre, Orthocentre
A triangle has four classic centres, each the meeting point of three special lines. NTPC and CBT-2 questions test the angle each centre makes with two vertices, the 2 : 1 split of a median at the centroid, and the radii of the incircle and circumcircle in equilateral and right triangles.
1. Centres Box
| Centre | Meeting point of | Key fact |
|---|---|---|
| Centroid G | Medians | Divides each median 2 : 1 from the vertex; the six small triangles have equal area |
| Incentre I | Internal angle bisectors | Equidistant from the sides (inradius r); ∠BIC = 90° + ∠A/2 |
| Circumcentre O | Perpendicular bisectors of the sides | Equidistant from the vertices (circumradius R); ∠BOC = 2∠A (acute ∠A) |
| Orthocentre H | Altitudes | ∠BHC = 180° − ∠A (acute triangle) |
| Triangle | Position of centres | Radii |
|---|---|---|
| Equilateral, side a | All four coincide | R = a/√3, r = a/(2√3), so R = 2r |
| Right-angled, legs a, b, hypotenuse c | O at the midpoint of the hypotenuse; H at the right-angle vertex | R = c/2, r = (a + b − c)/2 |
| Any triangle | r = Area/s (s = semi-perimeter); R = abc/(4 × Area) |
✗ ∠A = 70°, so ∠BIC = 2 × 70° = 140° | ✓ That is the circumcentre rule; incentre gives 90° + 35° = 125°
✗ The centroid is the midpoint of a median | ✓ It divides the median 2 : 1, so AG = 2/3 of the median
हिंदी नोट: केन्द्रक माध्यिकाओं को 2 : 1 में बाँटता है। अन्तःकेन्द्र पर ∠BIC = 90° + ∠A/2, परिकेन्द्र पर ∠BOC = 2∠A और लम्बकेन्द्र पर ∠BHC = 180° − ∠A। समकोण त्रिभुज का परिकेन्द्र कर्ण के मध्य बिंदु पर होता है।
Exam Pointer: These patterns had no verified NTPC shift in our check and are tagged syllabus-based. Centres of a triangle are a named CBT-2 geometry topic and the four angle rules settle almost every question; expect them inside harder shifts.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Centroid: medians and areas
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. The median AD of a triangle is 18 cm and G is the centroid. Find AG and GD.
AG : GD = 2 : 1, so AG = 12 cm and GD = 6 cm.
Answer: 12 cm and 6 cm
EXAMATLAS LEVEL
Q. G is the centroid of triangle ABC, whose area is 60 cm², and E is the midpoint of AC. Find the areas of triangles BGC and AGE.
The medians cut the triangle into six equal small triangles of 10 cm² each. BGC is two of them: 20 cm². AGE is one: 10 cm².
Answer: 20 cm² and 10 cm²
Pattern 2: Incentre: angle and inradius
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. I is the incentre of triangle ABC with ∠A = 70°. Find ∠BIC.
∠BIC = 90° + 35° = 125°.
Answer: 125°
EXAMATLAS LEVEL
Q. I is the incentre of triangle ABC with ∠BIC = 117° and ∠B = 60°. Find ∠C, and ∠AIB.
90° + ∠A/2 = 117° gives ∠A = 54°, so ∠C = 180° − 54° − 60° = 66°. ∠AIB = 90° + ∠C/2 = 90° + 33° = 123°.
Answer: 66°; 123°
Pattern 3: Circumcentre: angle at the centre
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. O is the circumcentre of an acute triangle ABC with ∠A = 50°. Find ∠BOC.
∠BOC = 2 × 50° = 100°.
Answer: 100°
EXAMATLAS LEVEL
Q. O is the circumcentre of triangle ABC and ∠OBC = 25°. Find ∠BAC, given that the triangle is acute.
OB = OC, so ∠OCB = 25° and ∠BOC = 130°. Then ∠BAC = 130°/2 = 65°.
Answer: 65°
Pattern 4: Orthocentre: angle between altitudes
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. H is the orthocentre of an acute triangle ABC with ∠A = 65°. Find ∠BHC.
∠BHC = 180° − 65° = 115°.
Answer: 115°
EXAMATLAS LEVEL
Q. H is the orthocentre of an acute triangle ABC and ∠BHC is three times ∠A. Find ∠A, and the circumcentre angle ∠BOC for the same triangle.
180° − ∠A = 3∠A gives ∠A = 45°. Then ∠BOC = 90°.
Answer: 45°; 90°
Pattern 5: Inradius and circumradius of special triangles
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the circumradius and inradius of an equilateral triangle of side 6√3 cm.
R = 6√3/√3 = 6 cm and r = R/2 = 3 cm.
Answer: 6 cm and 3 cm
EXAMATLAS LEVEL
Q. A right triangle has legs 9 cm and 12 cm. Find its circumradius, its inradius, and the ratio of the areas of its circumcircle and incircle.
The hypotenuse is 15 cm, so R = 7.5 cm. r = (9 + 12 − 15)/2 = 3 cm (check: Area/s = 54/18 = 3). Area ratio = (7.5/3)² = 25 : 4.
Answer: 7.5 cm, 3 cm; 25 : 4
60-Second Revision
- Centroid: 2 : 1 on each median; six equal small triangles.
- ∠BIC = 90° + A/2, ∠BOC = 2A (acute A), ∠BHC = 180° − A (acute triangle).
- Equilateral: R = a/√3 = 2r.
- Right triangle: R = hypotenuse/2, r = (a + b − c)/2.