Altitude of a triangle
A mason holds a plumb-line from the top of a triangular roof gable straight down to the floor. The straight drop from the peak to the base is an altitude of the triangle.
The perpendicular segment from a vertex of a triangle to the opposite side is its altitude (शिरोलंब). Every triangle has three altitudes.
Rule 1: Recall and drawing altitudes — Recall: angle bisectors meet at the incentre I, and perpendicular bisectors of the sides meet at the circumcentre S. In ΔABC, seg AP ⊥ BC, so AP is the altitude on base BC. To draw the altitudes of ΔXYZ, use a set-square to drop XR ⊥ YZ, then YQ ⊥ XZ and ZP ⊥ XY.
In the figure : In the figure the angle bisectors meet at the incentre I, and the perpendicular bisectors of the sides meet at the circumcentre S.
In the figure : In the figure altitude: the perpendicular from a vertex to the opposite side. Here seg AP ⊥ BC.
In the figure : In the figure the altitudes of ΔXYZ: XR ⊥ YZ, YQ ⊥ XZ and ZP ⊥ XY.
- Easy : Easy: A triangle has 3 vertices, so 3 altitudes.
- Medium : Medium: If the base is 10 cm and the area is 40 cm2, the altitude on it is 2 × 4010 = 8 cm.
- Exam-type : Exam: In an isosceles triangle the altitude from the apex also bisects the base.
An altitude makes 90° with the opposite side.
Joining a vertex to the midpoint and calling it an altitude. That is a median.
Memory clue: "शिरोलंब = लंब, 90°".
Spot the difference
| Segment | Property |
|---|---|
| Altitude | ⊥ opposite side |
| Number in a triangle | 3 |
| Tool | Set-square |
Practice questions
- 1What is an altitude?
- 2How many altitudes does a triangle have?
- 3What angle does an altitude make with the base?
- 4Which tool draws an altitude?
- 5Base 12 cm, area 30 cm2: altitude?
- 6How is an altitude different from a median?
- 7In ΔABC, AP ⊥ BC. Which is the base?
- 8In which triangle do all the altitudes equal the medians?
Show answers (try them yourself first)
- 1The perpendicular from a vertex to the opposite side.
- 2Three.
- 390°.
- 4A set-square.
- 52 × 3012 = 5 cm.
- 6Altitude goes at 90°; median goes to the midpoint.
- 7BC.
- 8In an equilateral triangle.
Orthocentre
Stand three poles at the corners of a triangle and tie a rope from each pole straight to the opposite wall at 90°. All three ropes cross at one single knot.
The three altitudes are concurrent (संगामी); the point where they meet is the orthocentre (लंबकेंद्र), denoted O.
Rule 1: Finding O — In ΔXYZ the altitudes XR, YQ and ZP pass through one point. That point is O. To find O, draw any two altitudes; the third passes through the same point (use it as a check).
In the figure : In the figure the three altitudes pass through one point, the orthocentre O.
- Easy : Easy: Two altitudes are enough to locate O.
- Medium : Medium: The third altitude is a check: it must pass through O.
- Exam-type : Exam: Concurrent means passing through one common point.
Altitudes meet at the orthocentre O.
Confusing O with the centroid G. O is for altitudes, G is for medians.
Memory clue: "O = Orthocentre = शिरोलंब".
Spot the difference
| Point | Made by |
|---|---|
| Orthocentre O | Altitudes |
| Centroid G | Medians |
| Incentre I | Angle bisectors |
Practice questions
- 1What does concurrent mean?
- 2What is the orthocentre?
- 3Which letter denotes the orthocentre?
- 4How many altitudes do we need to find O?
- 5Which point comes from the angle bisectors?
- 6Which point comes from the perpendicular bisectors of the sides?
- 7Draw an acute triangle: where is O?
- 8Why is the third altitude a good check?
Show answers (try them yourself first)
- 1Passing through one common point.
- 2The meeting point of the three altitudes.
- 3O.
- 4Two.
- 5The incentre.
- 6The circumcentre.
- 7Inside the triangle.
- 8Because all three must meet at one point.
Location of the orthocentre
A tall tent pole sits in the middle of a tent, but for a very flat tent the poles lean outside. In the same way, where the altitudes meet depends on the shape of the triangle.
The position of the orthocentre O depends on the type of triangle: right → at the right-angle vertex, acute → inside, obtuse → outside.
Rule 1: Three cases — Right angled (sides 3, 4, 5): O is the vertex of the right angle. Acute angled (angles 60°, 70°, 50°): O is inside. Obtuse angled (one angle 110°): O is outside; the altitudes themselves do not meet, but the lines containing them (extended, dashed) meet at O.
In the figure : In the figure right angled: O is the vertex of the right angle. Obtuse angled: O is outside the triangle. Acute angled: O is inside the triangle.
In the figure : In the figure in an obtuse angled triangle the altitudes themselves do not meet, but the lines containing them (dashed) meet at O outside the triangle.
- Easy : Easy: Angles 40°, 60°, 80° are all acute, so O is inside.
- Medium : Medium: Angles 25°, 35°, 120°: one obtuse angle, so O is outside.
- Exam-type : Exam: Sides 5, 12, 13: 25 + 144 = 169, so it is right angled and O is at the vertex between 5 and 12.
Right: vertex. Acute: inside. Obtuse: outside (extend the altitudes).
Looking for O inside an obtuse triangle. Extend the altitudes with dashed lines instead.
Memory clue: "काटकोन-कोपरा, लघु-आत, विशाल-बाहेर".
Spot the difference
| Triangle | Position of O |
|---|---|
| Right angled | At the right-angle vertex |
| Acute angled | Inside |
| Obtuse angled | Outside |
Practice questions
- 1Where is O in a right angled triangle?
- 2Where is O in an acute angled triangle?
- 3Where is O in an obtuse angled triangle?
- 4Angles 50°, 60°, 70°: where is O?
- 5Angles 30°, 40°, 110°: where is O?
- 6Sides 6, 8, 10: where is O?
- 7Why do we extend the altitudes of an obtuse triangle?
- 8In a right triangle, two altitudes are the two legs. Why?
Show answers (try them yourself first)
- 1At the vertex of the right angle.
- 2Inside.
- 3Outside.
- 4Inside.
- 5Outside.
- 6Right angled (36 + 64 = 100), so at the vertex between 6 and 8.
- 7The altitudes themselves do not meet; their lines do, outside.
- 8Each leg is perpendicular to the other, so it is the altitude on it.
Median and centroid
A school cook wants to cut a triangular flatbread so that each piece is exactly balanced. She joins each corner to the middle of the opposite side. These lines are medians.
A median (मध्यगा) joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid (केंद्रक) G, always inside the triangle.
Rule 1: Drawing medians — In ΔHCF, D is the midpoint of CH, so FD is a median. To draw the medians of ΔABC, mark midpoints P, Q, R of AB, BC, AC and draw CP, AQ, BR. They are concurrent at G.
In the figure : In the figure median: the segment from a vertex to the midpoint of the opposite side. D is the midpoint of CH (equal ticks), so FD is a median.
In the figure : In the figure the medians AQ, BR and CP (P, Q, R are the midpoints of AB, BC, AC) are concurrent at the centroid G.
In the figure : In the figure the centroid G is in the interior of every triangle: right angled, obtuse angled and acute angled.
In the figure : In the figure a cardboard triangle balances on a pin placed at its centroid G.
- Easy : Easy: A triangle has 3 medians.
- Medium : Medium: The midpoint of a 14 cm side divides it into 7 cm and 7 cm.
- Exam-type : Exam: The centroid is always inside, unlike the orthocentre which can be outside.
Median → midpoint. Centroid G is the balance point and is always inside.
Calling the altitude a median. The median goes to the midpoint, not at 90°.
Memory clue: "मध्यगा = मध्यबिंदू, G = तोल".
Spot the difference
| Segment | Goes to |
|---|---|
| Altitude | Opposite side at 90° |
| Median | Midpoint of opposite side |
| Centroid G | Balance point inside |
Practice questions
- 1What is a median?
- 2What is the centroid?
- 3Which letter denotes the centroid?
- 4Can the centroid be outside a triangle?
- 5If D is the midpoint of CH = 18 cm, find CD.
- 6Why does a cardboard triangle balance at G?
- 7In an equilateral triangle, is each altitude also a median?
- 8How many medians do we draw to find G?
Show answers (try them yourself first)
- 1A segment from a vertex to the midpoint of the opposite side.
- 2The meeting point of the three medians.
- 3G.
- 4No, always inside.
- 59 cm.
- 6G is its balance (centre of mass) point.
- 7Yes.
- 8Two are enough.
Property of the centroid
If you balance a ruler on a finger, the ruler tilts if the finger is not at the centre. A triangle balances at its centroid, and the centroid splits each median into a 2-part piece and a 1-part piece.
The centroid divides each median in the ratio 2 : 1, the longer part towards the vertex: AG : GR = 2 : 1.
Rule 1: Using 2 : 1 — Vertex part = 23 of the median and side part = 13. If AG = 12 then GR = 6. If GP = 5 then CG = 10. If BG = 8 then BQ = 12 (8 + 4). If the whole median is 24, then AG = 16 and GR = 8.
In the figure : In the figure the centroid divides each median in the ratio 2 : 1. If l(AG) = 14 then l(GR) = 7: the part from the vertex is twice the part from G to the side.
In the figure : In the figure on every median the part from the vertex to G is 2 units and the part from G to the side is 1 unit.
In the figure : In the figure if the whole median l(AR) = 30 cm, then l(AG) = 23 × 30 = 20 cm and l(GR) = 13 × 30 = 10 cm.
- Easy : Easy: GR = 3 means AG = 6.
- Medium : Medium: Whole median 27 cm: AG = 18 cm and GR = 9 cm.
- Exam-type : Exam: AG = 14 cm gives GR = 7 cm and the whole median AR = 21 cm.
Vertex side : base side = 2 : 1. Total parts = 3.
Putting 2 on the part near the side. The longer part is near the vertex.
Memory clue: "शिरोबिंदूकडे 2, बाजूकडे 1".
Spot the difference
| Given | Find | Answer |
|---|---|---|
| GR = 5 | AG | 10 |
| AG = 12 | Whole median | 18 |
| Whole median 30 | AG, GR | 20, 10 |
Practice questions
- 1In what ratio does G divide a median?
- 2Which part is longer?
- 3GR = 4. Find AG.
- 4AG = 10. Find GR.
- 5Whole median = 45 cm. Find AG.
- 6BG = 6. Find the whole median BQ.
- 7AP = 6 and G is the centroid on AP. Find AG and GP.
- 8Fraction of a median from the vertex to G?
Show answers (try them yourself first)
- 12 : 1.
- 2The one towards the vertex.
- 38.
- 45.
- 530 cm.
- 69.
- 7AG = 4, GP = 2.
- 823.
Quick revision
- Altitude: perpendicular from a vertex to the opposite side; the altitudes meet at the orthocentre O.
- Orthocentre: right angled triangle → the vertex of the right angle; obtuse angled → exterior; acute angled → interior.
- Median: vertex to the midpoint of the opposite side; the medians meet at the centroid G, which divides each median in the ratio 2 : 1.
Quick revision
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