krit.club logo

Geometry - Medians and Altitudes of a Triangle

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Every triangle has exactly three medians, which are always concurrent at a single point called the Centroid (GG). The centroid divides each median in the ratio 2:12:1 from the vertex.

A triangle ABC with medians intersecting at centroid G. D is the midpoint of BC.
•

An altitude is a perpendicular segment from a vertex to the line containing the opposite side. The length of the altitude is the height of the triangle. The three altitudes of a triangle meet at a point called the Orthocenter (HH).

Triangle with a vertical line representing the altitude from the top vertex to the base.
•

In an equilateral triangle, the medians and altitudes are identical. This means the centroid and the orthocenter coincide at the same point.

•

For an obtuse-angled triangle, the orthocenter lies outside the triangle because two of the altitudes must be drawn to the extensions of the sides.

📐Formulae

Ratio of Centroid segments:AG=23AD and GD=13AD (where AD is the median and G is the centroid)\text{Ratio of Centroid segments}: AG = \frac{2}{3}AD \text{ and } GD = \frac{1}{3}AD \text{ (where } AD \text{ is the median and } G \text{ is the centroid)}

Area of a Triangle=12×Base×Altitude\text{Area of a Triangle} = \frac{1}{2} \times \text{Base} \times \text{Altitude}

In an equilateral triangle of side s, Length of Altitude/Median=32s\text{In an equilateral triangle of side } s, \text{ Length of Altitude/Median} = \frac{\sqrt{3}}{2}s

In △ABC, if AD is the median to BC, then BD=DC=12BC\text{In } \triangle ABC, \text{ if } AD \text{ is the median to } BC, \text{ then } BD = DC = \frac{1}{2}BC

💡Examples

Problem 1:

In △PQR\triangle PQR, PTPT is a median and GG is the centroid. If the length of the median PTPT is 15 cm15 \text{ cm}, find the lengths of the segments PGPG and GTGT.

Solution:

  1. We know that the centroid GG divides the median PTPT in the ratio 2:12:1 from the vertex PP.
  2. Therefore, PG=22+1×PT=23×15 cmPG = \frac{2}{2+1} \times PT = \frac{2}{3} \times 15 \text{ cm}.
  3. PG=2×5=10 cmPG = 2 \times 5 = 10 \text{ cm}.
  4. Similarly, GT=12+1×PT=13×15 cmGT = \frac{1}{2+1} \times PT = \frac{1}{3} \times 15 \text{ cm}.
  5. GT=5 cmGT = 5 \text{ cm}.
  6. Verification: PG+GT=10+5=15 cmPG + GT = 10 + 5 = 15 \text{ cm}, which matches the total length of the median.

Explanation:

The problem uses the property that the centroid divides the median into two parts in a 2:12:1 ratio, where the part connected to the vertex is longer.

Problem 2:

In a right-angled triangle XYZXYZ where ∠Y=90∘\angle Y = 90^{\circ}, the sides XY=6 cmXY = 6 \text{ cm} and YZ=8 cmYZ = 8 \text{ cm}. Identify the lengths of the altitudes from vertex XX to side YZYZ and from vertex ZZ to side XYXY. Also, name the orthocenter.

Solution:

  1. An altitude is a perpendicular from a vertex to the opposite side.
  2. In a right-angled triangle, the legs are perpendicular to each other.
  3. The altitude from XX to YZYZ is the side XYXY itself, so length =6 cm= 6 \text{ cm}.
  4. The altitude from ZZ to XYXY is the side YZYZ itself, so length =8 cm= 8 \text{ cm}.
  5. Since the altitudes XYXY and YZYZ meet at vertex YY, and the third altitude from YY to the hypotenuse XZXZ also passes through YY, the orthocenter is the vertex YY.

Explanation:

This example demonstrates that in a right-angled triangle, the two legs act as altitudes for each other, and the vertex containing the right angle is the orthocenter.

Problem 3:

In △ABC\triangle ABC, ADAD is a median and GG is the centroid. If GD=4 cmGD = 4 \text{ cm}, calculate the lengths of AGAG and the whole median ADAD.

Median AD divided by centroid G.

Solution:

  1. We know that the centroid GG divides the median ADAD in the ratio AG:GD=2:1AG : GD = 2 : 1.
  2. Given GD=4 cmGD = 4 \text{ cm}.
  3. Since AGGD=21\frac{AG}{GD} = \frac{2}{1}, we have AG=2×GD=2×4=8 cmAG = 2 \times GD = 2 \times 4 = 8 \text{ cm}.
  4. The total length of the median AD=AG+GD=8+4=12 cmAD = AG + GD = 8 + 4 = 12 \text{ cm}.

Explanation:

The centroid property states that the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint of the side.

Problem 4:

An isosceles triangle PQRPQR has PQ=PR=13 cmPQ = PR = 13 \text{ cm} and QR=10 cmQR = 10 \text{ cm}. Calculate the length of the altitude PMPM drawn from PP to QRQR.

Isosceles triangle PQR with altitude PM.

Solution:

  1. In an isosceles triangle, the altitude to the base also acts as a median. Therefore, MM is the midpoint of QRQR.
  2. QM=MR=12×QR=102=5 cmQM = MR = \frac{1}{2} \times QR = \frac{10}{2} = 5 \text{ cm}.
  3. In right-angled △PMQ\triangle PMQ, using Pythagoras theorem: PM2+QM2=PQ2PM^2 + QM^2 = PQ^2 PM2+52=132PM^2 + 5^2 = 13^2 PM2+25=169PM^2 + 25 = 169 PM2=144PM^2 = 144 $$PM = \sqrt{144} = 12 \text{ cm}$.

Explanation:

In isosceles triangles, the altitude from the vertex between equal sides bisects the base. This creates a right-angled triangle where the altitude can be found using the Pythagorean theorem.