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 that joins a vertex of the triangle to the midpoint of the opposite side. Every triangle has exactly three medians. Visually, if you have a triangle △ABC\triangle ABC, and DD is the midpoint of side BCBC such that BD=DCBD = DC, the segment ADAD is the median from vertex AA.

•

The point where all three medians of a triangle intersect is called the Centroid, usually denoted by the letter GG. A key property of the centroid is that it always lies inside the triangle, regardless of whether the triangle is acute, obtuse, or right-angled. Visually, the centroid GG looks like the balance point or the center of gravity of the triangle.

•

The centroid GG divides each median in the ratio 2:12:1. This means the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint of the opposite side. For a median ADAD with centroid GG, this is represented as AG:GD=2:1AG : GD = 2 : 1.

•

An altitude of a triangle is a perpendicular line segment drawn from a vertex to the opposite side (or the line containing the opposite side). The opposite side is then referred to as the base for that altitude. Visually, an altitude forms a 90∘90^{\circ} angle (a square symbol at the base) with the side it hits.

•

The point of intersection of the three altitudes of a triangle is called the Orthocenter, often denoted by HH. Unlike the centroid, the position of the orthocenter changes based on the type of triangle: it lies inside an acute-angled triangle, at the vertex containing the right angle in a right-angled triangle, and outside in an obtuse-angled triangle.

•

In an equilateral triangle, the medians and the altitudes are the same line segments. For an isosceles triangle, only the median drawn from the vertex between the equal sides to the base coincides with the altitude to that base.

•

The three medians divide the triangle into six smaller triangles of equal area. Even though these six triangles might look different in shape, their calculated areas are identical.

📐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.