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The Triangle and its Properties - Medians and Altitudes of a Triangle

Grade 7CBSE

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. Visually, if you have a triangle ABC\triangle ABC and you find the exact middle point DD on the side BCBC, the line ADAD is the median. It essentially bisects the side into two equal parts such that BD=DCBD = DC.

Every triangle has exactly three medians, one from each vertex. These three medians always intersect at a single point inside the triangle called the Centroid. Visually, the centroid can be thought of as the 'balancing point' or center of gravity of the triangle.

An altitude of a triangle is a perpendicular line segment drawn from a vertex to the line containing the opposite side. Visually, an altitude represents the 'height' of the triangle. It always meets the opposite side (or its extension) at a right angle, which is 9090^{\circ}.

A triangle has three altitudes, one for each vertex. The point where all three altitudes meet is called the Orthocenter. Unlike the centroid, the orthocenter's location changes visually based on the triangle type: it stays inside for acute triangles, sits at the right-angle vertex for right triangles, and falls outside for obtuse triangles.

In an obtuse-angled triangle, two of the altitudes lie outside the triangle. To visualize this, you must extend the base lines of the triangle outward so that the perpendicular line from the opposite vertex can meet them.

In an isosceles triangle, the median and altitude drawn from the vertex connecting the two equal sides are the same line segment. In an equilateral triangle, all three medians are identical to the three altitudes, meaning the Centroid and the Orthocenter are the exact same point at the very center of the triangle.

📐Formulae

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

If AD is a median to side BC, then BD=DC=12BC\text{If } AD \text{ is a median to side } BC \text{, then } BD = DC = \frac{1}{2} BC

Centroid Ratio: AG=2×GD (where G is the centroid on median AD)\text{Centroid Ratio: } AG = 2 \times GD \text{ (where } G \text{ is the centroid on median } AD \text{)}

💡Examples

Problem 1:

In ABC\triangle ABC, ADAD is the median to the side BCBC. If the length of BCBC is 14 cm14 \text{ cm}, find the length of BDBD.

Solution:

  1. Understand that a median connects a vertex to the midpoint of the opposite side. Since ADAD is the median, DD is the midpoint of BCBC.\n2. By the property of medians, BD=12×BCBD = \frac{1}{2} \times BC.\n3. Substitute the given value: BD=12×14BD = \frac{1}{2} \times 14.\n4. BD=7 cmBD = 7 \text{ cm}.

Explanation:

Because the median bisects the side it is drawn to, we simply divide the total length of side BCBC by 2 to find the length of the segment BDBD.

Problem 2:

Find the area of a triangle where the base is 10 cm10 \text{ cm} and the corresponding altitude is 6 cm6 \text{ cm}.

Solution:

  1. Use the area formula: Area=12×base×altitudeArea = \frac{1}{2} \times \text{base} \times \text{altitude}.\n2. Substitute the given dimensions: Area=12×10×6Area = \frac{1}{2} \times 10 \times 6.\n3. Perform the calculation: Area=5×6=30Area = 5 \times 6 = 30.\n4. The final area is 30 cm230 \text{ cm}^2.

Explanation:

The altitude of a triangle acts as its height. By multiplying the base by the altitude and then taking half of that product, we determine the total space enclosed by the triangle.