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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

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A median of a triangle is a line segment connecting a vertex to the midpoint of the opposite side. Every triangle has exactly three medians, which all intersect at a single point called the centroid.

A triangle ABC with a median AD connecting vertex A to the midpoint D of side BC.
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An altitude of a triangle is the 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 intersect at a point called the orthocenter.

A triangle PQR with an altitude PM dropping perpendicularly to the side QR.
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In an isosceles triangle, the median and altitude from the vertex (joining the equal sides) to the base are the same line segment. In an equilateral triangle, all medians are also altitudes.

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Altitudes can lie inside or outside the triangle. In an obtuse-angled triangle, two of the altitudes lie outside the triangle, while in a right-angled triangle, the two legs themselves act as altitudes.

📐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.
  2. By the property of medians, BD=12×BCBD = \frac{1}{2} \times BC.
  3. Substitute the given value: BD=12×14BD = \frac{1}{2} \times 14.
  4. 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}.
  2. Substitute the given dimensions: Area=12×10×6Area = \frac{1}{2} \times 10 \times 6.
  3. Perform the calculation: Area=5×6=30Area = 5 \times 6 = 30.
  4. 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.

Problem 3:

In △PQR\triangle PQR, SS is the midpoint of QRQR. Name the line segments PSPS and PTPT if PTPT is perpendicular to QRQR.

Triangle PQR showing altitude PT and median PS.

Solution:

  1. Since SS is the midpoint of the side QRQR and it is connected to the opposite vertex PP, the segment PSPS is the median.
  2. Since PTPT is perpendicular to QRQR (indicated by the 90∘90^{\circ} angle), the segment PTPT is the altitude.

Explanation:

By definition, a median joins a vertex to the midpoint of the opposite side, while an altitude is the perpendicular distance from a vertex to the opposite side.

Problem 4:

Given an obtuse-angled triangle XYZXYZ where ∠Y>90∘\angle Y > 90^{\circ}, draw the altitude from vertex XX to the side YZYZ. Does it lie inside the triangle?

Obtuse triangle XYZ with altitude XP drawn outside the triangle to the extension of YZ.

Solution:

  1. To draw the altitude from XX, we must extend the side ZYZY to a point PP outside the triangle.
  2. We then draw a perpendicular from XX to the line YPYP.
  3. The segment XPXP is the altitude. It lies outside the triangle.

Explanation:

In obtuse triangles, altitudes from the acute-angled vertices fall on the extension of the opposite sides and thus lie in the exterior of the triangle.