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A Tale of Three Intersecting Lines - Constructions Related to Altitudes of Triangles

Grade 7CBSE

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

🔑Concepts

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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). Every triangle has exactly three altitudes.

Triangle ABC with altitude AD perpendicular to base BC
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The three altitudes of a triangle always intersect at a single point called the orthocenter. In an acute-angled triangle, the orthocenter lies inside the triangle.

An acute triangle showing three intersecting altitudes at the orthocenter H
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In a right-angled triangle, two of the altitudes are the sides forming the right angle. The orthocenter is exactly at the vertex containing the right angle.

Right-angled triangle where the right-angle vertex is the orthocenter
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In an obtuse-angled triangle, two of the altitudes lie outside the triangle. The orthocenter is located outside the triangle.

Obtuse triangle with altitudes meeting at an external orthocenter

📐Formulae

Area of △=12×base×heightArea\ of\ \triangle = \frac{1}{2} \times \text{base} \times \text{height}

In △ABC, if AD⊥BC, then AD is the altitude.\text{In } \triangle ABC, \text{ if } AD \perp BC, \text{ then } AD \text{ is the altitude.}

ha=2×Areaah_a = \frac{2 \times Area}{a}

💡Examples

Problem 1:

In △PQR\triangle PQR, the base QR=14 cmQR = 14\text{ cm} and the corresponding altitude PS=8 cmPS = 8\text{ cm}. Find the area of the triangle. If another altitude RTRT is drawn to side PQPQ which measures 16 cm16\text{ cm}, find the length of RTRT.

Solution:

  1. Calculate Area using QRQR: Area=12×14×8=56 cm2Area = \frac{1}{2} \times 14 \times 8 = 56\text{ cm}^2 2. Use the Area to find altitude RTRT where base is PQPQ: 56=12×16×RT56 = \frac{1}{2} \times 16 \times RT 56=8×RT56 = 8 \times RT RT=568=7 cmRT = \frac{56}{8} = 7\text{ cm}

Explanation:

The area of a triangle remains constant regardless of which side is chosen as the base. We first find the area using the known base and altitude, then use that area to solve for the unknown altitude.

Problem 2:

Determine the position of the orthocenter for △XYZ\triangle XYZ where ∠Y=90∘\angle Y = 90^\circ.

Solution:

In a right-angled triangle, the altitudes from the two acute vertices are the legs of the triangle itself.

  • Altitude from XX to YZYZ is XYXY.
  • Altitude from ZZ to XYXY is ZYZY.
  • Altitude from YY to XZXZ is a perpendicular segment YMYM. All three meet at vertex YY.

Explanation:

For any right-angled triangle, the vertex where the right angle is formed serves as the orthocenter.

Problem 3:

A student calculated the areas of two triangular parts of a metal sheet. The first area is 45 cm245\text{ cm}^2 and the second is 38 cm238\text{ cm}^2. What is the total area?

Solution:

45+3883\begin{array}{r} 45 \\ + 38 \\ \hline 83 \end{array} The total area is 83 cm283\text{ cm}^2.

Explanation:

Basic addition of areas calculated from base and altitude measurements.

Problem 4:

In △LMN\triangle LMN, ∠M=120∘\angle M = 120^\circ. If we draw altitudes from vertices LL and NN to the opposite sides, where will they meet? Illustrate with a diagram.

Obtuse triangle LMN showing external intersection of altitudes at H

Solution:

  1. Since ∠M=120∘\angle M = 120^\circ, the triangle is obtuse-angled.
  2. In an obtuse triangle, altitudes from the acute angles (LL and NN) fall on the extensions of the opposite sides (MNMN and LMLM).
  3. These altitudes, when extended, meet at the orthocenter HH located outside the triangle.
  4. Therefore, they meet at a point in the exterior region of the triangle.

Explanation:

For obtuse triangles, the altitudes from the two acute vertices must be drawn to the exterior lines containing the bases. Their point of intersection (orthocenter) always lies outside the triangle.

Problem 5:

Consider an isosceles △ABC\triangle ABC where AB=AC=10 cmAB = AC = 10\text{ cm} and BC=12 cmBC = 12\text{ cm}. Find the length of the altitude ADAD from AA to BCBC and then find the area of the triangle.

Isosceles triangle ABC with altitude AD and dimensions labeled

Solution:

  1. In an isosceles triangle, the altitude to the base bisects the base. So, BD=DC=122=6 cmBD = DC = \frac{12}{2} = 6\text{ cm}.
  2. In right-angled △ABD\triangle ABD, by Pythagoras theorem: AD2+BD2=AB2AD^2 + BD^2 = AB^2 AD2+62=102AD^2 + 6^2 = 10^2 AD2+36=100AD^2 + 36 = 100 AD2=64  ⟹  AD=8 cmAD^2 = 64 \implies AD = 8\text{ cm}.
  3. Area of △ABC=12×base×height\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} Area=12×12×8=48 cm2Area = \frac{1}{2} \times 12 \times 8 = 48\text{ cm}^2.

Explanation:

The altitude in an isosceles triangle acts as a median for the non-equal side, allowing us to use the Pythagoras theorem to find its height and subsequently the area.