Altitude of a Triangle

Chapter 41

Concept Explanation:

The Altitude of a Triangle is a perpendicular segment from a vertex of the triangle to the line containing the opposite side (also called the base). The altitude forms a right angle with the base, and each triangle has three possible altitudes, one from each vertex.

Altitudes play a significant role in determining the area of a triangle, as the area is calculated using the formula:

$\text{Area of a triangle} = \frac{1}{2} \times \text{Base} \times \text{Height}$

Here, the height is the length of the altitude.

*** QuickLaTeX cannot compile formula:
\begin{tikzpicture}\draw<a href="0,0">thick</a> -- (4,0) node[right] {Base};\draw<a href="0,0">thick</a> -- (2,3) node[above] {Vertex};\draw<a href="2,3">thick</a> -- (4,0);\draw<a href="2,3">dashed</a> -- (2,0) node[below] {Altitude};\draw<a href="2,0">thick</a> -- (2,0.2);\end{tikzpicture}

*** Error message:
Error: Cannot create svg file

In the diagram, the dashed line represents the altitude of the triangle from the vertex to the base.

Common Mistakes:

Misidentifying the Altitude: Some students confuse the altitude with the median or angle bisector. The altitude always forms a right angle with the base.
Choosing Incorrect Base: When determining the altitude, ensure the correct base is selected based on the vertex from which the altitude is drawn.

Helpful Tips:

Use Right-Angle Detection: Always look for the perpendicular sign (a right angle) to identify the altitude correctly.
Visualize the Triangle: In obtuse triangles, the altitude from a vertex may fall outside the triangle, making visualization key.

Hard Questions:

Q1: In a right triangle with base 6 cm and height (altitude) 8 cm, what is the area of the triangle?

Step-by-Step Solution:

The area of a triangle is calculated using the formula:
$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$
Substitute the given values:
$\text{Area} = \frac{1}{2} \times 6 \, \text{cm} \times 8 \, \text{cm} = 24 \, \text{cm}^2$

Answer:

The area of the triangle is $24 \, \text{cm}^2$ .

Q2: In an equilateral triangle with side length 10 cm, find the length of the altitude.

Step-by-Step Solution:

An equilateral triangle can be split into two 30-60-90 triangles. The altitude is the side opposite the ( 60^\circ ) angle.
In a 30-60-90 triangle, the altitude is given by:
$\text{Altitude} = \frac{\sqrt{3}}{2} \times \text{Side length}]</li>   <li>Substitute the side length:[latex]\text{Altitude} = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \approx 8.66 \, \text{cm}]</li> </ol>   Answer:   <ul class="wp-block-list"> <li>The length of the altitude is approximately ( 8.66 \, \text{cm} ).</li> </ul>   <hr class="wp-block-separator has-alpha-channel-opacity"/>   <strong>Q3:</strong> In a triangle, the base is 12 cm and the area is ( 36 \, \text{cm}^2 ). Find the height (altitude).   <h4 class="wp-block-heading">Step-by-Step Solution:</h4>   <ol class="wp-block-list"> <li>The area of a triangle is given by:[latex]\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$
Rearranging to solve for the height:
$\text{Height} = \frac{2 \times \text{Area}}{\text{Base}} = \frac{2 \times 36}{12} = 6 \, \text{cm}$

Answer:

The height (altitude) is $6 \, \text{cm}$ .

Q4: In an isosceles triangle with equal sides of length 13 cm and base of length 10 cm, find the altitude.

Step-by-Step Solution:

The altitude splits the isosceles triangle into two right triangles. Half of the base is $5 \, \text{cm} ).</li>   <li>Use the Pythagorean theorem to find the altitude ( h ):[latex]h^2 + 5^2 = 13^2$
$h^2 = 169 - 25 = 144$
$h = \sqrt{144} = 12 \, \text{cm}$

Answer:

The altitude is $12 \, \text{cm}$ .

Q5: In a triangle with vertices ( A(0, 0) ), ( B(6, 0) ), and ( C(3, 4) ), find the altitude from ( C ) to the base ( AB ).

Step-by-Step Solution:

The base ( AB ) lies on the x-axis, so the altitude is the y-coordinate of point ( C ).
The altitude is the distance from ( C(3, 4) ) to the line ( AB ):
$\text{Altitude} = 4 \, \text{units}$

Answer:

The altitude is $4 \, \text{units}$ .

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