proof for heron's formula for equilateral triangle.

we know that area of a triangle is = 1/2 x base x height now lets us take an equilateral triangle ABC with side = m units and altitude AD to base BC.let AD = n units. now in right angled triangle ABD , AB = m , BD = m/2. now by Pythagoras therom AD-square + BD-square = AB-square i.e. n/2-square + m/2 square = m- square i.e (n-square +m-square )/4 = m-square i.e n-square + m-square = 4(m-square) i.e n-square = 3(m-square) i.e n = m x root 3 now area of triangle ABC = 1/2 x base x height = 1/2 x m/2 x m x root 3 = root 3/4 x m-square.

Note by Ksg Sarma
3 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...