# Find area of a 'curved triangle'

• May 16th 2009, 06:37 PM
BG5965
Find area of a 'curved triangle'
http://i301.photobucket.com/albums/n...965/Chook1.jpg

Background
Originally, there was a pen of radius 10cm for her animals to safely roam around in (Figure 1).

She found that the animals often destroyed the vegetation on the floor before it could regrow. She decided to split the pen into 3 equal lengths of AB, BC and CD (Figure 2).

She found she could join them together to create an equilateral triangle that had a side length of 10cm within the 'curved triangle'. (Figure 3).

Questions
1) What is the area of the whole 'curved triangle'?

2) She found she could enclose it in a square, with the sides of the curved triangle just touching the sides of the square. She found that any square she drew around this curved triangle was the same size. Assuming this is true, what is the side length of the square?

http://i301.photobucket.com/albums/n...965/Chook2.jpg
• May 17th 2009, 12:01 AM
earboth
Quote:

Originally Posted by BG5965
http://i301.photobucket.com/albums/n...965/Chook1.jpg

Background
Originally, there was a pen of radius 10cm for her animals to safely roam around in (Figure 1).

She found that the animals often destroyed the vegetation on the floor before it could regrow. She decided to split the pen into 3 equal lengths of AB, BC and CD (Figure 2).

She found she could join them together to create an equilateral triangle that had a side length of 10cm within the 'curved triangle'. (Figure 3).

Questions
1) What is the area of the whole 'curved triangle'?

2) She found she could enclose it in a square, with the sides of the curved triangle just touching the sides of the square. She found that any square she drew around this curved triangle was the same size. Assuming this is true, what is the side length of the square?

to #1: The radius of the arcs didn't change. Therefore you'll get an equilateral triangle with the side-length of 10.

$a_{\Delta}=\dfrac14 \cdot r^2 \cdot \sqrt{3}$

To this equilateral triangle are attached 3 congruent circle segments. The equilateral triangle with one attached segment (painted grey) forms one sixth of a circle with radius 10. Therefore the area of the complete shape is:

$a = \dfrac14 \cdot r^2 \cdot \sqrt{3} + 3\cdot \left(\dfrac16 \cdot \pi r^2 - \dfrac14 \cdot r^2 \cdot \sqrt{3} \right) = \dfrac12 r^2(\pi - \sqrt{3})$

to #2:
Two adjacent sides of the square are tangents to adjacent arcs(with radius 10) while the opposite sides of the square pass through vertices of the curved triangle. The distance between a vertex of the curved triangle to the corresponding tangent side of the square must be a radius too. Therefore the square has a side-length of r.
• May 17th 2009, 05:11 AM
smile4life
square root
hi, why is there a square root of 3? Is it because of the three segments of the circle? $
a_{\Delta}=\dfrac14 \cdot r^2 \cdot \sqrt{3}
$
• May 17th 2009, 05:59 AM
earboth
Quote:

Originally Posted by smile4life
hi, why is there a square root of 3? Is it because of the three segments of the circle? $
a_{\Delta}=\dfrac14 \cdot r^2 \cdot \sqrt{3}
$

If you want tocalculate the area of an equilateral triangle you need the base and the height of the triangle.
Take the left triangle AFC. Use Pythagorean theorem:

$a^2 = h^2+\left(\dfrac a2 \right)^2~\implies~ h^2 = a^2-\dfrac14 a^2 = \dfrac34 a^2$

Therefore $h = \dfrac12 a \sqrt{3}$

The area of the triangle is then:

$A = \dfrac12 \cdot \underbrace{a}_{base} \cdot \underbrace{\dfrac12 \cdot a \cdot \sqrt{3}}_{height} = \dfrac{a^2}4 \sqrt{3}$