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## Geometry (all content)

### Course: Geometry (all content)>Unit 14

Lesson 10: Area of inscribed triangle

# Area of inscribed equilateral triangle

A worked example of finding the area of an equilateral triangle inscribed within a circle who's area is known. This video uses Heron's formula and some trigonometry. Created by Sal Khan.

## Want to join the conversation?

• What is Heron's formula, and why didn't they show it to us before? I'm so confused... • I don't think you even need trig to solve the problem.
All you have to do is follow the video until around and from there, use the rules of the side of a 30:60:90 right triangle. You can find the side a/2 and the line from the center of the circle to where it reaches side a perpendicularly. With these you can find the eight of the triangle and the base. Then you can just subtract the area of the triangle to the circle. • why not use area of a triangle = 1/2 * base * height ?
would be easier than Heron's Formula • There may be multiple ways to tackle math problems. In fact I had to use the triangle area formula, law of cosines and knowledge of inscribed angles because this is the first time I hear about Heron's Formula.

I can only suggest you to try to solve any problem using everything you know so far and afterwards watch the rest of the video. Maybe there will be a better and easier way that you can learn and use next time.

• Do I have to read Euclid's book on geometry to understand geometry better?
(1 vote) • isn't there an easier way? area of triangle = 3(1/2 abSinC) (draw radii from the origin) = 3(1/2(2*2 Sin 120)) = 3(2Sin120) = 5.196, then the area of the circle minus this = 4pi-5.196 = 7.370
(1 vote) • You are correct. But, here's how I would have done it:
Area outside the triangle = πr² - ¼ a²√3
Because the area of an equilateral triangle is ¼ a²√3
Since a = r√3 also stated as a² = 3r²
Substituting, πr² - ¾r²√3
Since r = 2, we get 4π - 3√3 = 7.370
Of course my way does require knowing that a² = 3r² for an inscribed equilateral triangle (though it isn't too hard to derive if you didn't know that)
• how would i be able to find the area if the circle was in a triangle • Your question is probably about finding the area of an equilateral triangle with an inscribed circle given the circle's radius. This turns out to be very similar to Sal's question!

You can draw an equilateral triangle inside the circle, with vertices where the circle touches the outer triangle. Now, you know how to calculate the area of that inner triangle from Sal's video. Specifically, this is 3/4 * r^2 * sqrt(3). (When r=2 like in the video, this is 3 * sqrt(3).)

The outer triangle is simply 4 of these triangles (ASA postulate). So, the big triangle's area is 3 * r^2 * sqrt(3). (When r=2, this is 12 * sqrt(3).)
• Does anybody know what to do if it's the opposite (circle transcribed in an equilateral triangle)? thanks! • Yes you can. For example, if you know either the length of the (outside) equilateral triangle... or the radius of the (inside) circle, you can figure out the other. Let's take the example that you know the radius of the (inside) circle. This is a bit more challenging to do without video, but if you understood the video, you could probably follow:

R = radius of circle (known)
x = length of equilateral triangle
Draw the figure (outside equilateral triangle with inscribed circle).
Put a dot in the center of the circle.

Draw a line from the center of the circle to a point where the circle barely touches the triangle. This forms a right angle with that line, and that length = R (radius).

Also draw a line from the center to a nearby apex of the triangle (on the same triangle line edge). You have just formed an inside triangle. You know one of the angles (90 degrees). The other two are 60 and 30 degrees. You know this because the equilateral triangle angle was 60 degrees, and you just bisected it.
Just like the example in the video, the distance from the center to the outside triangle apex is 2*R, and the distance of the distance of the base of the inside triangle is SQRT(3)R, and by definition that distance is x/2 (half the length of the outside triangle edge).
Thus x/2 = SQRT(3)*R
x = 2*SQRT(3)*R
You can now use the formulas shown in the video for the area of the triangle and the circle.
If I did my math correctly...
Area = (SQRT(3)*x^2) / 4 - pi * R^2 ... then enter formula for 'x' above...
= (SQRT(3) * (2*SQRT(3)*R)^2) / 4 - pi * R^2
= SQRT(3) * 3 * R^2 - pi * R^2
or = (3
SQRT(3) - pi) * R^2
• Is this a general result? Like, if you take a circle of any radius and inscribe an equilateral triangle, would the area outside the triangle (but inside the circle) be `4π - 3*sqrt(3)`?

I asked because my class 10 finals are coming up and this could be asked as a 1-mark question and it would have been really helpful to just memorize this. Thank you in advance. • Why at is srt of 3 over 2 equal to a/4? • HOW Can we do the opposite..
i.e. an equilateral triangle of side 24 cm has a circle inscribed ..find the remaining are after circle is cut from the triangle • 1. Draw the figure and label the sides of the equilateral triangle in this case 24.
2. Connect a radii to any part of the circumference of the circle that is touching the triangle.
3. Connect another line segment to an apex of the same side. (For simplicity draw the line segments to the base, so connect the radii you are drawing to the base of the triangle where it is being touched by the circumference of the circle)
4. A right triangle has been formed
5. This a 30 - 60 - 90 triangle because you bisected one of the original angles of the equilateral triangle
6. The radii you drew is the short side of the triangle you created.
7. So you know the longer leg is r * SQRT(3)
8. Now draw the height of the equilateral triangle. This cuts the equilateral triangle in half.
9. If you drew the 30-60-90 triangle on the base of the equilateral triangle you can see the longer leg of the 30-60-90 triangle is the same as half the base of the equilateral triangle. Half the base of the equilateral triangle is 12. So r * SQRT(3) = 12
10. So r = 12 / SQRT(3)
11. If you rationalize the fraction you get r = 12 * SQRT(3) / 3
12. So r = 4 * SQRT(3)
13. The area of the circle is (4 * SQRT(3)^2 * pi which is also 48 pi
14. So the area of the circle is 48 pi you need to find the area of the equilateral triangle.
15. Using Heron's formula the area is 24^2 * SQRT(3)/4
16. So the area is 144 * SQRT(3)
17. So when you take off the circle the area is 144 * SQRT(3) - 48 pi
(1 vote)