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## Geometry (all content)

### Course: Geometry (all content) > Unit 4

Lesson 5: Angle bisectors# Inradius, perimeter, & area

Showing that area is equal to inradius times semiperimeter. Created by Sal Khan.

## Want to join the conversation?

- So just to clarify from3:24the triangles height is R?(6 votes)
- for each small triangle in the first triangle, the height is R and the base is one of the sides. But we don't know the height of the first triangle, so we want to find the height of the smaller triangles inside it.(2 votes)

- So is the formula "r * p / 2" another way to find the area of a triangle?(3 votes)
- Yes, if you know r and p.

It is more useful to determine the inradius r from the area and the perimeter.(4 votes)

- According to my textbook you can find the inradius 'r' of a triangle with formula r=A/s, where A=area of triangle and s= semiperimeter, Is this true??(2 votes)
- Yes because if r * s (aka Perimeter/2) is equal to A, then A/r=s, and A/s=r(3 votes)

- What would we use this in our everyday lives (besides teaching others it because it is just cool to teach people about things they don't know)? I don't really know why we would need this in our everyday lives...(1 vote)
- Well, if you're working on anything that's triangular shaped (maybe a table, maybe a patio, maybe a field, maybe even a city!) and you want to place something in the center so that is equally distanced and easy to access from all sides of the triangle - they do, or at least did, this a lot with churches and city halls - you're gonna be using the incenter! Lots of stuff are triangles just because they're so easy to work with and have a lot of different utilities!(5 votes)

- This video references a concept at6:00(Pythagorean Theorem) not addressed yet as if I follow the knowledge map. Should I be following another path to get to this concept?(2 votes)
- In my opinion, it is better to follow the playlists in order on each subject than to use the knowledge map.(3 votes)

- If the inradius divides the sides of the triangle (ab,bc,ca) into the ratios m1:n1, m2:n2, m3:n3 respectively, how can I find the lengths of the sides ab, bc and ca? The area is also not provided.

:((3 votes) - So, would the diameter of that circle be called the indiameter?(3 votes)
- Well I don't think there's something like indiameter and the diameter isn't included in any of the properties (acc to me) so forget about the diameter(1 vote)

- Suggestion: perhaps add some simple questions or reviews to the more empty spaces in the triangle theory? It doesent have to be difficult, it just gets slightly tedious, and I feel that I retain information much better that way.

The theories expressed here are great, and Sahls speech is very condensed- its just the gamified learning that really "sets it in stone" for me mentally.(2 votes) - could someone please explain that for me?(2 votes)
- This whole video is technically a proof for the formula 1/2rp. If you take half of the inradius and multiply it by the perimeter, you would be able to find the area of the triangle. To find the inradius you must find the point of intersection betwwen all three angle bisectors of the triangle.(2 votes)

- If we equate area = s.r with the heron's formula we'll get r = √{(s-a)(s-b)(s-c)/s} is this always true(2 votes)

## Video transcript

We're told the triangle ABC
has perimeter P and inradius r and then they want us
to find the area of ABC in terms of P and r. So we know that the
perimeter is just the sum of the sides
of the triangle, or how long a fence
would have to be if you wanted to go
around the triangle. And let's just remind
ourselves what the inradius is. If we take the angle
bisectors of each of these vertices-- each of these
angles right over here. So bisect that right over
there and then bisect that right over there. This angle is going to
be equal to that angle. This angle is going to
be equal to that angle and then this angle is going to
be equal to that angle there. And the point where those angle
bisectors intersect, that right over there, is our
incenter and it is equidistant from
all of the three sides. And the distance from those
sides, that's the inradius. So let me draw the inradius. So when you find the distance
between a point and a line, you want to drop
a perpendicular. So this length right over
here is the inradius. This length right over
here is the inradius and this length right
over here is the inradius. And if you want, you could
draw an incircle here with the center at the
incenter and with the radius r and that circle would
look something like this. We don't have to necessarily
draw it for this problem. So you could draw a circle
that looks something like that. And then we'd call
that the incircle. So let's think about how we can
find the area here, especially in terms of this inradius. Well, the cool thing
about the inradius is it looks like the
altitude-- or this looks like the altitude for this
triangle right over here, triangle A. Let's
label the center. Let's call it I for incenter. This r right over here is
the altitude of triangle AIC. This r is the altitude
of triangle BIC. And this r, which we didn't
label, that r right over there is the altitude of triangle AIB. And we know-- and
so we could find the area of each
of those triangles in terms of both
r and their bases and maybe if we sum up the
area of all the triangles, we can get something in terms of
our perimeter and our inradius. So let's just try to do this. So the area of the entire
triangle, the area of ABC, is going to be
equal to-- and I'll color code this-- is going to
be equal to the area of AIC. So that's what I'm
shading here in magenta. It's going to be equal to
the area of AIC plus the area of BIC, which is this
triangle right over here. Actually let me do that
in a different color. I've already used the blue. So let me do that in orange. Plus the area of BIC. So that's this area
right over here. And then finally plus the area--
I'll do this in a, let's see, I'll use this pink color--
plus the area of AIB. That is the area AIB. Take the sum of the areas
of these two triangles, you got the area of
the larger triangle. Now AIC, the area
of AIC, is going to be equal to 1/2
base times height. So this is going to be 1/2. The base is the length
of AC, 1/2 AC times the height-- times this
altitude right over here, which is just going
to be r-- times r. That's the area of AIC. And then the area of BIC is
going to be 1/2 the base, which is BC, times the
height, which is r. And then plus the area of
AIB, this one over here, is going to be 1/2
the base, which is the length of this side
AB, times the height, which is once again r. And over here, we can factor out
a 1/2 r from all of these terms and you get 1/2 r times
AC plus BC plus AB. And I think you see
where this is going. Plus-- now that's a different
shade of pink-- plus AB. Now what is AC plus BC plus AB? Well that's going to
be the perimeter, P, if you just take the
sum of the sides. So that is the perimeter of P
and it looks like we're done. The area of our triangle ABC
is equal to 1/2 times r times the perimeter, which is
kind of a neat result. 1/2 times the inradius times
the perimeter of the triangle. Or sometimes you'll see
it written like this. It's equal to r times P
over s-- sorry, P over 2. And this term right over here,
the perimeter divided by 2, is sometimes called
the semiperimeter. And sometimes it's denoted
by s so sometimes you'll see the area is
equal to r times s, where s is the semiperimeter. It's the perimeter divided by 2. I personally like it
this way a little bit more because I remember
that P is perimeter. This is useful because obviously
now if someone gives you an inradius and a perimeter,
you can figure out the area of a triangle. Or if someone gives you
the area of the triangle and the perimeter, you
can get the inradius. So if they give you
two of these variables, you can always get the third. So for example, if this was
a triangle right over here, this is maybe the most famous
of the right triangles. If I have a triangle that
has lengths 3, 4, and 5, we know this is
a right triangle. You can verify this from
the Pythagorean theorem. And if someone
were to say what is the inradius of this
triangle right over here? Well we can figure out
the area pretty easily. We know this is
a right triangle. 3 squared plus 4 squared
is equal to 5 squared. So the area is going to be
equal to 3 times 4 times 1/2. So 3 times 4 times
1/2 is 6 and then the perimeter here
is going to be equal to 3 plus 4, which
is 7, plus 5 is 12. And so we have the area. So let's write this area is
equal to 1/2 times the inradius times the perimeter. So here we have 12 is equal
to 1/2 times the inradius times the perimeter. So we have-- oh
sorry, we have 6. Let me write this in. The area is 6. We have 6 is equal to 1/2
times the inradius times 12. And so in this situation,
1/2 times 12 is just 6. We have 6 is equal to 6r. Divide both sides by 6,
you get r is equal to 1. So if you were to draw
the inradius for this one, which is kind of a neat result. So let me draw some
angle bisectors here. This 3, 4, 5 right triangle
has an inradius of 1. So this distance
equals this distance, which is equal to
this distance, which is equal to 1, which is
kind of a neat result.