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Area of Equilateral Triangle

```Date: 12/06/2002 at 20:27:34
From: Javier Vacio
Subject: Equilateral Area Problem

Hi,

I really need help with this problem. You have an equilateral triangle
ABC. Extend a line from vertex B to some point near the center of the
triangle. The length of this line is 3. Now extend another line from
vertex A to that point. The length of this side is 4. Now extend
another point form vertex C to that same point. The length of this
side is 5.

What is the exact area of equilateral triangle ABC?

Thank you.
```

```
Date: 12/07/2002 at 01:18:46
From: Doctor Greenie
Subject: Re: Equilateral Area Problem

Hello, Javier -

Let P be the point where the three segments BP, AP, and CP, of
lengths 3, 4, and 5 respectively, meet.

Draw triangle ABQ congruent to triangle ACP  (i.e., make a "copy" of
triangle ACP and rotate it about vertex A until vertex C coincides
with vertex B; under this rotation let point Q be the image of point
P).

Then AQ=4 and BQ=5. Since triangle ABC is equilateral, angle CAB is
60 degrees; since triangles ABQ and ACP are congruent, angle PAQ is
also 60 degrees. Then triangle PAQ is an isosceles triangle
(AP=AQ=4) with a vertex angle of 60 degrees; triangle PAQ is
therefore equilateral, so PQ=4, and angles APQ and AQP are each 60
degrees.

With PQ=4, triangle BPQ is a right triangle, because BP=3 and BQ=5.

Then angle APB is 60+90=150 degrees; and we have the following from
the law of cosines:

(AB)^2 = 3^2 + 4^2 - 2(3)(4)cos(150deg)

= 25 - 24(-sqrt(3)/2)

= 25 + 12sqrt(3)

The area of an equilateral triangle with side s is (s^2)sqrt(3)/4, so
the area of triangle ABC is

area = (25+12sqrt(3))(sqrt(3))/4

= (25sqrt(3) + 36)/4

When I first encountered this problem, many years ago, I obtained this
result using coordinate geometry; the algebra to obtain the result by
that method would have required a response 10 times as long as this
one and involving much uglier computations. I was not at all happy
with that solution; on about the 8th or 9th time I looked at the
problem (probably two years after I first saw it) I stumbled onto the
solution provided above.

I hope this helps.  Please write back if you have any further

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/10/2002 at 19:52:43
From: Javier Vacio
Subject: Mensa: Triangle Area Problem

Hi,

Is there a geometric solution to this problem?

Thank you!
```

```
Date: 12/16/2002 at 06:01:07
From: Doctor Floor
Subject: Re: Equilateral Triangle Area Problem

Hi, Javier,

Let me try to offer you a solution. Instead of the lengths 3, 4, and
5, I used the distances d1, d2, and d3 from C, B, and A, respectively,
to a fourth point E in the interior of the equilateral triangle ABC.

First, we rotate the figure of ABC and E about C through 60 degrees,
resulting in an equilateral triangle BHC with G congruent to ABC with
E. In particular we see that CE is rotated through 60 degrees to CG,
which shows that CEG is an equilateral triangle, so that EG = CE = CG
= d1. This triangle I will now refer to as EQUI(d1): an equilateral
triangle with side d1:

Now we draw some simple conclusions:

* The point E divides the original triangle into three triangles,
which I have made into a red one AEC, a yellow one AEB, and a white
one BEC.

* The area of the white and red triangles together is the same as
the area of BEG and CEG together.

* BEG is the triangle with sides equal to the given d1, d2, d3,
which I will from now refer to such a triangle as T.

* We conclude that the white and red triangles together have an area
equal to [T] + [EQUI(d1)] (where [x] is the area of x).

With similar reasoning we can draw similar conclusions about the
areas of the white and yellow triangle together, and about the areas
of the red and yellow triangle together.

Adding the areas of white+red, white+yellow, and red+yellow gives
2*[ABC]. It also gives 3*[T]+[EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)].

So, if we read in areas, we have:

[ABC] = 1.5*[T]+ ([EQUI(d1)]+[EQUI(d2)]+[EQUI(d3)])/2.

Is this enough for a purely geometrical solution of your problem?

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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