# Thread: Area of equilateral triangle given known distances of a point from the sides.

1. ## Area of equilateral triangle given known distances of a point from the sides.

yet another question for someone willing to help for I am tired wasting four long pages....
Q.There is a point inside an equilateral triangle which is at distances
1, 2 and 3 from the three sides. The area of the triangle is
A
not uniquely determinable
B 6 root3
C 6
D 12 root3

2. Label triangle A,B,C ; P = point inside
Place D on AB, E on BC, F on AC : PD = 2, PE = 3, PF = 1

Join PA, PB, PC

anglePAF = A ; then anglePAD = 60-A
sin(60-A) / sin(A) = 2

anglePBD = B ; then anglePBE = 60-B
sin(60-B) / sin(B) = 3/2

anglePCF = C ; then anglePCE = 60-C
sin(60-C) / sin(C) = 3

Plus if sides of triangle ABC = a, then area = 3a.

Hope that helps you continue on page 5

3. Hello findmehere.genius
Originally Posted by findmehere.genius
yet another question for someone willing to help for I am tired wasting four long pages....
Q.There is a point inside an equilateral triangle which is at distances
1, 2 and 3 from the three sides. The area of the triangle is
A
not uniquely determinable
B 6 root3
C 6
D 12 root3
See the attached diagram.

Quadrilateral $\displaystyle ANXM$ is cyclic, since $\displaystyle \angle ANX = \angle AMX = 90^o$. So $\displaystyle \angle NMX = \angle NAX = \theta$, say. (Angles in same segment.)

Similarly $\displaystyle \angle NLX = \angle NBX = \phi$, say.

Now in $\displaystyle \triangle NMX, NM^2 = 2^2+3^3 - 2.2.3.\cos120^o = 19$ (Cosine Rule)

$\displaystyle \Rightarrow NM= \sqrt{19}$

So $\displaystyle \frac{\sin\theta}{2}=\frac{\sin120^o}{\sqrt{19}}$ (Sine Rule)

$\displaystyle \Rightarrow \sin\theta = \sqrt{\frac{3}{19}}$

$\displaystyle \Rightarrow \tan\theta = \frac{\sqrt3}{\sqrt{19-3}}=\frac{\sqrt3}{4}$

$\displaystyle \Rightarrow AN = \frac{2}{\tan\theta}$ (from $\displaystyle \triangle ANX$)

$\displaystyle = \frac{8}{\sqrt3}$

Similarly $\displaystyle AQ= \sqrt7$, $\displaystyle \sin\phi= \sqrt{\frac37}$, and $\displaystyle \tan\phi = \frac{\sqrt3}{2}$

$\displaystyle \Rightarrow NB = \frac{2}{\tan\phi}=\frac{4}{\sqrt3}$

$\displaystyle \Rightarrow AB = \frac{8}{\sqrt3}+\frac{4}{\sqrt3}=\frac{12}{\sqrt3 }=4\sqrt3$

So area $\displaystyle \triangle ABC = \tfrac12AB.AC\sin60^o=\tfrac12.4\sqrt3.4\sqrt3.\tf rac{\sqrt3}{2}=12\sqrt3$

Grandad

4. Originally Posted by findmehere.genius
yet another question for someone willing to help for I am tired wasting four long pages....
Q.There is a point inside an equilateral triangle which is at distances
1, 2 and 3 from the three sides. The area of the triangle is

A
not uniquely determinable

B 6 root3

C 6

D 12 root3

In an equilateral triangle,
if you know the perpendicular distances to all three sides (which is given),
then that sum will equal the altitude (h) of the triangle (which is 1+2+3 = 6).

The base (or 1 side of the triangle) is $\displaystyle 2 h \tan(30deg)$

$\displaystyle \tan(30deg) = \dfrac{ \sqrt{3} } {3}$

Thus:
$\displaystyle \text{Area} = h^2 \cdot \dfrac{ \sqrt{3}} {3}$ = $\displaystyle 6^2 \cdot \dfrac{ \sqrt{3}} {3} \,\, = \,\, 12 \cdot \sqrt{3}$

as shown previously.

.

### there is a point inside an equilateral triangle which is at a distance 1,2 and 3 from three sides. The area of the triangle is - not uniquely determinable, 6,

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