# Thread: Pinding perimeter of triangle in surd form

1. ## Pinding perimeter of triangle in surd form

The points A and B lie on the line y=12x / 5
The point D divides the interval AB into the ratio AD : DB = 3 : 2
C is the point (24, -10)
Area of triangle is 845 square units.

I have found that

CD = 26 (I am assuming D is perpendicular C)

by using | ax + by + c| / sqrt(a^2 + b^2) [I first put the line equitation into general form then plugged away]

AB = 65

From 1/2 x AB x CD = 845

I do not know how to find the perimeter of this triangle, as shown below:

Any help would be great. Exam tomorrow

2. Hello RAz
Originally Posted by RAz
The points A and B lie on the line y=12x / 5
The point D divides the interval AB into the ratio AD : DB = 3 : 2
C is the point (24, -10)
Area of triangle is 845 square units.

I have found that

CD = 26 (I am assuming D is perpendicular C)

by using | ax + by + c| / sqrt(a^2 + b^2) [I first put the line equitation into general form then plugged away]

AB = 65

From 1/2 x AB x CD = 845

I do not know how to find the perimeter of this triangle, as shown below:

Any help would be great. Exam tomorrow
Your working is fine so far, on the assumption that $CD$ is perpendicular to $AB$.

You now need to find the distances $AD$ and $BD$, which you can do like this:

$ADB=3:2 " alt="ADB=3:2 " />

$\Rightarrow AD:AB = 3: 5$

$\Rightarrow AD = \tfrac35AB = 39$

$\Rightarrow BD = 26$

Now use Pythagoras' Theorem on $\triangle ADC: AC^2 = 39^2+26^2$

$= 13^2(3^2+2^2)=13^2\times13$

$\Rightarrow AC = 13\sqrt{13}$

and on $\triangle BDC: BC^2=26^2+26^2=26^2\times 2$

$\Rightarrow BC = 26\sqrt2$

So the perimeter of $\triangle ABC = 65+13\sqrt{13}+26\sqrt2 = 13(5+\sqrt{13}+2\sqrt2)$