In triangle \(ABC\) , \(AB = 5\) ,\(BC = 8\) , and the length of median \(AM\) is 4. Find \(AC\).

tertre
Mar 31, 2018

#1**+3 **

Here's one way to do this, tertre.....but maybe not the most elegant !!!!

Let B = (0,0) C = (8,0)

Since AM is a median drawn to BC....then M is (4,0)

Now.....construct a circle with a radius of 5 centered at the origin

The equation of this circle is

x^2 + y^2 = 25 (1)

And construct a cirrcle with a radius of 4 centered at M

The equation of this circle is

(x - 4)^2 + y^2 = 16 (2)

Subtract (2) from (1) and we have that

x^2 - (x - 4)^2 = 9 simplify

x^2 - x^2 + 8x - 16 = 9

8x = 25

x = 25/8

This is the x coordinate of A

To find the y coordinate, we have

(25/8)^2 + y^2 = 25

625/64 + y^2 = 25

y^2 = 25 - 625/64

y^2 = 1600/64 - 625/64

y^2 = 975/64 take th positive root

y = sqrt (975)/ 8

So A = (25/8, sqrt (975/8)

So....using the distance formula AC =

sqrt [ (8 - 25/8 )*2 + 975/64 ] =

sqrt [ ( 39^2) / 64 + 975 / 64 ] =

sqrt (39^2 + 975) / 8 =

sqrt (2496) / 8 =

sqrt (2^6 * 3 * 13) / 8 =

sqrt (64 * 3 * 13 ) / 8

8sqrt (39) / 8

sqrt (39)

Here's a pic :

CPhill
Mar 31, 2018

#3**+1 **

Here is yet another simple method:

In the diagram of CPhill, triangle ABM is an isosceles triangle with legs of 4, 4, 5.

Using the Law of Cosines, angle ABM =51.3178, angle BAM =51.3178, angle AMB =77.3644

Triangle ABC now has legs of 5, 8, AC, with angle ABC =51.3178

Again, using Law of Cosines, AC =sqrt(39) = ~6.245....etc.

Guest Mar 31, 2018

edited by
Guest
Mar 31, 2018

edited by Guest Mar 31, 2018

edited by Guest Mar 31, 2018