Can anyone solve this? By the way, they mean AB=9, BC=10, and AC=13. Answer: 10

dgfgrafgdfge111 Mar 14, 2019

#1**+2 **

Here's how I did this.....there are probably more efficient ways.....

Let B = (0,0) C = (10, 0)

Construct a circle centered at B with a radius of 9

Its equation will be x^2 + y^2 = 81

Then....construct a circle at C with a radius of 13

Its equation is (x - 10)^2 + y^2 = 169

These will intersect at A

We can find the x coordinate of A thusly

Subtract the first equation from the second and we have that

(x - 10)^2 - x ^2 = 88 simplify

x^2 - 20x + 100 - x^2 = 88

-20x = -12

x = 12/20 = 3/5

And its y coordinate is

(3/5)^2 + y^2 = 81

(9/25) + y^2 = 2025 /25

y^2 = [ 2025 - 9] / 25 take the positive root

y = √[2016] / 5

Draw a perpendicular to BC from A to meet BC at E

So....EAD will be a right triangle

With AE = √[2015 ] /5

And ED = 5 - 3/5 = 22/5

These will form the legs of EAD with AD the hypotenuse

So....

AE^2 + ED^2 = AD^2

2016/25 + 484/25 = AD^2

2500 / 25 = AD^2

100 = AD^2

10 = AD

CPhill Mar 15, 2019

#2**+1 **

Well, calc wasn't exactly allowed... Is there a way to do it without a calc?

dgfgrafgdfge111
Mar 16, 2019