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# 2017 NS 25

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

Mar 14, 2019
edited by dgfgrafgdfge111  Mar 14, 2019
edited by dgfgrafgdfge111  Mar 15, 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 = √ / 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

So....   Mar 15, 2019
edited by 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
edited by dgfgrafgdfge111  Mar 16, 2019
#3
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I didn't use a calculator.....the final square root is easy to evaluate....   CPhill  Mar 16, 2019
#4
+2

Okay then. Thanks!!!

dgfgrafgdfge111  Mar 16, 2019
edited by dgfgrafgdfge111  Mar 16, 2019