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# Geometry

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Find BC.

Apr 15, 2022

#1
+14231
+1

Find BC.

Hello Guest!

$$f(x)=\sqrt{22^2-x^2}\\ g(x)=\sqrt{13^2-x^2}+13\\ \sqrt{22^2-x^2}=\sqrt{13^2-x^2} +13\\ 22^2-x^2=13^2-x^2+2\cdot 13\cdot \sqrt{13^2-x^2} +13^2\\ 13^2-x^2= (\frac{22^2-13^2-13^2}{26})^2 \\ x=\sqrt{13^2- (\frac{22^2-13^2-13^2}{26})^2}$$

$$x=11.725\\ \overline{BC}=2x$$

$$\overline{BC}=23.449$$

!

Apr 16, 2022
#2
+124696
+1

13  is the circumradius  = C

C  = 13  =    product  of the sides / [  4 Area ]

Let x =  BC

13  =           (22 * 22 * x)

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sqrt  [  (22 + 22 + x ) ( 22 + 22 - x) ( x + 22 - 22) (x + 22 - 22) ]

13  =         [ 22 * 22 * x  ]

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sqrt  [ ( 44 + x)  ( 44 - x)  (x^2) ]

13 =     [ 22 * 22 * x ]

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x * sqrt [ (44 + x) * (44 - x) ]

sqrt [ (44 + x) (44 - x) ] =    484   /  13

sqrt [ 1936 - x^2 ]  =  484/13       square both sides

1936  -  x^2   =    (484)^2 / 169

x^2  =   1936   -  (484)^2 /  169

x^2  = 92928  / 169

x = sqrt [ 92928 / 169 ]  ≈  23.449  = BC

Apr 16, 2022