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# Hard geometry

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In acute triangle $ABC,$ $\angle A = 45^\circ.$  Let $D$ be the foot of the altitude from $A$ to $\overline{BC}.$  if $BD = 2$ and $CD = 4,$ then find the area of triangle $ABC.$

Jan 4, 2024

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Since A = 45.....then B + C = 135  ⇒   C = 135 -B

tan B =  AD / BD = AD/ 2   ⇒  AD =  2tan B

tan C =  AD/CD =  AD/ 4   ⇒  AD =  4tan C

2 tan B  = 4 tan C    →  tan B = 2 tan C

So

tan B  =  2 tan (135 - B)

tan B  =  2 [ tan 135 - tan B  ] /  [ 1 + tan 135 *tan B]

tan B  = 2  [ -1 - tan B ] /  [ 1 -tan B ]

tan B [ 1 - tan B ]  =  -2 [ 1 +tan B ]

tan B [ tan B - 1 ]  = 2 [ 1 + tan B]

tan^2 B - tan B = 2 + 2tan B

tan^2 B - 3tan B - 2 =  0         let x  = tan B

x^2 - 3x   = 2

x^2 - 3x  + 9/4 =  2  + 9/4

(x -3/2)^2  = 17/4        take the positive root....the negative root produces a negative angle for B

x - 3/2  = sqrt (17) / 2

x  =  [ 3 +sqrt (17 ]  / 2  = tan B

So

tan B =  AD / BD =  AD / 2

So

AD / 2 =  [ 3 + sqrt (17) ] / 2

AD =  [3 + sqrt (17)]

[ ABC ]   =   (1/2)BC * AD  =   (1/2)(BD +CD) ( AD) =  (1/2) (6) ( 3 + sqrt (17) )  =

3 ( 3 + sqrt (17) )  =

9 + 3sqrt (17)

Jan 4, 2024