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# hard geo

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ABCD is a unit square.  E is the center of the square, and CF is tangent to the semicircle with diameter AD.  Find the area of triangle DEF. Dec 22, 2020

#1
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ABCD is a unit square.  E is the center of the square, and CF is tangent to the semicircle with diameter AD.  Find the area of triangle DEF.

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Let G be a midpoint on AD.

Angles ADF and DCG are congruent.

DF = 2( sin∠DCG * CD) = 0.894427191

DE = 1/2(BD) = 0.707106781

∠EDF = 45º -  ∠ADF = 18.43494882º

Using the law of cosines, we can calculate the third side of the triangle DEF.

Using Heron's formula we can find the area of a triangle DEF.

[DEF] = 1/10 squared units

Dec 22, 2020
#2
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Thanks, jugoslav.....here's another approach

Note that CF  = 1     and  if we let A =(0,0).....then E  =(1/2, 1/2)

Since CF  and CD  are tangents drawn from an exterior point, the are equal

If we let the equation of the semi-circle be a circle with a radius of 1/2  centered at (1/2,0)

And let us  construct a circle centered at C =  (1,1) with a radius of 1

These two circles will intersect at F which will let us solve this easily

Equation of  semi-circle  is    (x-1/2)^2  + y^2  = (1/2)^2   ⇒   x^2 - x + 1/4 + y^2  = 1/4 ⇒

x^2 - x + y^2    = 0     (1)

Equation of the  circle  centered at C is    (x -1)^2  + (y - 1)^2  = 1^2  ⇒ x^2 - 2x + 1 + y^2 - 2y + 1   =  1 ⇒

x^2 - 2x + y^2  -2y  + 1    =  0     (2)

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

x + 2y -1   =  0

x  = 1 - 2y       (3)

Sub this into (1) to find the  y  coordinate  of F

(1 -2y)^2  - (1 -2y)  + y^2   =  0

4y^2  - 4y  + 1  - 1 + 2y + y^2   = 0

5y^2 -2y  =  0

y ( 5y - 2)    = 0

We  seek to solve this

5y  - 2    =  0

5y  =2

y = 2/5

And  using (3)    x   =1 - 2 (2/5)   =   1/5

So  the coordinates of F  are  (1/5, 2/5)

We can  now find the  three sides of triangle DEF

DE  =  sqrt [  (1/2 -1)^2  + (1/2 - 0)^2 ]  =  sqrt  [ 1/4 + 1/4] =  sqrt (1/2)  =  1/sqrt (2)

EF = sqrt [ (1/5 - 1/2)^2 + ( 2/5 - 1/2)^2 ]  = sqrt [ (3/10)^2  + (1/10)^2 ]  = 1/sqrt (10)

DF =  sqrt [ ( 1 - 1/5)^2  +  ( 2/5 - 0)^2 ] =  sqrt  [ (4/5)^2 + (4/5)^2 ]  = 2/sqrt (5)

Using the Law of Cosines

FE^2  = DE^2  + DF^2  - 2 ( DE * DF)cos (EDF)

1/10 = 1/2  + 4/5  - 2 ( (1/sqrt (2)  * 2/sqrt (5) ) *cos (EDF)

(1/10 - 1/2 -4/5)

_____________    = cos EDF  =  3/sqrt (10)

4/sqrt (10)

And sin  EDF  =  sqrt (1 - (3/10)^2 )  =    sqrt  ( 1 - 9/10)  = sqrt (1/10)

So  area of DEF  =   (1/2)DE * DF sin (EDF)  = ((1/2) (1/sqrt (2) * 2/sqrt (5)) ( 1 /sqrt (10)  =

(1/sqrt (10) )^2   =   1/10   Dec 23, 2020