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.
+1641 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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let G be a midpoint on AD.
Quadrilateral CDGF is a kite.
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
+129907 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
