+5 1.A,B,C,D, and E are points on a circle of radius 2 in counterclockwise order. We know AB=BC=DE=2 and CD = EA. Find [ABCDE]
Enter your answer in the form x + y sqrt(z) in simplest radical form.
2. ABCDEF is a regular hexagon with area 1. The intersection of ACE and BDF is a smaller hexagon. What is the area of the smaller hexagon?
+129873 Here's the first one:
See the way I've set it up
angle ABC = 120° ......so the area of triangle ABC = (1/2)*2^2 sin (120) = √3 units^2
And ACDE forms an isosceles trapezoid
Let the bottom base be DE = 2
A = (0,2) and C = (-√ (3) , -1) E = (2, 0) D = ( 1, -√3)
And AC forms the top base = √ [ 3 + 9 } = √12 units
Call G the mid-point of AC = (-√3/2 , 1/2 )
And F is the midpoint of DE = ( 3/2, -√3/2 )
And FG is the height of the trapezoid = √ [ ( 3 + √3)^2 + (1 + √3)^2 ] / 2 =
√ [ 4 ( 4 + 2√3) ]/ 2 = √ (4 + 2√3) = √[ 1 + √3 ]^2 = 1 + √3 units
So.....the area of the trapezoid is
(1/2) (height) (sum of the bases) =
(1/2) ( 1 + √3) [ 2 + √12 ] =
(1/2) (1 + √3) [ 2 + 2√3 ] =
( 1 + √3)^2 = [ 4 + 2√3 ] units^2
So.....the total area is
[ √3 + 4 + 2√3 ] = 4 + 3√3 units^2
+129873 2)
Here's a pic of the way I set this up :
The radius of the larger hexagon can be found as follows :
Area = 1 ....so....
3 r^2 sin (60) = 1
3r^2 * √3/2 = 1
r^2 = 2 / √27
r = √ [ 2 / √27 ]
The y coordinate of B is r * sin (60°) = √ [ 2 / √27 ] * [ √3 / 2 ]
And the y coordinate of F is r * sin (-60°) - √ [ 2 / √27 ] * [ √3 / 2 ]
And the distance between them, BF, is just 2 * √ [ 2 / √27 ] * [ √3 / 2 ] =
√ [ 2 / √27 ] * √3 =
√ [ 6 / √27]
Note that because they span equal arcs, chords AC and BF are equal
And by SSS, triangle CBA is congruent to triangle BAF and both are isosceles
And angle ACB = angle BAC = angle BAF = angle FBA
So angle FBA = angle BAC....so BH = HA
And angle CBA = angle BAE = 90°
So...subtracting equal angles , then angle HBI = angle HAM
And angle BHI = angle AHM
So...by SAS, triangle BHI is congruent to triangle AHM
But since HI is a transversal that cuts parallels DB and AE....then angle HAM = angle HIB
But angle HAM = angle HBI......so angle HBI = angle HIB
And angle HIB = angle HMA.....so angle HAM = angle HMA
So HM = AH and HI = HB
So.....HM = HI = AH = HB
So .....since triangles BHI and AHM are congruent.....then HM = HB
And by similar logic, we can prove that HM = MF
So..... HB = HM = HF
So....side HM of hexagon HIJKLM is 1/3 of BF
So....the area of the hexagon is
3 * [ 6 / √27] / 9 * √3 / 2 =
√3 / √27 =
√3 / [ 3 √3] =
1/3 units^2
