Reply
Tue 17 Oct, 2006 07:15 pm
I think I got this answer right, but just wanted to see what the rest of the mathematics communtiy got for an answer. Here's the question...
Given a square ABCD of side length 1, with E on CD and F in the interior of the square so that EF is perpendicular to DC and AF is congruent to BF is congruent to EF, find the area of the quadrilateral ADEF.
1 - 1/2 the area of the triangle ABF
I'ma take a stab at this...
Triangle ABE is equilateral as per AF = BF = EF, point F being the incenter the triangle.
Thus:
Angle ABF = 30 degrees (Angle ABE is 60 degrees, bisected by segment BF).
Angle BAF = 30 degrees
Angle AFB = 120 degrees
Using sine law: (a / sin A = b / sin B),
where a = length of side BF,
A = Angle BAF, (30)
b = side AB (1), and
B = Angle AFB (120),
we get a = sqrt(3/2)
Next, get area of trapezoid AFED.
FE = AF = sqrt (3/2)
DE = 0.5
AD = 1
Area (AFED) = (0.5 (1 + sqrt(3/2))) / 2
Of course, I'm no math whiz...
AF = BF = EF = 5/8 (Pythagorean theorem on small triangle with legs 1/2 and 1-EF, and hypotenuse AF).
Quad is a trapezoid with area 1/2 * (1 + EF)/2.
Since EF is 5/8, area is 13/32.
Oops! Saw one major error on my computation...
xwraith wrote:
we get a = sqrt(3/2)
a should be 0.5(sqrt(3/2))
My final answer (I think) is:
(0.5 (1 + (1 / 2) (sqrt(3 / 2)))) / 2
= 0.4030931
markr wrote:AF = BF = EF = 5/8 (Pythagorean theorem on small triangle with legs 1/2 and 1-EF, and hypotenuse AF).
Hmm... I don't think so... this assumes that AFB is a right triangle and it's clearly not. Point F has to lie somewhere above the center of the square in order for EF to be congruent to segments AF and BF, making AFB an isosceles, obtuse triangle. Therefore, in your computation using the Pythagorean theorem, it will be incorrect to assume that the length both legs of the small triangle will be equal to 0.5
markr wrote:AF = BF = EF = 5/8
The length of line AF = BF, but there is no constraint that it is equal to EF. Thus there are infinite possible areas for AFED depending on the location of F.
stuh505 wrote:markr wrote:AF = BF = EF = 5/8
The length of line AF = BF, but there is no constraint that it is equal to EF. Thus there are infinite possible areas for AFED depending on the location of F.
From the original post:
"Given a square ABCD of side length 1, with E on CD and F in the interior of the square so that EF is perpendicular to DC and
AF is congruent to BF is congruent to EF."
xwraith wrote:markr wrote:AF = BF = EF = 5/8 (Pythagorean theorem on small triangle with legs 1/2 and 1-EF, and hypotenuse AF).
Hmm... I don't think so... this assumes that AFB is a right triangle and it's clearly not. Point F has to lie somewhere above the center of the square in order for EF to be congruent to segments AF and BF, making AFB an isosceles, obtuse triangle. Therefore, in your computation using the Pythagorean theorem, it will be incorrect to assume that the length both legs of the small triangle will be equal to 0.5
The lengths of the legs are 1/2 and 1-EF. The small triangle is half of ABF. Continue the line segment EF until it intersects AB. Call the point of intersection G. The small triangle is AGF.
I asked a friend (who's waaay better at math than me) and he arrived at the same solution as you, markr. He just went through a lot more steps.
1. Find length of BE through Pythagorean theorem = 1.118
2. Find angle BEC using sine law = 63.44 degrees
3. Find angle AEB using congruent triangles and supplementary angles = 53.123 degrees
4. Find angle FEB by bisecting angle AEB = 26.56 degrees
5. Find angle BFE using the definition of an isosceles triangle = 126.88 degrees
6. Find length of side FE using sine law = 0.625
7. Find area of trapezoid AFED = 0.40625 sq.units
Your friend made it look difficult. No trig is required.
Stumble It!
Tweet This
Bookmark on Delicious
Share on Facebook
Share on MySpace