A triangle ABC is inscribed within a square by joining one corner of the square with the midpoints of?

Question

the two sides not containing the corner. If the original square has edgelength x, express the area of triangle ABC as a function of x.
- Picture (http://img80.imageshack.us/img80/756/graphoh5.jpg)

I don't know where to start with this at all. Will someone please tell me their process for solving this.

Thank You!

I fixed the link http://img80.imageshack.us/img80/756/graphoh5.jpg

Answer 1

square area: x^2

area of each of the two congruent triangles: 1/2 * x/2 * x = x^2/4

area of the small triangle: 1/2 * x/2 * x/2 = x^2/8

therefore, area of the required triangle: x^2 - x^2/4 - x^2/8 = 5x^2/8

Answer 2

The lenght of longer sides of the triange can be found because they are the hypotenuse of the right angle triangle with sides of length x and x/2.
side^2 = x^2 + (x/2)^2
side = [x^2+(x^2/4)]^(1/2)
side=[x^2 *(1+1/4)]^(1/2)
side=x*(5/4)^(1/2)
the short side of the isoselece triangle is the hypotenuse of the triangle with both sides of lenth x/2. Find this length, then multiply the answer by the length of the side i calculated, then multiply by 1/2.
Rock on

Answer 3

simply start with the area of the square, and subtract the triangles that are not triangle ABC
f(x)=x^2-2(x*x/2*1/2)-(1/2*x/2*x/2)
its easier if you simplify it from here

Answer 4

First you have to put the sides of the triangle as a function of "x".

x^2 + (x/2)^2 = (AB)^2
x^2 + ((x^2)/4) = (AB)^2
(4x^2 + x^2)/4 = (AB)^2
(5x^2)/4 = (AB)^2
AB = sqrt((5x^2)/4)
AB = (x/2)sqrt(5)

(x/2)^2 + (x/2)^2 = (BC)^2
((x^2)/4) + ((x^2)/4) = (BC)^2
(x^2 + x^2)/4) = (BC)^2
(2x^2)/4 = (BC)^2
BC = (x/2)sqrt(2)

Since we know this is an isosceles triangle

AB = AC

Area of an isosceles triangle
K = b sqrt(4a^2 -b^2)/4

K = (xsqrt(2))/2 * sqrt(4((x/2)sqrt(5))^2 - ((x/2)sqrt(2))^2)/4
K = (xsqrt(2))/2 * sqrt(4((5x^2)/4) - (1/2)x^2)/4
K = (xsqrt(2))/2 * sqrt(5x^2 - (1/2)x^2)/4
K = ((xsqrt(2))/8) * sqrt((10x^2 - x^2)/2)
K = ((xsqrt(2))/8) * sqrt((9x^2)/2)
K = ((xsqrt(2))/8) * 3xsqrt(1/2)
K = ((xsqrt(2))/8) * (3xsqrt(2))/2
K = (3x^2 * 2)/16
K = (3x^2)/8
K = (3/8)x^2

ANS : (3/8)x^2

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Another way to do this is to find the area of each other triangle besides ABC

x^2 - ((2(x * (x/2))/2) + (((x/2)^2)/2 =
x^2 - (((x^2)/2) + (((x^2)/4)/2)) =
x^2 - (((x^2)/2) + (((x^2)/4)/(2/1)))
x^2 - (((x^2)/2) + (((x^2)/4)*(1/2)))
x^2 - (((x^2)/2) + ((x^2)/8))
x^2 - ((4x^2 + x^2)/8)
x^2 - ((5x^2)/8)
(8x^2 - 5x^2)/8
(3x^2)/8
(3/8)x^2

Either way

ANS : (3/8)x^2