# Length of Line in a Square

Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is one and area $XYZ = \frac{2}{3}$ what is the length of $YZ$.

I worked the area of trapezium $ABYZ$ to equal $YZ$:

$\text{Area of square not covered by triangle} = (1-YZ)(1)$

reason ($\text{area of rectangle}=\text{Length}\times \text{Breadth}$)

Therefore $\text{area of trapezium} = 1-(\text{Area of rectangle}) = 1-(1-YZ) = YZ$.

$\text{Area of trapezium} =\left(\frac{a+b}{2}\right)h$ Therefore: $YZ = \frac{(1+YZ)}{2} (h=1)$

$YZ = 1$.

Where did I go wrong.

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Here is your mistake: You worked out the area of the trapezoid by the formula for the area of the rectangle. However, a trapezoid is not a rectangle, so the area of the trapezoid is not equal to YZ.

You were in the right direction though. One way to go about it would be to subtract from 1 the area of the triangles ADY and BCZ... Try it :)

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Area of trapezium = [(a+b)/2]*h What is wrong with that? –  fosho Sep 26 '12 at 8:32
You wrote that uncovered area = (1-YZ)(1). This is in fact twice what you needed. –  Karolis Juodelė Sep 26 '12 at 8:33
You are right in that part. But you have a mistake in figuring that the area of the square not covered by triangles is 1 - YZ. The two triangles don't add together to form a rectangle; the angles are bent wrong. –  Yoni Rozenshein Sep 26 '12 at 8:34
Oh i see what you are saying... thanks –  fosho Sep 26 '12 at 8:35

$AB=BC=CD=DA=1$

Let, $|YZ| = x$ and Length of perpendicular drawn from X to YZ = $u$

So, $\frac{1}{2}\cdot(x)\cdot(u) = \frac{2}{3}$ .................(Eq. 1)

and using similarity rule of two triangles $\triangle XYZ$ and $\triangle ABX$,

$\frac{u}{1+u}=\frac{x}{1}$....................................(Eq. 2)

From the above two equations, you can find $|YZ|$

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I SAID DO NOT give answer –  fosho Sep 26 '12 at 8:42
Ohh. I'm sorry. –  Sumit Bhowmick Sep 26 '12 at 8:45