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The shape in question

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 = 2/3 what is the length of YZ.

I worked the area of trapezium ABYZ to equal YZ: Area of square not covered by triangle = (1-YZ)(1) reason (rectangle Length*Breadth) Therefore area of trapezium = 1-(Area of rectangle) = 1-(1-YZ) = YZ. Area of trapezium = [(a+b)/2]*h Therefore: YZ = (1+YZ)/2 (h=1) YZ = 1.

Where did i go wrong.

EDIT please do not give the answer

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2 Answers 2

up vote 1 down vote accepted

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

enter image description here

$AB=BC=CD=DA=1$

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

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