# Thread: Fractions of area - the pained pane

1. ## Fractions of area - the pained pane

Last question i need help with:

THE PAINED PANE:
A triangular piece of glass (labeled XYC) is cut from a rectangular pane (labled ABCD)

What fraction of the area of ABCD is XYC?

What I know:
Points X and Y are the midpoints of the two sides of the rectangular pane.

Area of Triangle XBY is:

$Area \ XBY = \frac{1}{2}.\frac{1}{2}x.\frac{1}{2}y$

$Area \ XBY = \frac{1}{8}xy$

Area of Trapezium AXCD

$Area \ AXCD = \frac{1}{2}(y+\frac{1}{2}y)x$

$Area \ AXCD = \frac{3}{4}xy$

That means that the total area that HASN'T been cut out is...

$AreaXBY + Area AXCD=\frac{7}{8}xy$

The area of the WHOLE rectangle is $Total \ Area=xy$

That means that the fractional area of triangle cut out is...

$xy-\frac{7}{8}xy=\frac{1}{8}xy$

That means that the area of XYC is $\frac{1}{8}$ of Area ABCD

This is the first time I've helped someone so I hope its correct!

3. maybe u misread me but the triangle is XYC.

Ps. i really appreciate your help thankies

4. That's the one I worked out! =P

Well by subtracting the other two areas around it

5. ok.. but it seems confusing Your answer is correct and I know that but I don't know how u got it. can you explain plz? thankies

6. Hello, anthmoo!

Let $L \,= \,AB \,= \,CD$, the length of the rectangle.
Let $W \,= \,AD \,= \,BC$, the width of the rectangle.

Tip your head to the left and look at $\Delta XYC.$

. . The base is: $CY \,=\,\frac{1}{2}BC \,=\,\frac{1}{2}W$
. . Its height is: $XB \,=\,\frac{1}{2}AB\,=\,\frac{1}{2}L$

The area of $\Delta XYC$ is: . $\frac{1}{2}(CY)(XB)\:=\:\frac{1}{2}\left(\frac{1}{ 2}W\right)\left(\frac{1}{2}L\right) \:=\:\frac{1}{8}(LW)$

The area of the triangle is one-eighth of the area of the rectangle.

7. Originally Posted by puppy_wish
ok.. but it seems confusing Your answer is correct and I know that but I don't know how u got it.
I was scared of this...ok:

Work through my solution step by step by looking at the diagram...

Step 1: I found the area of triangle XBY (the triangle surrounded by the points X, B and Y). This area is $\frac{1}{8}$ the area of the whole thing (this is not XYC but the areas are awkardly the same...).

Step 2: I found the area of the trapezium AXCD (you see I'm finding the area of everything EXCEPT the area of XYC). This turned out to be $\frac{3}{4}$ the area of the whole thing. This can be said as $\frac{6}{8}$ the area of the whole thing, to make it easier later.

Step 3: I added the two areas together which is $\frac{1}{8} + \frac{6}{8} = \frac{7}{8}$. That is the area of EVERYTHING but the pane of glass we are cutting out.

Step 4: We can now work out the area of pane of glass by $\frac{8}{8} - \frac{7}{8} = \frac{1}{8}$. That means that fraction area of the pain of glass is $\frac{1}{8}$ that of the rectangular area of the pane of glass.

8. THANK YOU SO MUCH BOTH OF YOU! This really helped me. THANK YOU AGAIN, you guys really rok!!