Triangle $XYZ$ is inside square $KLMN$ shown below so that its vertices each lie on three different sides of the square. It is known that: $\bullet$ The area of square $KLMN$ is $1$. $\bullet$ The vertices of the triangle divide three sides of the square up into these ratios: $KX : XL = 3 : 2$ $KY : YN = 4 : 1$ $NZ : ZM = 2 : 3$ What is the area of the triangle $XYZ$? (Note that the sketch is not drawn to scale).
Problem
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Tags: geometry, areas, square
franzliszt
30.12.2020 20:42
Scale everything up by a factor of $5$ and then use coordinates with $N=(0,0)$. Then $Z=(2,0),Y=(0,1),X=(3,5)$. Now Shoelace and scale back down.
XxSnowWolfxX
31.12.2020 04:17
Answer: 11/50?
My thought process: I listed out the given information and used the ratio to find all the lengths of the square. I proceeded to try to find the length of $XY$ and $YZ$ and somehow use that information to find the area of $XYZ$. I soon realized that this way wasn't going to work and tried to find the area of $XYZ$ by subtracting the non-shaded areas from the area of the square. I wrote the equation of $[\triangle XYZ]$= $[KLMN]-[KYX]-[YNZ]-[XLMZ]$. I had to find the area of $XLMZ$ to actually make this equation work. So, I decided to dissect $XLMZ$ into a rectangle and a triangle called $XLMA$ and $XAZ$ (I made point $A$). It was pretty easy to solve from there since we can find out the length of $ZA$ by subtracting $ZM$ by $XL$. So I got the final answer of 11/50.
btw sorry if my handwriting is messy @below thanks!
rocketsri
31.12.2020 07:03
Notice, we have $Z = \left( \frac{2}{5} , 0 \right) , Y = \left( 0 , \frac{1}{5} \right) , X = \left( \frac{3}{5} , 1 \right)$. So, by Shoelace, the area of the triangle is $$\frac{ \frac{3}{5} \cdot \frac{1}{5} + 0 \cdot 0 + \frac{2}{5} \cdot 1 - \left ( 1 \cdot 0 + \frac{1}{5} \cdot \frac{2}{5} + 0 \cdot \frac{3}{5} \right)}{2} = \frac{ \frac{11}{25} }{2} = \boxed{ \frac{11}{50}.}$$