A geometry problem by Ajay Sambhriya

Geometry Level 2

The above shows a square A B C D ABCD of side length 12.

The points M M and N N are on the side length A B AB such that A M = M N = N B AM=MN = NB .

Find the ratio of areas between the triangle N O M NOM and the triangle D O C DOC .

1 9 \frac19 1 8 \frac18 1 3 \frac13 1 6 \frac16

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Ajay Sambhriya by Ajay Sambhriya , 27, India

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3 solutions

The areas of similar plane figures have the same ratio as the squares of two corresponding sides. Note that M N = 12 3 = 4 MN=\dfrac{12}{3}=4 . Therefore, we have

A M N O A D C O = 4 2 1 2 2 = 16 144 = \dfrac{A_{MNO}}{A_{DCO}}=\dfrac{4^2}{12^2}=\dfrac{16}{144}= 1 9 \boxed{\dfrac{1}{9}}

1 Helpful 1 Interesting 0 Brilliant 0 Confused
Ajay Sambhriya
Jan 5, 2017

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Fidel Simanjuntak
Jan 16, 2017

The ratio of the areas between two triangles is the ratio between the square of the length of their sides, so we have [ M N O ] [ D O C ] = 4 2 1 2 2 \frac{[MNO]}{[DOC]} = \frac{4^2}{12^2} which is equal to 1 9 \frac{1}{9} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You used an ASCII 2 superscript which isn't visible when rendered in LaTeX. You need to put a ^2 instead. (I went ahead and fixed this one up for you.)

Jason Dyer Staff - 4 years, 4 months ago

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Thank you, sir

Fidel Simanjuntak - 4 years, 4 months ago

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