# quotient between areas of triangles

i am confused for answer from the following link

http://www.naec.ge/images/doc/EXAMS/exams-2011-gat-5-ivlisi.pdf


problem 67,problem states that quadrangular ABCD has A and C vertex on y axis,and coordinates of B and D is given ,we are asked to find ratio between are ABC and ADC,from given figure it is clear that it will be ration between height of both triangle or 5.4/3 which is equal 1.8,but answer is different 1.5 why?thanks

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Sorry, the answer $1.5$ is right, $1.8$ is wrong.

Look at $\triangle ABC$, viewed as having base $AC$. Then its height is the perpendicular distance from $B$ to the $y$-axis. this perpendicular distance is the absolute value of the $x$-coordinate, namely $9$.

Why? Imagine dropping a perpendicular from $B$ to the $y$-axis. The length of the perpendicular is the distance you must travel in the $x$-direction to get to the $y$-axis. that's $9$.

Similarly, if you view $\triangle ADC$ as having base $AC$, the height is $6$.

The ratio of the areas is therefore $9/6$.

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You have the right idea, but you’re using the wrong heights. The side that the two triangles share is $\overline{AC}$. Measured from this side, the height of $\triangle ABC$ is 9, and the height of $\triangle ADC$ is 6, so the ratio is $\frac{9}{6} = 1.5$.

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great thanks for everyone i have found my mistake –  dato datuashvili Jul 13 '11 at 4:51

The area of ABC is $\frac{\overline{AC} \times 9}2$ and the area of $ABD = \frac{\overline{AC} \times 6}2$, which explains the ratio. You were using the $y$ coordinates instead of the $x$'s.

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