Geometric ratio

Geometry Level 3

The diagonals of $ABCD$ intersects at $O$ .
The ratio of the areas of the triangles $\bigtriangleup AOB : \bigtriangleup BOC : \bigtriangleup COD : \bigtriangleup DOA$ is $1:2:3:x$ . What is $x$ ?

1.5 3 2.5 4 2

1 solution

Ashraful Mahin
Jun 10, 2016

We are given that $\bigtriangleup COD:\bigtriangleup BOC=3:2$ .
Consider the base along line $BOD$ .
Since these two triangles have the same height (distance from C to BOD), hence the ratio of their areas is the ratio of their base.
Thus $DO:BO = 3:2$ .

Similarly, tirangles $AOD$ and $BOA$ have the same height (distance from A to BOD), hence the rato of their areas it the ratio of their base, so $x:1 = \bigtriangleup AOD:\bigtriangleup BOA=3:2$ . Thus, this gives $x = \frac{3 \times 1 }{ 2}$ .

FYI There is no need to use distinct Latex brackets for each term of the equation. You can simply put the Latex brackets around the entire equation. I've edited the problem for your reference.

Calvin Lin Staff - 4 years, 12 months ago

Oh thanks sir,that will very nice for me.

Ashraful Mahin - 4 years, 12 months ago

Geometric ratio

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