Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 3.8 AMC 10/12 Find the answer Italy

Problem:

In an isosceles trapezoid $ABCD$ with longer base $AB$ , the diagonals are divided by their point of intersection $O$ into parts proportional to the numbers 1 and 3. Knowing that the area of triangle $BOC$ is 15, what is the area of the entire trapezoid?

Pick one

Solution

Solution:

The answer is $\mathbf{( C )}$ . Clearly triangle $AOD$ is congruent to triangle $BOC$ , so it also has area 15.

Triangles $ODC$ and $OCB$ have the same height $CH$ , and since the base $OD$ of $ODC$ is $1/3$ of the base $OB$ of $BOC$ , the area of $ODC$ is $1/3$ of the area of $BOC$ , that is $15/3 = 5$ .

Triangles $OAB$ and $AOD$ have the same height $AK$ , and since the base $OB$ of $OAB$ is three times the base $OD$ of $AOD$ , the area of $OAB$ is three times the area of $OAD$ , that is $3 \cdot 15 = 45$ .

The area of trapezoid $ABCD$ will therefore be $15 + 5 + 15 + 45 = 80$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.