2012 AMC 10A Problem 15

Question

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of

\triangle ABC?

\dfrac16

\dfrac15

\dfrac29

\dfrac13

\dfrac{\sqrt{2}}{4}

Place

A

and

B

on coordinate axes

Find the intersection of the two drawn lines

We can use coordinate geometry to figure out where the intersection of the two lines occurs.

Let

A

be the origin and

B = (1, 0).

Then the slope of the line through

A

is

-\dfrac{1}{2},

which makes the equation of the line

y = -\dfrac{1}{2}x.

The slope of the line through

B

is

2.

The

y

-intercept is

-2.

This makes the equation of this line

y = 2x - 2.

Equating the equations, we get

2x - 2 = -\dfrac{1}{2}x

x = \dfrac{4}{5}.

This makes the

y

-coordinate of

C

-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}.

The area of triangle

ABC

is then

\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}.

Thus, B is the correct answer.

Problem 15 in other years