In the given figure, the vertices
of triangle 𝐴𝐵𝐶 are 𝐴 four, six; 𝐵 one, five; and 𝐶 seven, two. A line segment 𝐷𝐸 is drawn to
intersect 𝐴𝐵 and 𝐴𝐶 at 𝐷 and 𝐸, respectively, such that 𝐴𝐷 over 𝐴𝐵 is
equal to 𝐴𝐸 over 𝐴𝐶, which is equal to one over three. Calculate the area of triangle
𝐴𝐷𝐸 and compare it with the area of triangle 𝐴𝐵𝐶.
Let’s look at two different methods
to solve this problem. The first one we’ll call the box
method. This will be helpful if we don’t
have a right triangle and we’re only given the vertices of a triangle. The first thing we do is sketch a
coordinate plane and then grab these points accurately. 𝐵 is located at point one, five
and 𝐴 at four, six, 𝐶 at seven, two.
We then notice that the figure we
were given is not drawn to scale or to the correct shape of our triangle. But back to the box method, we
wanna create a rectangle around this triangle. To find the area of triangle
𝐴𝐵𝐶, we take the area of the rectangle and then we find the area of these three
right triangles. And we subtract the areas of the
three right triangles from the area of the rectangle. But how do we do this?
We see that the distance from point
𝐵 to point 𝐶 on the 𝑥-axis is six units and the distance from point 𝐶 to point
𝐵 on the 𝑦-axis is three units. Let’s just move this formula down
to give us a little bit more room. Back to the area of the red right
triangle, the area of this triangle will be equal to one-half six times three:
one-half its height times its base.
And for the yellow triangle, the
distance from 𝐶 to 𝐴 on the 𝑦-axis is four units and the distance from 𝐴 to 𝐶
on the 𝑥-axis is three units, which means the area of the yellow triangle equals
one-half times three times four. We’ll rewrite that with
multiplication just to make it clear.
And finally, we repeat the same
process for the pink triangle. The distance from 𝐴 to 𝐵 on the
𝑦-axis is one unit and the distance from 𝐴 to 𝐵 on the 𝑥-axis is three
units. The area of the pink triangle is
one-half times three times one.
We can’t forget the area of the
rectangle. This rectangle has a length of six
and a width of four. The area of the rectangle is found
by multiplying six times four. Now, we’re ready to solve. Six times four is 24 minus one-half
times six times three which equals nine minus one-half times three times four which
equals six. One-half times three times one
equals three over two or one and a half. 24 minus nine minus six, this part
To subtract three-halves from nine,
we need to write nine with the denominator of two. The whole number nine written as a
fraction with the denominator of two is 18 over two. 18 over two minus three over two
equals 15 over two. The area of triangle 𝐴𝐵𝐶 equals
15 over two units squared.
We’re still interested in the area
of the smaller triangle, triangle 𝐴𝐷𝐸. And we want to set up a
relationship between the area of the small interior triangle, triangle 𝐴𝐷𝐸, and
the large outside triangle, triangle 𝐴𝐵𝐶. These two triangles share an
angle. On either side of this shared
angle, both of their sides are in the same proportion.
And that means triangle 𝐴𝐷𝐸 is
similar to triangle 𝐴𝐵𝐶. The ratio of the areas of two
similar triangles is their ratio squared. And that means the area of 𝐴𝐷𝐸
over the area of 𝐴𝐵𝐶 will be one over nine. We can use this to solve for the
area of 𝐴𝐷𝐸. We know that the area of our larger
triangle is 15 over two. And we’re missing the area of our
By cross multiplying 15 over two
times one, we see that 15 over two equals nine times the area of the smaller
triangle. After that, I wanna get rid of the
nine on the right side. And I can do that by multiplying
both sides of the equation by one over nine. On the right side of the equation,
we’re left with 𝑆. The numerator of the other side is
one times 15 and the denominator is nine times two. The area of the smaller triangle is
15 over 18. Both the numerator and the
denominator are divisible by three. 15 divided by three is five. 18 divided by three equals six.
The area of the smaller triangle,
triangle 𝐴𝐷𝐸, is five-sixths units squared.
Our question was asking for two
things: the area of triangle 𝐴𝐷𝐸, which we found five-sixths units squared, and
it wanted us to compare this with the area of triangle 𝐴𝐵𝐶. The comparison looks like this: the
area of the smaller triangle over the area of the larger triangle is in the ratio
one to nine.
But if you remember all the way at
the beginning of the problem, I said I was going to show you two different
methods. We looked at the box method. I wanna show you a formula for
finding the area of a triangle if you’re given coordinates.
The area 𝐴 𝑥 times 𝐵 𝑦 minus 𝐶
𝑦 plus 𝐵 𝑥 times 𝐶 𝑦 minus 𝐴 𝑦 plus 𝐶 𝑥 times 𝐴 𝑦 minus 𝐵 𝑦 over two
then take the absolute value. When it says 𝐴 sub 𝑥, it means
the 𝑥-coordinate of the 𝐴 vertex; for us that’s four. And then, we need to multiply it by
𝐵 𝑦; that’s the 𝑦-coordinate of the 𝐵 value and the 𝑦-coordinate of the 𝐶
vertex. We’ll continue with this
pattern. 𝐵 𝑥 equals one 𝐶 𝑦 equals two
minus six plus seven times six minus five all over two.
If we solve what’s in the
parentheses, five minus two equals three, two minus six equals negative four, six
minus five equals one. Now, we’ll do the
multiplication. Four times three equals 12. One times negative four equals
negative four. And seven times one equals
seven. 12 plus seven equals 19 minus four
equals 15. 15 over two take the absolute value
which equals 15 over two.
Using a different method, we’ve
confirmed that the area of the larger triangle is 15 over two units. Once you found the area of the
larger triangle using this formula, you could find the area of the smaller triangle
in the same way by setting up a proportion. You would still find that it was
one over nine. And then the area of the smaller
triangle would be five over six units squared.