Video Transcript
Given that line 𝑀𝑋 is parallel to
line 𝐾𝐷, which of the following has the same area as triangle 𝑁𝑀𝐾? (A) Triangle 𝐶𝑁𝐻, (B) triangle
𝐶𝑍𝐻, (C) triangle 𝐻𝑁𝑍, (D) quadrilateral 𝑍𝑂𝑋𝐻, or (E) quadrilateral
𝐻𝑁𝐾𝐶.
Let’s begin by highlighting the
shape we’re looking at. Triangle 𝑁𝑀𝐾 is highlighted
here. We’re looking for a shape that has
an equal area. So let’s recall how to find the
area of a triangle. For a triangle whose height is ℎ
units and base is 𝑏 units, the area of that triangle will be equal to one-half
times the base times the height. Recall that the base of a triangle
doesn’t necessarily need to be at the bottom. That just depends on its
orientation.
So here we can label the base as the
length of line segment 𝑀𝑁. The height is then the
perpendicular distance between line segment 𝑀𝑁 and the line that has point 𝐾 on
it. We’ve already been told that the
line passing through 𝑀𝑋 is parallel to the line passing through 𝐾𝐷. This means that the perpendicular
distance between the line 𝑀𝑋 and the line 𝐾𝐷 can be labeled as ℎ units. It doesn’t matter where on the line
we’re looking. This is really useful because we
now know that any triangle created here will have a height of ℎ units. The triangle 𝐶𝑍𝐻 would have a
perpendicular height of ℎ units, as would the height of the triangle 𝑋𝐷𝑂.
We’ve shown so far that triangles
𝑁𝑀𝐾, 𝐶𝑍𝐻, and 𝑋𝐷𝑂 have the same height, ℎ units. Looking closely, we see that the
bases of these triangles have been marked out with a dash mark. This means we can go further and
say that these three triangles have the same base, 𝑏 units. This means we’ve shown that the
area of these three triangles will be the same.
However, we need to go ahead and
consider the other two quadrilaterals that were listed here. 𝑍𝑂𝑋𝐻 I’ve highlighted here in
yellow and 𝐻𝑁𝐾𝐶 in green. These quadrilaterals have one pair
of parallel sides, which means they’re trapezoids. The area of a trapezoid is equal to
one-half base one plus base two times the height. While these trapezoids do share the
same height as the triangles we’ve already mentioned, that’s the perpendicular
distance between the lines 𝑀𝑋 and 𝐾𝐷, in order for one of these trapezoids to
have the same area as triangle 𝑁𝑀𝐾, we would have to be able to prove that the
two parallel bases are equal in length to the base 𝑀𝑁. And there’s nothing on this diagram
that gives us enough information or to indicate that this would be the case.
What we can be certain of is that
the area of triangle 𝑁𝑀𝐾 is equal to the other two triangles we’ve
considered. This is option (B) triangle
𝐶𝑍𝐻. With regard to the other two
triangles in this list, triangle 𝐶𝑁𝐻 and triangle 𝐻𝑁𝐶, it is true that these
two triangles would have the same height as our three triangles we’ve already
considered. However, we have no information
about the bases of these triangles. And therefore, we cannot claim that
they have the same area as triangle 𝑁𝑀𝐾. From this list, the only triangle
that certainly has the same area as triangle 𝑁𝑀𝐾 is triangle 𝐶𝑍𝐻.
