Video Transcript
A rectangular wall shown in the
diagram has a length 𝑙 equals 18 metres and a height ℎ equals 11 metres. The part of the wall shown in
orange is in the shade, and the part of the wall shown in white is in direct
sunlight. What is the area of the wall? Answer to the nearest square
metre. What is the area of the part of the
wall in direct sunlight? Answer to the nearest square
metre.
All right, so we’re looking at
these two separate questions. What is the area of the wall and
what is the area of the part of the wall in direct sunlight? Looking for a moment at our
diagram, we see the sketch of the wall. And we’re told that the orange
part, this wedge right there, is the part of the wall that’s in the shade, whereas
the white part, the part of the wall here shown in these dashed lines, is in direct
sunlight. Along with this, we’re told what
the values of the length and the height of the wall are. Based on all this, first off, we
want to solve for the area of this wall.
As we work to figure out what that
area is, a helpful clue is that we’re told the wall is rectangular and that it has
the length and height marked out on our sketch. This means when we calculate the
overall area of the wall, we’ll include the orange section, and we’ll also include
the white section, the sections and shade and in sunlight. It’s this total combined area that
makes up the area of the wall. Because our walls are rectangle, we
can recall the mathematical equation for the area of a rectangle. We’ll call it 𝐴 sub 𝑟. That’s often written as the base of
the rectangle 𝑏 multiplied by its height ℎ. Now in our case, the base 𝑏 is
represented by this variable 𝑙 which stands for length.
Because it’s just a different name
for the same thing, we could write that into our equation. The area of a rectangle is equal to
its length times its height. And let’s give this area a
name. Let’s call it capital 𝐴. Based on our equation, this area is
equal to the length of a rectangle multiplied by its height. And both those values are given to
us. The length is 18 metres and the
height is 11 metres. With those values substituted in,
our next step is to combine them through multiplication. Before we do that though, notice
that each one of these values has both a number as well as a unit attached to that
number. When we multiply the values
together, we can treat these two things separately. First, we can combine the numbers,
18 times 11, and then the units, metres times metres.
When we multiply 18 times 11,
that’s equal to 198. And then a metre multiplied by a
metre is a metre squared. And this then is the total area of
our rectangular wall. To the nearest square metre, it’s
198 square metres. Now, let’s move on to the second
part of our question, which asks about the area of the part of the wall that’s in
direct sunlight. Going back to our problem
statement, we’re told that the area that’s shown in white is the area in direct
sunlight. So that’s this area here on our
sketch. Now, because our wall is a
rectangle, that means that this angle right here is a 90-degree angle. All the interior angles of a
rectangle are. So that means that this area and
white, which we can see is triangular, is a right triangle. And not only that, but the length
of this triangle is equal to 𝑙. And its height is equal to ℎ.
At this point, we’ll want to recall
the mathematical equation for the area of a triangle. If we call that area 𝐴 sub 𝑡,
it’s equal to one-half the length of the triangle 𝑙 multiplied by its height ℎ. And notice that this equation for
the area of the triangle is the same as the equation for the area of the rectangle,
except for now multiplying by one-half. This means that there are two ways
we could go about answering this second question, about solving for the area of the
part of the wall in direct sunlight. One way would be to take the answer
we calculated earlier, 198 metres squared, and divide it by two.
Based on our equations for the area
of a triangle and the area of a rectangle, when they share a common exterior, as our
rectangle and triangle do in this case, then the triangle’s area as we saw is half
the rectangles area. And a second way to solve for the
triangle’s area is simply to use this equation here. If we go about answering the
question the second way, then we’ll be multiplying one-half by the length of our
triangle, 18 metres, and by its height, 11 metres. And if we separate out our numbers
and our units, then we have one-half times 18 times 11 all multiplied by a metre
times a metre. The number of part of it, one-half
times 18 times 11, is equal to 99. And then like we saw earlier, a
metre times a metre is a metre squared. And that is our answer. The area of the wall in direct
sunlight is 99 metres squared.
