Video Transcript
In this video, our topic is finding
the areas of rectangles and triangles. We’ll see how to calculate these
areas mathematically and will also learn how these two shapes, rectangles and
triangles, can represent many different physical scenarios. The place we can begin is in
recalling, perhaps from math class, the mathematical formula for the area of a
rectangle. When we have a shape that is a true
rectangle, that means that all the sides are at right angles to one another. And then if we label the height of
this rectangle ℎ and then say we call the base of this rectangle 𝑏, then the area
of the shape, we’ll call it 𝐴 sub 𝑟 is equal to the product of the base and the
height. The area is equal to 𝑏 times
ℎ.
On our diagram, this area
represents the entire interior space of this rectangle. So far, so good. But what if we then draw a line
from one corner of a rectangle to the other opposite that corner? So it would look something like
this. In that case, we can see that we’ve
created two identical triangles. This one right here and this one
right here. And the area of each one is just
half the area of the rectangle. If we focus on the lower triangle,
for example, we can see that it has the same height and the same base as the
rectangle whose area we calculated. But the equation for this
triangle’s area, and we’ll call it 𝐴 sub 𝑡, is not 𝑏 times ℎ, but it’s one-half
that, one-half the base of the triangle times its height.
Now, just as an interesting side
note, even if our triangle looked different than it does here, for example, say that
instead of having this, this, and this as the three legs in the triangle, it’s three
legs were here, here, and here. This new triangle we’ve drawn is
not a right triangle anymore. But nonetheless, it’s total area is
still one-half its base multiplied by its height, where the base of the triangle is
still 𝑏 and it’s total maximum height is still ℎ. Now to this point, we’ve been
treating the dimensions, the base, and the height of these shapes very
abstractly. We’ve assigned them these
variables, 𝑏 and ℎ, and we’ve left it at that. But in a real-life situation, these
variables would have numbers as well as units attached to them.
For example, let’s say that this
rectangle represented the area of a grassy field. And say that this field is 50
metres wide and is 25 metres deep. If we want to calculate the total
area of this field, the total area of the rectangle that represents it, we know
mathematically how to do that. But notice that when we plug in the
actual values for the base and the height of this rectangle, we’re actually going to
be multiplying twice. First, we’ll multiply the numbers
together, 50 and 25, and then we’ll also be multiplying the units, metres and
metres. And we can even think of doing this
separately, multiplying the numbers together first and then the units.
Out-front, we would have the
numbers 50 times 25 then followed up by the units, metres times metres. 50 times 25 is 1250 and a metre
times a metre is a metre squared. So then reading it altogether, the
total area of this field is 1250 square metres. And if we were to divide this
field, using a diagonal line that goes from one corner to an opposite corner, using
our equation for the area of a triangle, we could then calculate that area. The important thing to remember is
that when we use these equations and have real physical values to substitute in for
𝑏 and ℎ, we’ll want to be careful to combine not just the numbers but also the
units of those values.
Now so far, we’ve only considered
using these equations when the units of both the terms involved, the base and the
height, either of the rectangle or the triangle are the same. In this example here, the units for
both the base and the height are metres, 50 metres and 25 metres. But what if those units weren’t the
same? What if the height of our shape
represented one physical quantity and the base represented another? Let’s consider an example of that
happening.
Say that we have this ball and that
we roll the ball along the ground so that it moves in a constant speed of three
metres per second. And say further that we let this
ball roll for 10 seconds, and over that time it keeps the same speed. So let’s write that maximum time
down and we’ll call it 𝑡. Knowing all this, what we could do
is then plot the speed of this ball on an axis against the time during which it
travelled. Here’s what that could look
like. Say we have a two-dimensional plot
and on the vertical axis, we have speed in metres per second while on the horizontal
axis we plot time in seconds. Marking out the vertical axis,
let’s say that this represents one metre per second, this represents two, and this
represents three. And let’s say further that our
horizontal axis is marked out like this with time going from zero seconds up to 10
seconds.
At this point, we’re able to plot
the motion of this ball as it rolled up three metres per second for 10 seconds. We could say that the ball starts
rolling at a time of 𝑡 equals zero seconds. And then it moves along at a
constant speed of three metres per second for 10 seconds. Looking at this graph, we might
just be seeing a horizontal line representing our ball speed. But look at this. If we sketch in a dashed line
vertically upward from the maximum time at which our ball is rolling, then what we
now have, enclosed by our curve, this dashed, line and the two axes, is a
rectangle. We know it’s a rectangle because
all of the sides of this shape are at right angles to one another. If we look at the base and the
height of our rectangle, we see something interesting. The rectangle’s base, that’s this
section right here, has units of seconds and it’s 10 seconds long, whereas the
rectangles height, this space here, has units of metres per second.
So in the language of our
mathematical equations, where the area of a rectangle is represented by a base and a
height, we could say that our base is 10 seconds, but our height is three metres per
second. So then what will happen if we try
to calculate the area of this rectangle. Well before we do, let’s recall a
relationship between speed, time, and distance. If an object is moving at a
constant speed, we’ll call it 𝑠, then that speed is equal to the distance the
object travels divided by the time it took to travel that distance. And if we rearrange this equation,
if we multiply both sides of the equation by time 𝑡, then we see that that term
cancels out on the right-hand side, giving us the result that speed times time is
equal to distance.
Let’s keep that in mind as we move
ahead to solve for the area of our rectangle, what we’ve called capital 𝐴. Based on our formula, to calculate
this area, we’ll take our base, 10 seconds, and we’ll multiply it by our height of
three metres per second. Now, as we’ve seen, the units of
these two terms are not the same. But that’s okay. We’re multiplying them together and
the units multiply just like the numbers do. Let’s go about this multiplication
this way. First, we’ll combine the numbers
and then, separately, we’ll combine the units. So the area of a rectangle is equal
to 10 times three; that’s 30. And its units are equal to seconds
multiplied by metres per second. We can see then that the units of
seconds cancel out because they appear both in top and bottom, leaving us simply
with units of metres; that is units of distance.
Now, as we look back up to our
equation for distance, that is equal to speed times, time, we see that our result
makes sense. We’ve multiplied a speed, three
metres per second, by a time, 10 seconds. And in doing that, we’ve solved for
a distance 30 metres. And here’s the meaning of our
result. Here’s what this 30 metres
represents. Not only does it represent the
total area of this rectangle on our curve, but it also tells us the total distance
that the ball travelled over this 10-second interval. All this means that we were able to
solve for a meaningful physical quantity the total distance travelled by this object
by representing the objects motion on a graph and in calculating the area of that
shape.
And what’s perhaps even more
interesting, let’s say that the speed of our ball wasn’t constant. Let’s say that it varied over this
10-second interval. Specifically, let’s say that the
ball speed looked like this, where it started out at zero and then it increased up
to this maximum speed of three metres per second and then that its speed steadily
decreased back to zero at a time of 10 seconds. If the ball speed had varied like
this, we could still use this area method to solve for the total distance that it
travelled. In this instance though, we would
calculate the area of this triangle rather than the area of the rectangle we solved
for earlier. So it is possible for the base and
the height of our shapes to have different units. And in fact, when this happens,
that can help us answer questions about different physical scenarios. Let’s get a bit of practice with
these ideas now using an example.
A wall of a house has a length 𝑙
equals 12 metres and a height ℎ equals 5.5 metres, as shown in the diagram. What is the area of the wall to the
nearest square metre?
Okay, so here’s our wall, and we’re
told the length of this wall as well as its height. We’re told the length is 12 metres
and the height of the wall is five and a half metres. To the nearest square metre, we
want to solve for the area of this wall. If we label that area capital 𝐴,
then it represents the entire interior space of this rectangle. And we know that this is a
rectangle because, based on our diagram, all the sides of the shape are at right
angles to one another. In order to solve for this area of
the wall, let’s recall the mathematical formulation for the area of a rectangle. If we call that area 𝐴 sub 𝑟,
then it’s equal to the length of a rectangle 𝑙 multiplied by its height ℎ.
By the way, another way to write
this equation is 𝑏 times ℎ, where 𝑏 is the base of the rectangle. But here, since we have a length 𝑙
representing that same distance, we’ll use 𝑙 in our equation. So in general, the area of a
rectangle is equal to its length times its height. And for our rectangle, this wall,
that area is equal to the specific length and the specific height given multiplied
together. When we substitute in the given
values for 𝑙 and ℎ, we see that each one includes a number as well as a unit
attached to that number. When we multiply the length times
the height, we can deal with these two quantities separately. What we’ll do first is will
multiply the numbers together and then second will combine the units.
So then we can write the area this
way. It’s equal to 12 times 5.5. That’s the combination of the
numbers. And then the units are metres times
metres. If we multiply 12 by five and a
half, that’s equal to exactly 66. And then if we multiply a metre by
a metre, that’s equal to a metre squared. So we’ve combined both the numbers
to give us 66 and the units to give us metre squared. Our question asked for the area of
the wall to the nearest square metre. And looking at our results, we see
that it’s written that way. So then the area of this wall is 66
metres squared.
Let’s look now at a second example
exercise.
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.
Let’s summarize now what we’ve
learned about finding the areas of rectangles and triangles. Starting off, we saw that for a
rectangle with base 𝑏 and height ℎ, its area 𝐴 is equal to 𝑏 times ℎ. On the other hand, for a triangle
with base 𝑏 and height ℎ, its area is given is one-half 𝑏 times ℎ. And lastly, we saw that rectangle
and triangle area can be used to answer questions about physical processes, as we
saw in the case of the ball, rolling along at a steady speed for some amount of
time.
