### Video Transcript

In this video, we will learn how to find
the perimeters of composite figures. A composite figure is one which is made
up of two or more shapes. So, for example, on the screen here, we
have a composite figure composed of two rectangles. We can also create more complicated
composite figures involving triangles, semicircles, or other two-dimensional shapes, such as
parallelograms or trapezoids. For example, the one on the right of the
screen, which is composed of a triangle and a semicircle.

Letβs begin by reviewing some of the key
formulae we need for calculating the perimeters of two-dimensional shapes. And at this point, weβll remember also
that the perimeter of a shape is the distance all the way around its edge.

Firstly, a square. A square has four sides all of equal
length. So if we let the side length of the
square be π units, then the perimeter will be equal to four π units. Thatβs π plus π plus π plus s. Secondly, a rectangle. In general, this will have a length π
units and a width π€ units. So the perimeter, the distance all the
way around its edge, is two π plus two π€. Thirdly, for a general triangle, if we
label the three sides as π, π, and π units, then the perimeter will be equal to π plus
π plus π units. In the special case of equilateral
triangles, where all three sides are the same length, then the perimeter can be written as
three π , where π is the side length of the triangle.

Next, a circle. The perimeter, which we more usually call
the circumference, of a circle is given by ππ or two ππ, where π is the diameter of the
circle and π is the radius. And finally, for now, a semicircle whose
perimeter is made up of a curved portion or arc and a straight line. The curved part is half the circumference
of the full circle. So we can write this as π times the
radius. And the straight part is the diameter of
the circle, which we can write as two π, giving a perimeter of ππ plus two π.

Now that weβve reviewed the key formulae
weβll need, letβs have a look at some examples.

Determine the perimeter of the shape
below.

We can see that the shape weβve been
given is a composite figure. Itβs made up of these two rectangles
joined together, or perhaps these two rectangles here. We could also think of it as a larger
rectangle, which has had a smaller rectangle cut out of it. In any case, we need to determine its
perimeter.

Now, when calculating the perimeter of a
shape such as this one, itβs a good idea to start in one corner and trace our way all the
way around the edge of the shape to make sure we donβt miss out any of the lengths. Letβs start at point π΄. The perimeter will be equal to π΄π΅ plus
π΅πΆ plus πΆπ· plus π·πΈ plus πΈπΉ plus πΉπ΄. And thatβs all the lengths we need to
include as weβre now back at our starting point.

Weβve been given on the diagram the first
four of these lengths. They are five, seven, three, and three
centimeters, respectively. But we havenβt been given the lengths
πΈπΉ or πΉπ΄. We can work them out though. Firstly, πΈπΉ will be the difference
between the two vertical sides of this figure, π΄π΅ minus πΆπ·. Thatβs five minus three, which is equal
to two centimeters. πΉπ΄ will be the difference between the
horizontal sides of the figure. Thatβs π΅πΆ minus π·πΈ, seven minus
three, which is equal to four centimeters.

So we now have the lengths of all six
edges of our composite figure, and so we can add them together. Five plus seven plus three plus three
plus two plus four is equal to 24. The units for this perimeter, which is a
length, will be the same as the units for the individual lengths. So our answer is 24 centimeters.

Now notice that this perimeter is
actually the same as the perimeter of the full rectangle π΄π΅πΆπΊ if we hadnβt removed the
smaller rectangle πΉπΈπ·πΊ. And the reason for this is that πΈπ· is
the same as the length we removed, πΉπΊ, and πΈπΉ is the same as the length we removed,
π·πΊ.

So for this reason, we could actually
have calculated the perimeter of this particular composite figure using the formula twice
the length plus twice the width for the perimeter of the original rectangle π΄π΅πΆπΊ.

Letβs now consider an example involving
circles.

Determine the perimeter of the figure,
using 22 over seven to approximate π.

The figure weβve been given is a
composite figure. Itβs composed of two semicircles attached
to either side of a rectangle. Tracing our finger all the way around the
edge of the shape from a given point, we see that the perimeter is composed of a straight
edge; a semicircular arc; another straight edge, the same length as the first; and a second
semicircular arc, the same length as the previous one.

From the figure, we can see that the
length of each straight edge is 42 centimeters. So we can substitute these values
directly. But what about the length of these two
semicircular arcs? Well, together, these two arcs will form
a full circle. And we know the formula for calculating
the perimeter or circumference of a circle is π times its diameter. From the figure, we can see that the
diameter of our circle is 49 centimeters. So the circumference of the circle is
49π.

Now weβre told in the question that we
should use 22 over seven as an approximation for π. This gives 49 multiplied by 22 over
seven. And then we can cancel a factor of seven
in the denominator with a factor of seven in the numerator, giving seven multiplied by
22. We can work this out using any
multiplication method we choose. Here Iβve shown the column method, and it
gives 154. So the perimeter of the figure becomes 42
plus 154 plus 42. Thatβs 238. The units for the perimeter are the same
as the units given for the individual lengths in the question. So they are centimeters. Notice that because we used 22 over seven
as an approximation for π, we had no need to use a calculator in this question.

Letβs now consider another example with a
composite figure composed of a triangle and a semicircle.

Use 3.14 to approximate π and calculate
the perimeter of the figure.

In this question, weβve been asked to
calculate the perimeter of a composite figure which looks a little bit like an ice cream
cone. We have a semicircle which sits on top of
a triangle. Notice that the dividing line between
these two shapes β thatβs the third side of the triangle or the straight edge of the
semicircle β is not part of the perimeter because it isnβt part of the outside of the full
figure. The perimeter is composed of the
semicircular arc and two of the sides of the triangle.

We can see from the figure indicated by
these lines here that the triangle is equilateral. All of its sides are the same length. So the two straight edges are each 35
centimeters long. For the semicircular arc, we recall that
the circumference of a full circle is π times the diameter. So the length of the semicircular arc
will be half of this. Itβs ππ over two. The diameter of this circle is the same
as the side length of the triangle. Itβs 35 centimeters. So the semicircular arc length is 35π
over two or 35 over two π.

Now weβre told in the question that we
need to use 3.14 as an approximation for π. So our perimeter is 35 over two
multiplied by 3.14 plus 35 plus 35. To work out 35 over two multiplied by
3.14 without a calculator, we can first divide 3.14 by two to give 1.57 and then multiply 35
by 1.57 using any multiplication method weβre comfortable with. Here Iβve used the grid method, to find
that 35 multiplied by 1.57 is equal to 54.95. So we have 54.95 plus 70 β thatβs 35 plus
35 β which is equal to 124.95. And the units for this perimeter are the
same as the units for the lengths in the question. Theyβre centimeters.

Because we used 3.14 to approximate π
then, there was no need for a calculator in this question. Although we did have some reasonably
tricky decimal calculations to work out.

Letβs now consider a slightly more
complicated example where the entire figure, whose perimeter we want to calculate, is
composed of parts of circles.

The given figure shows two half circles
and two-quarters of another circle. Find the perimeter of the shaded region,
taking 3.14 as an approximate value for π.

In this question then, we have two
different sizes of circle. Tracing our finger or pen all the way
around the edge of the figure, we see that it is composed firstly of the arc of a
semicircle. Then the arc of a different-sized quarter
circle. Then the arc of a semicircle which is
congruent to the first. And finally the arc of a quarter circle
which is congruent to the other quarter circle.

In total then, what we have is the full
circumference of the smaller circle β thatβs the orange one β and half the circumference of
the larger circle β thatβs the pink one. We know that the circumference of a
circle can be found using the formula ππ, where π is the diameter of the circle. So we just need to determine the diameter
of each of these circles.

From the figure, we can see that the
smaller circle has a diameter of 39 centimeters. So its circumference and its contribution
to the perimeter of the full figure is 39π. When we think about the pink circle,
however, this measurement of 39 centimeters is the radius of this circle. So the diameter is twice that. Itβs 78 centimeters. The circumference of the full pink circle
then would be 78π. But remember, we only have half the
circumference. So the length of the semicircular arc is
78π over two.

In fact, we find then that the two values
are the same. Both the circumference of the full orange
circle and the length of the semicircular arc for the pink circle are 39π. In total then, we have an exact perimeter
of 78π. But looking back at the information given
in the question, weβre asked to use 3.14 as an approximate value for π.

We can use a column method to work out
314 multiplied by 78, giving 24492. And we then need to divide this value by
100 to give the answer to the decimal calculation 3.14 multiplied by 78. This gives a value of 244.92. And the units for this perimeter are the
same as the units given for the individual lengths in the question. They are centimeters.

Letβs consider one final example which
requires a bit more of a problem-solving approach.

The given figure shows circle π
inscribed in square π΄π΅πΆπ·. The area of the shaded region of the
figure is two and one-third square centimeters. Using the approximation π equals 22 over
seven, find the perimeter of the shaded region.

Now, the first thing we notice is that we
havenβt been given any measurements at all on the diagram. In fact, the only information weβve been
given is that the area of the shaded region is two and one-third square centimeters. This area will be equivalent to the area
of rectangle π΄πΈπΉπ· minus the area of the semicircle.

Letβs see if we can use this to work out
some information about the dimensions of either the square or the circle. Weβll begin by letting the radius of our
circle be equal to π. Now, this radius is half the diameter of
the circle. And as the diameter of the circle is the
same as the side length of the square, then the squareβs side length will be equal to two
π. This rectangle, π΄π·πΉπΈ, therefore has a
width of π units and has a length of two π units. Its area, using the formula length
multiplied by width for the area of a rectangle, is therefore equal to two π squared.

The area of the semicircle will be half
the area of a full circle of radius π. So thatβs ππ squared over two. And so we have an equation involving the
radius of our circle. Two π squared minus ππ squared over
two is equal to two and one-third. We can factor π squared from the terms
on the left-hand side, giving two minus π over two all multiplied by π squared equals two
and one-third.

Now, at this point, we remember weβve
been asked to use the approximation π equals 22 over seven. So π divided by two or π multiplied by
a half is the same as 22 over seven multiplied by a half, which is 11 over seven. At the same time, we can think of the
integer two as the fraction 14 over seven. So the terms inside our parentheses
become 14 over seven minus 11 over seven, which simplifies to simply three over seven or
three-sevenths.

We can then convert the mixed number on
the right-hand side of our equation, two and one-third, into an improper fraction. And itβs equal to seven over three. To solve this equation then, we need to
divide both sides by three-sevenths in order to leave π squared on its own on the left-hand
side. Giving π squared equals seven over three
divided by three-sevenths.

But we recall that to divide by a
fraction, we can instead multiply by its reciprocal. So dividing by three-sevenths is
equivalent to multiplying by seven-thirds. And we have that π squared is equal to
seven over three multiplied by seven over three. Thatβs 49 over nine. Or we can write seven over three
multiplied by seven over three as seven over three all squared.

If π squared is equal to seven over
three all squared, then to find the value of π, we square-root both sides of the
equation. And we use only the positive value as π
is a length. So we have that π is equal to seven over
three. And weβve found the radius of our
circle.

So now that we know the radius of the
circle, we also know the side length of the square. Itβs twice this value, which is 14 over
three. And now weβre able to calculate the
perimeter of the shaded region. Picking a point on the perimeter β so
Iβve chosen point πΉ β and then traveling around the shaded region, we see that its
perimeter is composed of πΉπ·, π·π΄, π΄πΈ, and then the semicircular arc. πΉπ· and π΄πΈ are each seven over three
centimeters, and π·π΄ is 14 over three centimeters.

We then recall that the circumference of
a full circle is two ππ. So the arc length of a semicircle is half
this. Itβs simply ππ. The length of the semicircular arc is
therefore π multiplied by seven over three.

Again though, we recall that we need to
use 22 over seven as our approximation for π. And then we can cancel a factor of seven
from the numerator and denominator of these fractions. Weβre left with seven over three plus 14
over three plus seven over three plus 22 over three. And as these fractions all have common
denominator of three, they sum to 50 over three. We can then convert this to a mixed
number. Itβs 16 and two-thirds. And the units for this perimeter will be
centimeters. So weβve completed the problem. The perimeter of the shaded region is 16
and two-thirds centimeters.

Letβs now summarize what weβve seen in
this video. Firstly, a composite figure is made up of
two or more two-dimensional shapes joined together. When finding the perimeter of a composite
figure, itβs a good idea to trace our finger or pen all the way around the edge of the shape
to make sure we include all of the necessary lengths. We also need to know off by heart and be
able to apply the standard formulae for finding the perimeter of two-dimensional shapes,
such as squares, rectangles, triangles, and circles.

And finally, for questions involving
circles or parts of circles, we can give our answers as rounded decimals if we have a
calculator. Or we can give our answers exactly as
multiples of π or use 22 over seven or 3.14 as approximations for π.