Video: Perimeter of a Composite Figure

In this video, we will learn how to find the perimeters of composite figures.

17:53

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.

Let’s begin by reviewing some of the key formulae we need for calculating the perimeters of two-dimensional shapes. And at this point, we’ll 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 𝑠 units, then the perimeter will be equal to four 𝑠 units. That’s 𝑠 plus 𝑠 plus 𝑠 plus s. Secondly, a rectangle. In general, this will have a length 𝑙 units and a width 𝑀 units. So the perimeter, the distance all the way around its edge, is two 𝑙 plus two 𝑀. Thirdly, for a general triangle, if we label the three sides as π‘Ž, 𝑏, and 𝑐 units, then the perimeter will be equal to π‘Ž plus 𝑏 plus 𝑐 units. In the special case of equilateral triangles, where all three sides are the same length, then the perimeter can be written as three 𝑠, where 𝑠 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 πœ‹π‘‘ or two πœ‹π‘Ÿ, where 𝑑 is the diameter of the circle and π‘Ÿ 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œ‹ times the radius. And the straight part is the diameter of the circle, which we can write as two π‘Ÿ, giving a perimeter of πœ‹π‘Ÿ plus two π‘Ÿ.

Now that we’ve reviewed the key formulae we’ll need, let’s have a look at some examples.

Determine the perimeter of the shape below.

We can see that the shape we’ve been given is a composite figure. It’s 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, it’s 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 don’t miss out any of the lengths. Let’s start at point 𝐴. The perimeter will be equal to 𝐴𝐡 plus 𝐡𝐢 plus 𝐢𝐷 plus 𝐷𝐸 plus 𝐸𝐹 plus 𝐹𝐴. And that’s all the lengths we need to include as we’re now back at our starting point.

We’ve been given on the diagram the first four of these lengths. They are five, seven, three, and three centimeters, respectively. But we haven’t been given the lengths 𝐸𝐹 or 𝐹𝐴. We can work them out though. Firstly, 𝐸𝐹 will be the difference between the two vertical sides of this figure, 𝐴𝐡 minus 𝐢𝐷. That’s five minus three, which is equal to two centimeters. 𝐹𝐴 will be the difference between the horizontal sides of the figure. That’s 𝐡𝐢 minus 𝐷𝐸, 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 𝐴𝐡𝐢𝐺 if we hadn’t removed the smaller rectangle 𝐹𝐸𝐷𝐺. And the reason for this is that 𝐸𝐷 is the same as the length we removed, 𝐹𝐺, and 𝐸𝐹 is the same as the length we removed, 𝐷𝐺.

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 𝐴𝐡𝐢𝐺.

Let’s now consider an example involving circles.

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

The figure we’ve been given is a composite figure. It’s 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œ‹ 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œ‹.

Now we’re told in the question that we should use 22 over seven as an approximation for πœ‹. 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 I’ve shown the column method, and it gives 154. So the perimeter of the figure becomes 42 plus 154 plus 42. That’s 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œ‹, we had no need to use a calculator in this question.

Let’s now consider another example with a composite figure composed of a triangle and a semicircle.

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

In this question, we’ve 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€” that’s the third side of the triangle or the straight edge of the semicircle β€” is not part of the perimeter because it isn’t 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œ‹ times the diameter. So the length of the semicircular arc will be half of this. It’s πœ‹π‘‘ over two. The diameter of this circle is the same as the side length of the triangle. It’s 35 centimeters. So the semicircular arc length is 35πœ‹ over two or 35 over two πœ‹.

Now we’re told in the question that we need to use 3.14 as an approximation for πœ‹. 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 we’re comfortable with. Here I’ve 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€” that’s 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. They’re centimeters.

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

Let’s 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œ‹.

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€” that’s the orange one β€” and half the circumference of the larger circle β€” that’s the pink one. We know that the circumference of a circle can be found using the formula πœ‹π‘‘, where 𝑑 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œ‹. 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. It’s 78 centimeters. The circumference of the full pink circle then would be 78πœ‹. But remember, we only have half the circumference. So the length of the semicircular arc is 78πœ‹ 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œ‹. In total then, we have an exact perimeter of 78πœ‹. But looking back at the information given in the question, we’re asked to use 3.14 as an approximate value for πœ‹.

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.

Let’s consider one final example which requires a bit more of a problem-solving approach.

The given figure shows circle 𝑀 inscribed in square 𝐴𝐡𝐢𝐷. The area of the shaded region of the figure is two and one-third square centimeters. Using the approximation πœ‹ equals 22 over seven, find the perimeter of the shaded region.

Now, the first thing we notice is that we haven’t been given any measurements at all on the diagram. In fact, the only information we’ve 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 𝐴𝐸𝐹𝐷 minus the area of the semicircle.

Let’s see if we can use this to work out some information about the dimensions of either the square or the circle. We’ll begin by letting the radius of our circle be equal to π‘Ÿ. 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 square’s side length will be equal to two π‘Ÿ. This rectangle, 𝐴𝐷𝐹𝐸, therefore has a width of π‘Ÿ units and has a length of two π‘Ÿ units. Its area, using the formula length multiplied by width for the area of a rectangle, is therefore equal to two π‘Ÿ squared.

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

Now, at this point, we remember we’ve been asked to use the approximation πœ‹ equals 22 over seven. So πœ‹ divided by two or πœ‹ 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 it’s equal to seven over three. To solve this equation then, we need to divide both sides by three-sevenths in order to leave π‘Ÿ squared on its own on the left-hand side. Giving π‘Ÿ 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‘Ÿ squared is equal to seven over three multiplied by seven over three. That’s 49 over nine. Or we can write seven over three multiplied by seven over three as seven over three all squared.

If π‘Ÿ squared is equal to seven over three all squared, then to find the value of π‘Ÿ, we square-root both sides of the equation. And we use only the positive value as π‘Ÿ is a length. So we have that π‘Ÿ is equal to seven over three. And we’ve 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. It’s twice this value, which is 14 over three. And now we’re able to calculate the perimeter of the shaded region. Picking a point on the perimeter β€” so I’ve chosen point 𝐹 β€” and then traveling around the shaded region, we see that its perimeter is composed of 𝐹𝐷, 𝐷𝐴, 𝐴𝐸, and then the semicircular arc. 𝐹𝐷 and 𝐴𝐸 are each seven over three centimeters, and 𝐷𝐴 is 14 over three centimeters.

We then recall that the circumference of a full circle is two πœ‹π‘Ÿ. So the arc length of a semicircle is half this. It’s simply πœ‹π‘Ÿ. The length of the semicircular arc is therefore πœ‹ multiplied by seven over three.

Again though, we recall that we need to use 22 over seven as our approximation for πœ‹. And then we can cancel a factor of seven from the numerator and denominator of these fractions. We’re 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. It’s 16 and two-thirds. And the units for this perimeter will be centimeters. So we’ve completed the problem. The perimeter of the shaded region is 16 and two-thirds centimeters.

Let’s now summarize what we’ve 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, it’s 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œ‹ or use 22 over seven or 3.14 as approximations for πœ‹.

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