### Video Transcript

The line segment 𝐴𝐵 is a chord of

length 17 centimeters with a central angle of 155 degrees. Find the area of the major circular

segment giving the answer to the nearest square centimeter.

Let’s begin by sketching this

out. We’re told that the line segment

𝐴𝐵 is a chord of a circle with a length of 17 centimeters. If we add the center of the circle

𝑂, we’re told that the central angle is 155 degrees as shown. We can deduce that the line

segments 𝑂𝐴 and 𝑂𝐵 are the radii of our circle. This means the line segment 𝑂𝐴

must be equal in length to the line segment 𝑂𝐵. Let’s call them both 𝑥

centimeters.

Now the question’s asking us to

find the area of the major circular segment. The minor circular segment is

shaded in orange. So it follows that the major

circular segment is everything else in the circle. And so we’re actually trying to

find the area of this shape. Now, if we look at the shape, we

see it’s made up of two composite shapes. We have the triangle 𝐴𝑂𝐵 and

then we have a sector of a circle. We can find the angle of this

sector by subtracting 155 degrees from 360 degrees since angles around a point sum

to 360. And that tells us the angle of our

sector is 205 degrees. And so if we can find the area of

our triangle and the area of this sector, the combined area will tell us the area of

the major circular segment.

Now we’re going to use the

trigonometric formula for area of a triangle. That’s a half 𝑎𝑏 sin 𝑐. And we also know that for a sector

of a circle with radius 𝑟 and an angle of 𝜃, its area is 𝜃 over 360 times 𝜋𝑟

squared. Essentially, it’s a proportion of

the area of the whole circle. Now we do have a little bit of a

problem. We don’t actually know the radius

of our circle. We’ve called that 𝑥. We can however calculate it using

the information about triangle 𝐴𝑂𝐵. It’s a non-right-angled triangle,

so we can use the cosine rule to find the length 𝑥. The cosine rule says that 𝑎

squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴.

Now we do need to redefine this

since our angle is neither 𝐴 or 𝐵; it’s a vertex. Let’s call that 𝐶 rather than

𝑂. And so 𝑐 squared equals 𝑎 squared

plus 𝑏 squared minus two 𝑎𝑏 cos 𝐶. We’ll substitute everything we know

about our triangle into this formula. And we get 17 squared equals 𝑥

squared plus 𝑥 squared minus two 𝑥 squared cos 155 degrees. 17 squared is 289. And we’ll simplify a little by

adding 𝑥 squared and 𝑥 squared. And then we see we can factor the

right-hand side by taking out a common factor of two 𝑥 squared.

We then divide both sides by one

minus cos of 155 degrees, divide through by two, and then finally find the square

root of 289 over two times one minus cos of 155 degrees. That gives us a value of 8.706 and

so on. And we now have everything we need

to calculate the area of the major circular segment. Of course, for accuracy, going

forward, we will use this value of 8.706. And we may even go back to 𝑥

squared equals 289 over two times one minus cos of 155 for ease.

We begin by finding the area of the

triangle. We will use the formula a half 𝑎𝑏

sin 𝐶. And so that’s a half times 8.706

times 8.706 or 8.706 squared times sin of 155. That’s 16.017 and so on. Then the area of our sector is 205

over 360 times 𝜋 times 8.70 squared, which is 135.605 and so on. The area of our major circular

segment is the sum of these values. So it’s 16.017 plus 135.605. That gives us 151.62 or 152 correct

to the nearest whole value. And so, given the information about

our sector, the area of the major circular segment is 152 square centimeters.