### 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.

You are watching: Question Video: Finding the Area of a Major Circular Segment. Info created by GBee English Center selection and synthesis along with other related topics.