Question Video: Finding the Area of a Major Circular Segment Mathematics

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.