𝐴𝐵 is a chord of length 55 centimeters with a central angle of 95 degrees. Find the area of the minor circular segment, giving the answer to one decimal place.
Let’s begin with a sketch. We have the chord 𝐴𝐵 which is length 55 centimeters. And then the angle at the centre of the circle is 95 degrees. I have marked the centre of the circle as the point 𝑂. A chord divides a circle into two parts, a major and a minor segment. And it’s the minor segment shaded in orange that we’ve been asked to calculate the area of.
We can calculate the area of the minor segment as the difference between two other areas, the area of sector 𝐴𝑂𝐵 minus the area of triangle 𝐴𝑂𝐵. The area of the sector 𝐴𝑂𝐵 will be found by finding the area of the full circle, 𝜋𝑟 squared, and then multiplying by the portion of the circle that this sector represents. So that’s 95 over 360, as there’re 360 degrees in a full circle and this sector has 95 of them.
For the area of the triangle 𝐴𝑂𝐵, we need to recall the formula a half 𝑎𝑏 sin 𝑐. Which tells us how to calculate the area of a triangle, if we know two sides and the included angle. The angle that we know in this triangle is 95 degrees. And the two sides that enclose this angle are both the radius of the circle. So the area of triangle 𝐴𝑂𝐵 is given by a half 𝑟 squared sin 95 degrees.
So we have the calculation that we need in order to find the area of the minor segment. But the problem is this, both parts of the calculation use the radius of the circle, 𝑟, which at this stage, we don’t know. We know instead the length of the chord 𝐴𝐵. So we need to think about how we can calculate the radius from the information we’ve been given. Let’s consider dividing triangle 𝐴𝑂𝐵 up. We know that a line drawn from the centre of a circle to a chord will be a perpendicular bisector of that chord. Therefore, the chord will be cut in half, bisected, as will the central angle. And we’ll form two identical right-angled triangles.
Let’s look at one of these triangles more closely. As the chord has been bisected, then the length of the line from 𝐵 to where the chord meets the perpendicular bisector is half of 55. It’s 27.5 centimeters. The angle at the centre of the circle is half of 95 degrees. It’s 47.5 degrees. As this triangle is right-angled, we can apply trigonometry within this triangle in order to calculate the length of the radius of the circle. I begin by labelling the three sides of the right-angled triangle in relation to the 47.5-degree angle. We have the opposite, the adjacent, and the hypotenuse. The two sides which are going to be involved in the calculation are the side whose length I know, which is the opposite side, and the side whose length I wish to calculate, which is the hypotenuse.
Recalling the acronym SOHCAHTOA, this tells me that it’s the sine ratio that I need to use to calculate it. The definition of the sine ratio is that sine of an angle 𝜃 is equal to the length of the opposite side divided by the length of the hypotenuse. This means that in this triangle, sin of 47.5 degrees is equal to 27.5 over 𝑟. And now, I have an equation that I can solve in order to find the radius of the circle.
To solve this equation, I need to multiply both sides by 𝑟, to bring 𝑟 out of the denominator of our fraction and then divide both sides by sine of 47.5 degrees. This tells me that 𝑟 is equal to 27.5 over sin of 47.5 degrees. Evaluating this on my calculator and making sure that my calculator is in degree mode, I had that the radius of the circle is 37.29939. Now, we’re going to need to use this value in the next stage of the calculation. So I suggest you write it down to a large number of decimal places or you save it in your calculator memory. Remember, the reason we needed to know the radius of the circle was so that we could substitute it into our calculation for the area of the minor segment. I’m now going to delete the working out for 𝑟 so that I’ve got space to do the next stage of the calculation. So if you need to jot anything down, pause the video now and do so.
So, we found that the radius of the circle was equal to this decimal 37.29939. And now, substitute this for 𝑟 into our calculation for the area of the minor segment. Evaluating the two parts of the area on my calculator and again making sure it’s in degree mode, I have that the area of the sector is 1153.3857 and the area of the triangle is 692.9754. Subtracting gives 460.410 and the decimal continues. The question asked to give the answer to one decimal place. So finally, I need to round this value. The area of the minor circular segment, to one decimal place, is 460.4 centimeters square.