Question Video: Using Inscribed Angles in a Semi-Circle to Find the Area of a Triangle | Nagwa Question Video: Using Inscribed Angles in a Semi-Circle to Find the Area of a Triangle | Nagwa

Question Video: Using Inscribed Angles in a Semi-Circle to Find the Area of a Triangle Mathematics • Third Year of Preparatory School

Given that 𝐴𝐶 = 8 cm and the radius = 8 cm, find the area of △𝐴𝐵𝐶 rounded to the nearest integer.

04:07

Video Transcript

Given that 𝐴𝐶 equals eight centimeters and the radius equals eight centimeters, find the area of triangle 𝐴𝐵𝐶 rounded to the nearest integer.

And then we have a diagram which shows a circle with a triangle inscribed within it. We notice that line 𝐴𝐵 passes through the center of the circle, and so line 𝐴𝐵 is the diameter. So before we work out the area of the triangle we’ve been given, let’s identify what extra information we can calculate given the information we’ve been given. We are told that 𝐴𝐶 equals eight centimeters and that the radius is equal to eight centimeters. This means line segments 𝐵𝑀 and 𝑀𝐴 are both eight centimeters.

Now, in fact, since point 𝑀 is the center of the circle and 𝐶 lies on the circumference, we see that 𝑀𝐶 also forms the radius of our circle. So 𝑀𝐶 is also eight centimeters. And this is really useful since we know that all angles in an equilateral triangle, that is, a triangle where all three sides are of equal length, are 60 degrees. But this doesn’t necessarily help us find the area of triangle 𝐴𝐵𝐶. Remember, the formula that we can use to find the area of a triangle is a half times base times height. So if we could work out the length of the base and the height of this triangle 𝐴𝐵𝐶, we’d be able to find its area.

Now, in fact, we can quite quickly work out which sides of our triangle represent the base and the height. The base and the height must be perpendicular to one another. So we’re interested in sides 𝐴𝐶 and 𝐵𝐶. But how do we know that? Well, we know that the angle subtended by the diameter is 90 degrees. Our angle 𝐵𝐶𝐴 is indeed subtended from the diameter, so 𝐵𝐶𝐴 is 90 degrees. So we can define 𝐴𝐶 to be the base of our triangle and that is eight centimeters in length. But then this means that 𝐵𝐶 is the height of our triangle. So what is the length of 𝐵𝐶?

Well, now that we know we have a right triangle and we know a couple of the angles given in this triangle, we can work out the length of 𝐵𝐶 using right-triangle trigonometry. Here is our triangle 𝐴𝐵𝐶 with the right angle at 𝐶. We calculated that the measure of angle 𝐴 is 60 degrees. And of course, we’re trying to find the length of 𝐵𝐶, so let’s define that to be equal to 𝑥 centimeters. This side of the triangle lies directly opposite the included angle of 60 degrees, whilst the side 𝐴𝐶 is adjacent to the angle. And so we need to identify the trigonometric ratio that links the opposite side with the adjacent, that is, the tangent ratio. tan 𝜃 is opposite over adjacent.

In this case then, tan of 60 is 𝑥 divided by eight. But of course, tan of 60 is one of our exact value ratios. It’s the square root of three, so we get root three equals 𝑥 divided by eight. And if we multiply through by eight, we find that 𝑥 is equal to eight root three, and that’s great. We now know the length of 𝐵𝐶, which we said was the height of our triangle. It’s eight root three centimeters. So the area of our triangle is one-half times eight times eight root three. In exact form, that gives us 32 root three square centimeters. But If we enter this into our calculator, correct to the nearest integer, that’s 55 square centimeters. And so the area of triangle 𝐴𝐵𝐶 correct to the nearest integer is 55 square centimeters.

Now, it’s worth noting that we didn’t actually need to use right-triangle trigonometry. Instead, if we’d recognize that 𝐴𝐵 is the diameter and therefore 16 centimeters in length and we already knew that triangle 𝐴𝐶𝐵 was a right triangle at 𝐶, we could have used the Pythagorean theorem to calculate the length of 𝐵𝐶. In that case, we once again would have got eight root three, giving us a final area of 55 square centimeters. Either method is perfectly valid.

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