# Question Video: Finding the Area of a Triangle Given a Circle Inscribed in a Quadrilateral Mathematics

In the figure, a circle with radius 6 cm is inscribed in quadrilateral 𝐴𝐵𝐶𝐷. If 𝐸 is the midpoint of line segment 𝐴𝐵, 𝐴𝐵 = 10 cm, and 𝐶𝐺 = 6 cm, find the area of triangle 𝐵𝐶𝑀.

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### Video Transcript

In the following figure, a circle with radius six centimeters is inscribed in quadrilateral 𝐴𝐵𝐶𝐷. If 𝐸 is the midpoint of line segment 𝐴𝐵, 𝐴𝐵 equals 10 centimeters, and 𝐶𝐺 equals six centimeters, find the area of triangle 𝐵𝐶𝑀.

Let’s begin by adding any given lengths to our diagram. First, we’re told that the radius of the circle is six centimeters. In other words, the length of any line segment that joins the center 𝑀 to a point on its circumference is six centimeters in length. For instance, line segment 𝑀𝐸 is six centimeters, but we could choose any point on the circumference of the circle. So, we could write line segment 𝑀𝐻 is six centimeters. Similarly, 𝑀𝐹 and 𝑀𝐺 will all be six centimeters in length.

Next, we’re given three further pieces of information. We’re told that the line segment 𝐴𝐵 is 10 centimeters in length. We’re also told that 𝐸 is the midpoint of line segment 𝐴𝐵. So, 𝐴𝐸 and 𝐸𝐵 must each be five centimeters in length. Then, we’re also told that the line segment between point 𝐶 and 𝐺 is six centimeters. Our job is to use this to find the area of triangle 𝐵𝐶𝑀. And of course, that’s this triangle here.

So, we begin by recalling that the area of a triangle is half the length of its base multiplied by its height. Remember, its base and its height must be perpendicular lengths. Now, this is really useful because we do know a little bit more about some of the line segments within this circle. We know that the tangent to a circle will meet the radius at 90 degrees. And we mentioned earlier that line segment 𝑀𝐹 is the radius of our circle. In fact, we can also see that line segment 𝐶𝐵, which meets the circle at point 𝐹, is the tangent. So, line segment 𝑀𝐹 must be perpendicular to line segment 𝐶𝐵. This means we know the perpendicular height of the triangle. If we call its base the line segment 𝐶𝐵, then the perpendicular height must be 𝑀𝐹, which is six centimeters.

So next, we need to find the length of line segment 𝐶𝐵. In order to find the length of line segment 𝐶𝐵, we need to quote another fact about tangents. In particular, if we have a pair of tangent segments that meet at a point outside the circle — in here, that’s the tangent segments 𝐸𝐵 and 𝐹𝐵 which meet at point 𝐵 — then they must be equal in length. So, the length of line segment 𝐸𝐵 must be equal to the length of line segment 𝐹𝐵. And remember, we said that since line segment 𝐴𝐵 was 10 centimeters and point 𝐸 is exactly halfway along the line segment 𝐴𝐵, then both 𝐴𝐸 and 𝐸𝐵 are five centimeters. So, line segment 𝐹𝐵 must also be five centimeters.

And of course, if we look carefully, we see we can repeat this by looking at line segments 𝐹𝐶 and 𝐶𝐺. 𝐹𝐶 and 𝐶𝐺 must be equal in length. So, line segment 𝐹𝐶 is six centimeters. And so, we’re now able to find the length of line segment 𝐶𝐵. It will be the sum of these two measurements. It will be six centimeters plus five centimeters, which is, of course, equal to 11 centimeters. We now have the length of the base of our triangle and its perpendicular height. Its base is 11 centimeters, whilst its height is six centimeters.

So, the area of triangle 𝐵𝐶𝑀 is a half times 11 times six. One-half times six is, of course, equal to three. And so, the area is 11 times three, which is 33 or 33 square centimeters. Given the information about our circle and the quadrilateral 𝐴𝐵𝐶𝐷, the area of 𝐵𝐶𝑀 is 33 square centimeters.