Video Transcript
Let the line segment 𝐴𝐵 be the
diameter of a circle and the line 𝐵𝐶 a tangent to the circle at 𝐵. Given that the area of the circle
is 240.25𝜋 and the measure of angle 𝐴𝐶𝐵 is 45 degrees, calculate the area of
triangle 𝐴𝐵𝐶 to the nearest hundredth.
Let’s begin by sketching the
problem. We have a circle in which the line
segment 𝐴𝐵 is a diameter. There’s then a tangent to the
circle 𝐵𝐶 such that the measure of angle 𝐴𝐶𝐵 is 45 degrees. We’re asked to calculate the area
of the triangle formed by the points 𝐴, 𝐵, and 𝐶.
Now, we know that the area of a
triangle can be calculated as half of its base multiplied by its perpendicular
height. We haven’t been given any lengths
in the diagram. So let’s think about how we can
calculate them. The first thing to note is that
triangle 𝐴𝐵𝐶 is in fact a right triangle at 𝐵. We know this because we have a key
result, which states that a tangent to a circle is perpendicular to the radius and,
hence, the diameter at the point of contact. So the line 𝐵𝐶 is perpendicular
to the line segment 𝐴𝐵.
So, as we have a pair of
perpendicular sides in this triangle, we can calculate its area as one-half
multiplied by 𝐵𝐶 multiplied by 𝐴𝐵. We can also observe that this
triangle is in fact isosceles, because using the angle sum in a triangle, the
measure of the third angle, angle 𝐵𝐴𝐶, must also be 45 degrees. So, in fact, the sides 𝐴𝐵 and
𝐵𝐶 are of equal length.
The final piece of information
given that we haven’t used yet is that the area of the circle is 240.25𝜋. We know that the area of a circle
is calculated using the formula 𝜋𝑟 squared, where 𝑟 represents the radius of the
circle. So we can form an equation, which
will enable us to calculate the length of the circle’s radius and, hence, the length
of the circle’s diameter by doubling it.
Remembering that 𝐴𝐵 is a diameter
of this circle, this will then give us the lengths of both 𝐴𝐵 and 𝐵𝐶 to use in
calculating the area of the triangle. So, forming our equation then, we
have 𝜋𝑟 squared is equal to 240.25𝜋. We can immediately cancel a factor
of 𝜋 from both sides, leaving 𝑟 squared is equal to 240.25. We can solve for 𝑟 by square
rooting, giving 𝑟 equals the square root of 240.25, which is 15.5. The diameter of the circle is twice
this value, two times 15.5, which is 31 units.
So we now know that the length of
𝐴𝐵, and hence also the length of 𝐵𝐶, is 31 units. Substituting these two values into
our formula for the area of the triangle gives one-half multiplied by 31 multiplied
by 31, or a half multiplied by 31 squared. That’s 961 over two, or as a
decimal 480.5.
We’re asked to give the area to the
nearest hundredth. So we can add a zero in the second
decimal place. We weren’t given any units in the
question. So we can conclude that the area of
triangle 𝐴𝐵𝐶 to the nearest hundredth is 480.50 square units.
