In the figure below, if 𝑀𝐴 is
equal to 17.2 centimeters and 𝐴𝐵 is equal to 27.6 centimeters, find the length of
the line segment 𝑀𝐶 and the area of triangle 𝐴𝐷𝐵 to the nearest tenth.
Since 𝑀 is the center of the
circle and the line segment 𝑀𝐷 bisects the chord 𝐴𝐵 at 𝐶, we can apply the
chord bisector theorem. This states that if we have a
circle with center 𝑀 containing a chord 𝐴𝐵, then the straight line that passes
through 𝑀 and bisects 𝐴𝐵 is perpendicular to 𝐴𝐵. This means that the measure of
angle 𝑀𝐶𝐴 is 90 degrees. Since the chord 𝐴𝐵 has length
27.6 centimeters and we know that 𝐶 bisects this chord, 𝐴𝐶 is equal to 27.6
divided by two. This is equal to 13.8
We are also told that the radius
𝑀𝐴 is equal to 17.2 centimeters. We can therefore use the
Pythagorean theorem in the right triangle 𝑀𝐶𝐴 such that 𝑀𝐶 is equal to the
square root of 17.2 squared minus 13.8 squared. This is equal to 10.266 and so
on. And rounding to the nearest tenth,
the line segment 𝑀𝐶 is equal to 10.3 centimeters.
The second part of our question
asks us to calculate the area of triangle 𝐴𝐷𝐵. Clearing some space, we recall that
the area of any triangle is equal to the length of its base multiplied by the length
of its perpendicular height divided by two. We know that the base of our
triangle 𝐴𝐵 is equal to 27.6 centimeters. The perpendicular height 𝐶𝐷 will
be equal to the length of 𝑀𝐷 minus the length of 𝑀𝐶. 𝑀𝐷 is the radius of the circle,
and we know this is equal to 17.2 centimeters. Whilst we could use 10.3
centimeters for 𝑀𝐶, it is better for accuracy to use the nonrounded version: 𝑀𝐶
is equal to 10.266 and so on. Subtracting this from 17.2, we see
that the length of 𝐶𝐷 is 6.933 and so on centimeters.
We can now calculate the area of
triangle 𝐴𝐷𝐵 by multiplying this by 27.6 centimeters and then dividing by
two. This is equal to 95.683 and so
on. Once again, we need to round to the
nearest tenth, giving us an answer of 95.7 square centimeters.