An angle bisector of a triangle divides the opposite side of the triangle into segments 5 cm and 3 cm long. A second side of the
triangle is 7.6 cm long. Find the longest and shortest possible lengths of the third side of the triangle. Round answers to the nearest tenth of a centimeter.
a. 12.7 cm, 4.6 cm
b. 15 cm, 4.6 cm
c. 38 cm, 2 cm
d. 38 cm, 12.7 cm
a. 12.7 cm, 4.6 cm
b. 15 cm, 4.6 cm
c. 38 cm, 2 cm
d. 38 cm, 12.7 cm
1 answer:
There is a little-known theorem to solve this problem.
The theorem says that
In a triangle, the angle bisector cuts the opposite side into two segments in the ratio of the respective sides lengths.
See the attached triangles for cases 1 and 2. Let x be the length of the third side.
Case 1:
Segment 5cm is adjacent to the 7.6cm side, then
x/7.6=3/5 => x=7.6*3/5=4.56 cm
Case 2:
Segment 3cm is adjacent to the 7.6 cm side, then
x/7.6=5/3 => x=7.6*5/3=12.67 cm
The theorem can be proved by considering the sine rule on the adjacent triangles ADC and BDC with the common side CD and equal angles ACD and DCB.
The theorem says that
In a triangle, the angle bisector cuts the opposite side into two segments in the ratio of the respective sides lengths.
See the attached triangles for cases 1 and 2. Let x be the length of the third side.
Case 1:
Segment 5cm is adjacent to the 7.6cm side, then
x/7.6=3/5 => x=7.6*3/5=4.56 cm
Case 2:
Segment 3cm is adjacent to the 7.6 cm side, then
x/7.6=5/3 => x=7.6*5/3=12.67 cm
The theorem can be proved by considering the sine rule on the adjacent triangles ADC and BDC with the common side CD and equal angles ACD and DCB.
You might be interested in
If an expression is the difference of two squares, it will follow this format:
where a and b are integers or variables.
1)
Since the expression follows the format, it's a DoTS.
2) Since the binomial has an addition sign, it isn't a DoTS.
3)
The second term isn't an integer, so the expression isn't a DoTS.
4)
Since the expression follows the format, it's a DoTS.
Therefore, the first and fourth expressions are differences of two squares.
where a and b are integers or variables.
1)
Since the expression follows the format, it's a DoTS.
2) Since the binomial has an addition sign, it isn't a DoTS.
3)
The second term isn't an integer, so the expression isn't a DoTS.
4)
Since the expression follows the format, it's a DoTS.
Therefore, the first and fourth expressions are differences of two squares.