The perimeter of this triangle is 25 units. An angle bisector divides one side of the triangle into lengths of 2 and 3. Find the
lengths of the remaining two sides. EXPLAIN PLEASE.
1 answer:
Answer:
8 and 12
Step-by-step explanation:
Sides on one side of the angle bisector are proportional to those on the other side. In the attached figure, that means
AC/AB = CD/BD = 2/3
The perimeter is the sum of the side lengths, so is ...
25 = AB + BC + AC
25 = AB + 5 + (2/3)AB . . . . . . substituting AC = 2/3·AB. BC = 2+3 = 5.
20 = 5/3·AB
12 = AB
AC = 2/3·12 = 8
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<em>Alternate solution</em>
The sum of ratio units is 2+3 = 5, so each one must stand for 25/5 = 5 units of length.
That is, the total of lengths on one side of the angle bisector (AC+CD) is 2·5 = 10 units, and the total of lengths on the other side (AB+BD) is 3·5 = 15 units. Since 2 of the 10 units are in the segment being divided (CD), the other 8 must be in that side of the triangle (AC).
Likewise, 3 of the 15 units are in the segment being divided (BD), so the other 12 units are in that side of the triangle (AB).
The remaining sides of the triangle are AB=12 and AC=8.
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