geometry question from kangaroo math contest

Question

drawing

A rectangle has sides of length 6cm and 11cm. One long side is selected. The bisectors of the angles at either end of that side are drawn. These bisectors divide the other long side into 3 parts. What are the lengths of these parts? The options are

a) 1, 9, 1;

b) 2, 7, 2;

c) 3, 5, 3;

d) 4, 3, 4;

e) 5, 1, 5.

I drew a rectangle and a bisector coming out of each end at a 45 degree angle which divided the other long side into 3 parts but I don't know if I drew the bisectors right because I couldn't find a way that will for sure give the answer.

Answer 1

The bisectors make isosceles triangles. You should find they cross before they reach the other side of the rectangle.

Answer 2

$The \;bisectors\; make\; right\; angle\; isosceles\;triangles,\; vertex\; is\; inside\; rectangle, \;since\; 6>11/2,\; y \;is\; the\; height\; of\; the\; smaller\; one,\; x\; is\; a \;foot\; on \;the \;side\; to\; be\; solved. $

$y= \frac {x} {2} $

$Appropriate\; right\; angle \;triangles\; are\; similar,\; thus\;$

$\frac {x} {11}=\frac {y} {6-y}$

$\frac {x} {11}=\frac {\frac {x} {2}} {6-\frac {x} {2}}$

$x=1,\; other\; segments\; are\; 5\; each$

Answer 3

The answer is e) $5, 1, 5$

You can solve it with the following algebra.

Let $a$ be the first part of the segment. Let $b$ be the second part of the segment. Let $c$ be the third part of the segment.

$$a+b+c = 11, a+b = 6, b+c = 6$$

Therefore: $$a = 5, b = 1, c = 5$$