Answer:
a) [tex]x=2\dfrac{2}{3}\ units\\ \\y=5\dfrac{1}{3}\ units[/tex]
b) [tex]x=9\ units\\ \\y=5.4\ units[/tex]
Step-by-step explanation:
Q11 a) ABCD is a parallelogram, then [tex]AB=CD=12[/tex] units.
Consider triangles AED and BEF. In these triangles
[tex]\angle AED\cong \angle BEF[/tex] by reflexive property;
[tex]\angle DAE\cong \angle FBE[/tex] are corresponding angles formed when parallel lines BC and AD are intersected by transversal AE. By corresponding angles theorem, they are congruent.
By AA similarity theorem, [tex]\triangle AED\sim \triangle BEF[/tex]. Similar triangles have proportional corresponding parts, so
[tex]\dfrac{AE}{BE}=\dfrac{AD}{BF}\\ \\\dfrac{12+6}{6}=\dfrac{8}{x}\\ \\18x=48\ [\text{Cross multoply}]\\ \\x=\dfrac{48}{18}=\dfrac{8}{3}=2\dfrac{2}{3}\ units[/tex]
ABCD is a parallelogram, then [tex]BC=AD=8[/tex] units, so
[tex]y=8-2\dfrac{2}{3}=5\dfrac{1}{3}\ units.[/tex]
Q11b) ABCD is a parallelogram, then [tex]AB=DC=16[/tex] units. By segment addition postulate,
[tex]FA+FB=AB\\ \\10+FB=16\\ \\FB=6\ units[/tex]
Consider triangles EBF and ECD. In these triangles,
[tex]\angle BEF\cong \angle CED[/tex] by reflexive property
[tex]\angle EBF\cong \angle ECD[/tex] are corresponding angles formed when parallel lines AB and CD are intersected by transversal EC. By corresponding angles theorem, they are congruent.
So, [tex]\triangle FEB\sim \triangle DEC[/tex] by AA similarity theorem. Similar triangles have proportional corresponding parts, so
[tex]\dfrac{EF}{ED}=\dfrac{FB}{DC}=\dfrac{EB}{EC}\\ \\\dfrac{x}{x+15}=\dfrac{6}{16}=\dfrac{y}{y+9}[/tex]
From the first equality,
[tex]16x=6(x+15)\\ \\16x=6x+90\\ \\16x-6x=90\\ \\10x=90\\ \\x=9\ units[/tex]
From the second equality:
[tex]6(y+9)=16y\\ \\6y+54=16y\\ \\54=16y-6y\\ \\10y=54\\ \\y=5.4\ units[/tex]
