Question

Please answer asap with explanation ​

Answer 1

Answer:

a) $x=2\dfrac{2}{3}\ units\\ \\y=5\dfrac{1}{3}\ units$

b) $x=9\ units\\ \\y=5.4\ units$

Step-by-step explanation:

Q11 a) ABCD is a parallelogram, then $AB=CD=12$ units.

Consider triangles AED and BEF. In these triangles

$\angle AED\cong \angle BEF$ by reflexive property;

$\angle DAE\cong \angle FBE$ 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, $\triangle AED\sim \triangle BEF$. Similar triangles have proportional corresponding parts, so

$\dfrac{AE}{BE}=\dfrac{AD}{BF}\\ \\\dfrac{12+6}{6}=\dfrac{8}{x}\\ \\18x=48\ [\text{Cross multoply}]\\ \\x=\dfrac{48}{18}=\dfrac{8}{3}=2\dfrac{2}{3}\ units$

ABCD is a parallelogram, then $BC=AD=8$ units, so

$y=8-2\dfrac{2}{3}=5\dfrac{1}{3}\ units.$

Q11b) ABCD is a parallelogram, then $AB=DC=16$ units. By segment addition postulate,

$FA+FB=AB\\ \\10+FB=16\\ \\FB=6\ units$

Consider triangles EBF and ECD. In these triangles,

$\angle BEF\cong \angle CED$ by reflexive property

$\angle EBF\cong \angle ECD$ are corresponding angles formed when parallel lines AB and CD are intersected by transversal EC. By corresponding angles theorem, they are congruent.

So, $\triangle FEB\sim \triangle DEC$ by AA similarity theorem. Similar triangles have proportional corresponding parts, so

$\dfrac{EF}{ED}=\dfrac{FB}{DC}=\dfrac{EB}{EC}\\ \\\dfrac{x}{x+15}=\dfrac{6}{16}=\dfrac{y}{y+9}$

From the first equality,

$16x=6(x+15)\\ \\16x=6x+90\\ \\16x-6x=90\\ \\10x=90\\ \\x=9\ units$

From the second equality:

$6(y+9)=16y\\ \\6y+54=16y\\ \\54=16y-6y\\ \\10y=54\\ \\y=5.4\ units$