Answer:
The length of AE is 20 units.
Step-by-step explanation:
Given two segments AD and BC intersect at point E to form two triangles ABE and DCE. Side AB is parallel to side DC. A E is labeled 2x+10. ED is labeled x+3. AB is 10 units long and DC is 4 units long.
we have to find the length of AE
AB||CD ⇒ ∠EAB=∠EDC and ∠EBA=∠ECD
In ΔABE and ΔDCE
∠EAB=∠EDC (∵Alternate angles)
∠EBA=∠ECD (∵Alternate angles)
By AA similarity, ΔABE ≈ ΔDCE
therefore, [tex]\frac{AE}{ED}=\frac{AB}{CD}[/tex]
⇒ [tex]\frac{2x+10}{x+3}=\frac{10}{4}[/tex]
⇒ [tex]8x+40=10x+30[/tex]
⇒ [tex]x=5[/tex]
Hence, AE=2x+10=2(5)+10=20 units
The length of AE is 20 units.
