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, $\frac{AE}{ED}=\frac{AB}{CD}$

⇒ $\frac{2x+10}{x+3}=\frac{10}{4}$

⇒ $8x+40=10x+30$

⇒ $x=5$

Hence, AE=2x+10=2(5)+10=20 units

The length of AE is 20 units.

5 0

3 years ago

What's 2+2?<br> A) 5<br> B) 2<br> C) 4<br> D) 10

Triss [41]

Answer:

C) 4

Step-by-step explanation:

I hope this helps :)

3 0

2 years ago

Anyone know the answer the answer to this algebra 2 question? Anyone that gets it will give brainiest

viktelen [127]

the answer is x = 2, x = -5

3 0

2 years ago

Read 2 more answers