Question

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Answer 1

Answer:

First of all notice that triangles CDE and BAE are similar (all three angles there are equal).

This means that the ratio of corresponding sides in these triangles are equal (corresponding sides are opposite equal angles): CD/AB = ED/AE --> 9/3 = ED/AE --> 3 = ED/AE --> 3*AE = ED.

Now, since AD = AE + ED = 4, then AE + 3*AE = 4 --> AE = 1 (and ED = 3).

The area of triangle AEC = 1/2*(base)*(height) = AE*CD = 1/2*1*9 = 4.5.

Answer 2

Answer:

$A_{AEC}$ = 4.5 units²

Step-by-step explanation:

AB║CD ⇒ ∠ABE ≅ ∠DCE (alternate interior angles) and ∠AEB ≅ ∠DEC (vertical angles) ⇒ ΔABE ~ ΔDCE ⇒ $\frac{CD}{BA}$ = $\frac{9}{3}$ = 3

Let AE = x , then ED = 4 - x

$\frac{4-x}{x}$ = 3 ⇒ 3x = 4 - x ⇒ x = 1

AE = 1 and DE = 3

$A_{ACD}$ = $\frac{(4)(9)}{2}$ = 18

$A_{ECD}$ = $\frac{(9)(3)}{2}$ = 13.5

$A_{AEC}$ = $A_{ACD}$ - $A_{ECD}$ = 18 - 13.5 = 4.5 units²