Respuesta :

msm555

Answer:

Solution given:

∆ ABE is similar to ∆ ACD

since their side will be proportional

AB/AC=AE/AD

AB/(AB+BC)=AE/(AE+ED)

63/(63+7)=(7x-40)/(7x-40+18)

9/10=(7x-40)/(7x-22)

doing crisscrossed multiplication

9(7x-22)=10(7x-40)

63x-198=70x-400

400-198=70x-63x

202=7x

x=202/7 or 28.85units.

Answer:

[tex]x=28\frac{6}{7}[/tex]

[tex]x \approx 28.857[/tex]

Step-by-step explanation:

It is given that lines (BE) and (CD) are parallel, thus (<AEB) and (EDC) are congruent by alternate interior angles theorem. Moreover, (<A) is shared between the two triangles, therefore it is also congruent. Hence, triangles (ABE) and (ACD) are similar by (angle-angle) similarity.

Side (AD) is composed of segments (AE) and (ED), therefore one can find the total measure of segment (AD):

AD = AE + ED

AD = 7x - 40 + 18

AD = 7x - 22

Side (AC) is made of segments (AB) and (BC), thus one can find the total length of the side (AC) by adding these two segments:

AC = AB + BC

AC = 63 + 7

AC = 70

When two triangles are similar, the ratios of the sides are equal. Therefore, one can make the following statement:

[tex]\frac{AE}{AD}=\frac{AB}{AC}[/tex]

Substitute,

[tex]\frac{7x-40}{7x-22}=\frac{63}{70}[/tex]

Cross products,

[tex]70(7x-40)=63(7x-22)[/tex]

Distribute,

[tex]490x-2800=441x-1386[/tex]

Inverse operations,

[tex]490x-2800=441x-1386[/tex]

[tex]49x-2800=-1386[/tex]

[tex]49x=1414[/tex]

[tex]x=28\frac{6}{7}[/tex]