boyznthahood64
Answered

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If ABC ~ DEC, solve for x. The image is not drawn to scale.

A.
x = 2
B.
x = 3
C.
x = 4
D.
x = 13

If ABC DEC Solve For X The Image Is Not Drawn To Scale A X 2 B X 3 C X 4 D X 13 class=

Sagot :

[tex]If\ \Delta ABC\sim\Delta DEC\ then:\\\\\frac{19-2x}{11-x}=\frac{8+x}{6}\ where\ x\in(-8;\ 11)\ \ \ |cross\ multiply\\\\(11-x)(8+x)=6(19-2x)\\11(8)+11(x)-x(8)-x(x)=6(19)+6(-2x)\\88+11x-8x-x^2=114-12x\\-x^2+3x+88=114-12x\ \ \ |subtract\ 114\ from\ both\ sides\\-x^2+3x-26=-12x\ \ \ \ \ |add\ 12x\ to\ both\ sides\\-x^2+15x-26=0\ \ \ \ |change\ signs\\x^2-15x+26=0\\x^2-2x-13x+26=0\\x(x-2)-13(x-2)=0\\(x-2)(x-13)=0\iff x-2=0\ or\ x-13=0\\\\x=2\in(-8;\ 11)\ or\ x=13\notin(-8;\ 11)\\\\Answer:\boxed{\boxed{A.\ x=2}}[/tex]
JeanaShupp

Answer: A.    x = 2

Step-by-step explanation:

In the given picture we have [tex]\triangle{ABC}\sim\triangle{DEC}[/tex]

Since, we know that the corresponding sides in similar triangles are in proportion.

Therefore, we have

[tex]\dfrac{CD}{AC}=\dfrac{CE}{BC}\\\\\Rightarrow\dfrac{8+x}{6}=\dfrac{19-2x}{11-x}\\\\\Rightarrow\ (11-x)(8+x)=6(19-2x)\\\\\Rightarrow\ 88 + 3x -x^2=114-12x\\\\\Rightarrow x^2-15x+26=0\\\\\Rightarrow\ (x-13)(x-2)=0\\\\\Rightarrow\ x=13,2[/tex]

But x can not be 13 because BC=11-13=-2, which is not possible.

Therefore, the value of  x=2.

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