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ella [17]
3 years ago
12

The coordinates of Trapezoid EFGH are E(8, 8), F(8,10), G(1, 6), and H(4, 1). The image of EFGH under dilation is E’F’G’H’. If t

he coordinates of vertex E’ are (4, 4), what are the coordinates of vertex F’? A)(2, 6) B)(4, 6) C)(2, 5) D)(4, 5)
Mathematics
1 answer:
umka2103 [35]3 years ago
6 0

Answer:

D) (4,5)

Step-by-step explanation:

Vertex E was dilated by 1/2 because 8/2 = 4. Thus, you will divide all the other coordinates by the same scale factor.

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Since segments ST and PQ are parallel, triangles SRT and PRQ are similar due to the AAA postulate. In general, the ratio between the corresponding sides of two similar triangles is constant; therefore,

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Furthermore,

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Finding PR and RS,

\begin{gathered} ST=4+PS \\ SR=12 \\ PQ=15 \end{gathered}

Then,

\frac{12}{PR}=\frac{TS}{15}\begin{gathered} \Rightarrow\frac{12}{PR}=\frac{4+PS}{15} \\ \Rightarrow\frac{12}{PS+12}=\frac{4+PS}{15} \end{gathered}

Solving for PS,

\begin{gathered} 12\cdot15=(PS+12)(PS+4) \\ \Rightarrow180=PS^2+16PS+48 \end{gathered}

Solve the quadratic equation in terms of PS, as shown below

\begin{gathered} \Rightarrow PS^2+16PS-132=0 \\ \Rightarrow PS=\frac{-16\pm\sqrt[]{16^2-4(-132)}}{2}=\frac{-16\pm28}{2} \\ \Rightarrow PS=-22,6 \end{gathered}

And PS is a segment; therefore, it has to be positive.

Hence, the answer is PS=6

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