Question

1. solve for x and find the length of \\( \overline { f n } \\)

\\( x ^ { 2 } - 27 \\)
\\( 6 x \\)
\\( x = \\)
\\( 4 x + 12 \\)
\\( f n = \\)

1. given \\( \delta m n o \cong \delta p q r \\), solve for x and y.

\\( x = \\)
\\( y = \\)

Explanation:

Step1: Solve for \(x\) in problem 18

Since the two sides \(x^{2}-27\) and \(6x\) are equal (marked as congruent in the isosceles triangle), we set up the equation \(x^{2}-27 = 6x\).
Rearrange it to the quadratic form \(x^{2}-6x - 27=0\).
Factor the quadratic equation: \((x - 9)(x+ 3)=0\).
Using the zero - product property \(x-9 = 0\) or \(x + 3=0\), so \(x=9\) or \(x=-3\).
Since \(x\) represents a length, \(x>0\), so \(x = 9\).

Step2: Find the length of \(FN\) in problem 18

Substitute \(x = 9\) into \(4x + 12\).
\(FN=4x+12=4\times9+12=36 + 12=48\).

Step3: Solve for \(x\) and \(y\) in problem 19

Since \(\triangle MNO\cong\triangle PQR\), corresponding sides are equal.
We have \(4x-y=26\) and \(6x + 2y=32\).
From \(4x-y=26\), we can express \(y = 4x-26\).
Substitute \(y = 4x - 26\) into \(6x+2y=32\):

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Substitute \(x = 6\) into \(y=4x-26\), \(y=4\times6-26=24 - 26=-2\).

Answer:

For problem 18:
\(x = 9\)
\(FN=48\)

For problem 19:
\(x = 6\)
\(y=-2\)