Question

1. if ( ef = 5x + 14 ), ( fg = 13 ), and ( eg = 127 ), find the value of ( x ). (figure is not drawn to scale)
2. find the perimeter of ( \triangle abc ) with vertices ( a(-7, -5) ), ( b(1, -5) ), and ( c(-7, 1) ).

Explanation:

Question 9

Step1: Use segment addition postulate

By segment addition postulate, \(EG = EF+FG\). Given \(EF = 5x + 14\), \(FG=13\), and \(EG = 127\), we substitute into the equation: \(127=(5x + 14)+13\).

Step2: Simplify the right - hand side

Simplify \((5x + 14)+13\) to \(5x+27\). So the equation becomes \(127 = 5x+27\).

Step3: Solve for \(x\)

Subtract 27 from both sides: \(127-27=5x+27 - 27\), which gives \(100 = 5x\). Then divide both sides by 5: \(x=\frac{100}{5}=20\).

Step1: Find the length of \(AB\)

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(A(-7,-5)\) and \(B(1,-5)\), \(x_1=-7,y_1 = - 5,x_2=1,y_2=-5\). Then \(AB=\sqrt{(1-(-7))^2+(-5 - (-5))^2}=\sqrt{(8)^2+0^2}=8\).

Step2: Find the length of \(AC\)

For \(A(-7,-5)\) and \(C(-7,1)\), \(x_1=-7,y_1=-5,x_2=-7,y_2 = 1\). Then \(AC=\sqrt{(-7-(-7))^2+(1-(-5))^2}=\sqrt{0^2+(6)^2}=6\).

Step3: Find the length of \(BC\)

For \(B(1,-5)\) and \(C(-7,1)\), \(x_1 = 1,y_1=-5,x_2=-7,y_2=1\). Then \(BC=\sqrt{(-7 - 1)^2+(1-(-5))^2}=\sqrt{(-8)^2+(6)^2}=\sqrt{64 + 36}=\sqrt{100}=10\).

Step4: Calculate the perimeter

The perimeter \(P=AB + AC+BC\). Substitute \(AB = 8\), \(AC = 6\), and \(BC = 10\) into the formula: \(P=8 + 6+10=24\).

Answer:

D. 20

Question 10