Question

1. if \\( \overline{de} \\) is a midsegment of \\( \triangle abc \\), find \\( x \\).

equation:
\\( x = \\) \\( de = \\) \\( bc =

1. if \\( \overline{zy} \\) is the midsegment of \\( \triangle rqs \\), find \\( x \\), \\( zy \\) and \\( sq \\).

equation:
\\( x = \\) \\( zy = \\) \\( sq =

1. the measure of one base of an isosceles triangle is \\( (4x - 1)^\circ \\). if the non - base angle is \\( 62^\circ \\), what is the value of \\( x \\)?

equation:
\\( x =

1. \\( \overline{ad} \\) is the median of \\( \triangle abc \\). write an equation and solve for \\( x \\) to find the length of \\( \overline{bd} \\).

equation:
\\( x = \\) \\( bd =

Explanation:

Problem 19

Step1: Recall midsegment theorem

The midsegment of a triangle is parallel to the third side and half its length. So, \( DE=\frac{1}{2}BC \).

Step2: Substitute the given expressions

We have \( DE = x - 3 \) and \( BC=3x - 12 \). So the equation is \( x - 3=\frac{1}{2}(3x - 12) \).

Step3: Solve the equation

Multiply both sides by 2: \( 2(x - 3)=3x - 12 \)
Expand left side: \( 2x-6 = 3x - 12 \)
Subtract \( 2x \) from both sides: \( - 6=x - 12 \)
Add 12 to both sides: \( x = 6 \)

Step4: Find \( DE \) and \( BC \)

\( DE=x - 3=6 - 3 = 3 \)
\( BC = 3x-12=3\times6 - 12=18 - 12 = 6 \)

Step1: Recall midsegment theorem

The midsegment of a triangle is parallel to the third side and half its length. So, \( ZY=\frac{1}{2}SQ \).

Step2: Substitute the given expressions

We have \( ZY=x + 2 \) and \( SQ = 3x - 8 \). So the equation is \( x + 2=\frac{1}{2}(3x - 8) \).

Step3: Solve the equation

Multiply both sides by 2: \( 2(x + 2)=3x - 8 \)
Expand left side: \( 2x + 4=3x - 8 \)
Subtract \( 2x \) from both sides: \( 4=x - 8 \)
Add 8 to both sides: \( x = 12 \)

Step4: Find \( ZY \) and \( SQ \)

\( ZY=x + 2=12 + 2 = 14 \)
\( SQ=3x - 8=3\times12 - 8=36 - 8 = 28 \)

Step1: Recall properties of isosceles triangle

In an isosceles triangle, the base angles are equal, and the sum of angles in a triangle is \( 180^{\circ} \). Let the base angles be \( (4x - 1)^{\circ} \) each. So the equation is \( 2(4x - 1)+62 = 180 \).

Step2: Solve the equation

Expand: \( 8x-2 + 62=180 \)
Simplify: \( 8x + 60=180 \)
Subtract 60: \( 8x=120 \)
Divide by 8: \( x = 15 \)

Answer:

Equation: \( x - 3=\frac{1}{2}(3x - 12) \)
\( x = 6 \), \( DE = 3 \), \( BC = 6 \)

Problem 20