Question

hw #3 - drawing conclusions

1. be is a median of triangle abc. if ae=2x-4 and ac=6x-14, find the value of x and the length of ae and ac.
2. zy is a median of triangle mzb. if my=3x+15 and yb=7x-13, find the value of x.
3. in isosceles triangle dog, do ≅ og. if m∠d = 7x and m∠o = 40°, what is m∠g?
4. rt is an altitude of triangle hrb. if m∠htr = 12x - 18, find the value of x.

Explanation:

Problem 1:

Step1: Recall the median property

A median of a triangle divides the opposite side into two equal parts. So, \( AE = EC \) and \( AC = AE + EC = 2AE \).
Given \( AE = 2x - 4 \) and \( AC = 6x - 14 \), we have \( 6x - 14 = 2(2x - 4) \).

Step2: Solve the equation

Expand the right - hand side: \( 6x - 14 = 4x - 8 \).
Subtract \( 4x \) from both sides: \( 6x - 4x - 14 = 4x - 4x - 8 \), which gives \( 2x - 14 = - 8 \).
Add 14 to both sides: \( 2x - 14 + 14 = - 8 + 14 \), so \( 2x = 6 \).
Divide both sides by 2: \( x=\frac{6}{2}=3 \).

Step3: Find the lengths of \( AE \) and \( AC \)

For \( AE \): Substitute \( x = 3 \) into \( AE = 2x - 4 \), we get \( AE=2\times3 - 4=6 - 4 = 2 \).
For \( AC \): Substitute \( x = 3 \) into \( AC = 6x - 14 \), we get \( AC=6\times3 - 14 = 18 - 14 = 4 \).

Step1: Recall the median property

A median of a triangle divides the opposite side into two equal parts. So, \( MY = YB \).
Given \( MY = 3x + 15 \) and \( YB = 7x - 13 \), we set up the equation \( 3x+15 = 7x - 13 \).

Step2: Solve the equation

Subtract \( 3x \) from both sides: \( 3x - 3x+15 = 7x - 3x - 13 \), which gives \( 15 = 4x - 13 \).
Add 13 to both sides: \( 15 + 13 = 4x - 13 + 13 \), so \( 28 = 4x \).
Divide both sides by 4: \( x=\frac{28}{4}=7 \).

Step1: Recall the isosceles triangle property

In an isosceles triangle, if two sides are equal, the angles opposite them are equal. Given \( DO\cong OG \), then \( \angle D\cong\angle G \), so \( m\angle D=m\angle G = 7x \).
The sum of the interior angles of a triangle is \( 180^{\circ} \). So, \( m\angle D + m\angle O+m\angle G=180^{\circ} \).

Step2: Substitute the known values and solve for \( x \)

Substitute \( m\angle D = 7x \), \( m\angle O = 40^{\circ} \) and \( m\angle G = 7x \) into the angle - sum formula: \( 7x+40^{\circ}+7x = 180^{\circ} \).
Combine like terms: \( 14x+40^{\circ}=180^{\circ} \).
Subtract \( 40^{\circ} \) from both sides: \( 14x=180^{\circ}- 40^{\circ}=140^{\circ} \).
Divide both sides by 14: \( x = \frac{140^{\circ}}{14}=10^{\circ} \).

Step3: Find \( m\angle G \)

Since \( m\angle G = 7x \), substitute \( x = 10^{\circ} \), we get \( m\angle G=7\times10^{\circ}=70^{\circ} \).

Answer:

\( x = 3 \), \( AE = 2 \), \( AC = 4 \)

Problem 2: