Question

1. given \\(\overline{cd}\\) as shown with \\(\overline{ca} \cong \overline{bd}\\) and \\(cd = 98\\). if \\(cb = 3(x + 8)\\) and \\(ad = 4(x + 3)\\), find \\(ab\\).
2. given \\(t\\) to be the midpoint of \\(\overline{ab}\\) with \\(at = \frac{3}{4}x + 6\\) and \\(tb = \frac{1}{2}x + 9\\). find \\(ab\\)
3. given \\(\overline{ac}\\) with \\(ab = 2x - 4\\), \\(bc = 4x\\) and \\(ac = 2x + 12\\). find \\(ac\\).
4. given \\(\overline{mh}\\) as shown with \\(\overline{mt} \cong \overline{ah}\\), \\(mt = 5x - 8\\) and \\(ah = 3x + 12\\). if \\(at = 6\\), find \\(th\\).

Explanation:

Problem 7

Step1: Analyze segment relationships

Since \(\overline{CA} \cong \overline{BD}\), \(CA = BD\). Also, \(CB = CA + AB\) and \(AD = AB + BD\). So \(CB - AB = AD - AB\) implies \(CB = AD\) (wait, no, actually \(CB = CA + AB\) and \(AD = AB + BD\), and since \(CA = BD\), then \(CB = AD\) only if \(AB\) is same, but let's write expressions. Given \(CB = 3(x + 8)\) and \(AD = 4(x + 3)\), and \(CD = CB + BD = CB + CA\) (since \(CA = BD\)), also \(CD = 98\). Wait, \(CD = CA + AB + BD\), and \(CA = BD\), so \(CD = 2CA + AB\). Also, \(CB = CA + AB = 3(x + 8)\), \(AD = AB + BD = AB + CA = 4(x + 3)\). So \(CB = AD\)? Wait, \(CA + AB = AB + CA\), so that's always true, but we have \(CB = 3(x + 8)\) and \(AD = 4(x + 3)\), so set them equal? Wait, no, maybe \(CB + AD = CD + AB\)? Wait, let's draw the points: C---A---B---D. So \(CD = CA + AB + BD\), \(CB = CA + AB\), \(AD = AB + BD\). Since \(CA = BD\), let \(CA = BD = y\). Then \(CB = y + AB\), \(AD = AB + y\), so \(CB = AD\). Therefore, \(3(x + 8)=4(x + 3)\).

Step2: Solve for x

\(3(x + 8)=4(x + 3)\)
\(3x + 24 = 4x + 12\)
\(24 - 12 = 4x - 3x\)
\(x = 12\)

Step3: Find CB and AD

\(CB = 3(12 + 8)=3(20)=60\)
\(AD = 4(12 + 3)=4(15)=60\)

Step4: Find AB using CD

\(CD = CB + BD\), but \(BD = CA\), and \(CB = CA + AB\), so \(CD = (CA + AB) + CA = 2CA + AB\). Also, \(CB = CA + AB = 60\), so \(CA = 60 - AB\). Then \(CD = 2(60 - AB) + AB = 120 - AB\). But \(CD = 98\), so \(120 - AB = 98\)
\(AB = 120 - 98 = 22\)

Step1: Use midpoint property

Since \(T\) is the midpoint of \(\overline{AB}\), \(AT = TB\). So \(\frac{3}{4}x + 6=\frac{1}{2}x + 9\)

Step2: Solve for x

\(\frac{3}{4}x - \frac{1}{2}x = 9 - 6\)
\(\frac{3}{4}x - \frac{2}{4}x = 3\)
\(\frac{1}{4}x = 3\)
\(x = 12\)

Step3: Find AT (or TB)

\(AT=\frac{3}{4}(12)+6 = 9 + 6 = 15\)

Step4: Find AB

Since \(T\) is midpoint, \(AB = 2 \times AT = 2 \times 15 = 30\)

Step1: Segment addition postulate

For \(\overline{AC}\) with \(A---B---C\), \(AB + BC = AC\). So \((2x - 4) + 4x = 2x + 12\)

Step2: Solve for x

\(2x - 4 + 4x = 2x + 12\)
\(6x - 4 = 2x + 12\)
\(6x - 2x = 12 + 4\)
\(4x = 16\)
\(x = 4\)

Step3: Find AC

\(AC = 2x + 12 = 2(4) + 12 = 8 + 12 = 20\)

Answer:

22

Problem 8