Question

geometry
segment addition and subtraction postulate
find the length ind
1)

points a, b, and c

1. ( ac = 16 ) and ( ab = 11 ). find ( bc ).

find the length indicated.

1. find ( mk )

points a, b, c, and

1. ( bd = 14, bc = 6 ), and ( ad = 18 ). find ( ac ).

Explanation:

Step1: Use segment addition postulate

For collinear points \(A\), \(B\), \(C\) with \(B\) between \(A\) and \(C\), \(AC = AB+BC\). So \(BC=AC - AB\). Given \(AC = 16\) and \(AB = 11\), then \(BC=16 - 11\).

Step2: Calculate the value

\(BC=16 - 11=5\)

Step3: Use segment addition postulate for the second - part

Since \(AC = AB + BC\), then \(AB=AC - BC\). Given \(AC = 17\) and \(BC = 6\), so \(AB=17 - 6\)

Step4: Calculate the value

\(AB=17 - 6 = 11\)

Step5: Use segment addition postulate for \(MK\)

We know that \(NL=11\), \(NK = 15\), so \(LK=NK - NL=15 - 11 = 4\). And \(ML = 8\), then \(MK=ML+LK\)

Step6: Calculate \(MK\)

\(MK=8 + 4=12\)

Step7: Use segment addition postulate for \(DC\)

We know that \(FC=FE + ED+DC\). Given \(FE = 7\), \(ED = 4\), \(FC = 17\). So \(DC=FC-(FE + ED)\)

Step8: Calculate \(DC\)

\(DC=17-(7 + 4)=17 - 11 = 6\)

Step9: Use segment addition postulate for \(AC\) (first case)

We know that \(AD = AB+BD\), so \(AB=AD - BD\). Given \(AD = 18\), \(BD = 14\), then \(AB=18 - 14 = 4\). And \(AC=AB + BC\), given \(BC = 6\), so \(AC=4+6\)

Step10: Calculate \(AC\) (first case)

\(AC=4 + 6=10\)

Step11: Use segment addition postulate for \(AC\) (second case)

We know that \(AD = AB+BD\), so \(AB=AD - BD\). Given \(AD = 17\), \(BD = 10\), then \(AB=17 - 10 = 7\). And \(AC=AB + BC\), given \(BC = 1\), so \(AC=7 + 1\)

Step12: Calculate \(AC\) (second case)

\(AC=7+1 = 8\)

Answer:

1. \(BC = 5\)
2. \(AB = 11\)
3. \(MK = 12\)
4. \(DC = 6\)
5. \(AC = 10\)
6. \(AC = 8\)