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as shown in the figure, point o is the midpoint of $overline{ac}$, the …

Question

as shown in the figure, point o is the midpoint of $overline{ac}$, the diagonal of rectangle $abcd$. m is the midpoint of $overline{ad}$. if $om=\frac{5}{2}$ and $ob = \frac{13}{2}$, then the perimeter of $modc$ is ____. a 17 b 18 c 19 d 20

Explanation:

Step1: Use mid - point property in rectangle

In rectangle \(ABCD\), \(O\) is the mid - point of \(AC\) and \(M\) is the mid - point of \(AD\). Then \(OM\) is parallel to \(CD\) and \(OM=\frac{1}{2}CD\) (by the mid - point theorem in a triangle). Given \(OM = \frac{5}{2}\), so \(CD=5\).

Step2: Find the length of \(BC\)

Since \(O\) is the mid - point of \(AC\) and \(OB=\frac{13}{2}\), then \(AC = 2OB=13\) (diagonals of a rectangle are equal and bisect each other). In right - triangle \(ABC\) with \(AC = 13\) and \(AB = CD = 5\), by the Pythagorean theorem \(BC=\sqrt{AC^{2}-AB^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169 - 25}=\sqrt{144}=12\). Then \(AD = BC = 12\), and \(MD=\frac{1}{2}AD = 6\).

Step3: Find the length of \(OD\)

Since \(O\) is the mid - point of \(AC\) and \(AC = BD\) (diagonals of rectangle), \(OD=\frac{1}{2}BD=\frac{1}{2}AC=\frac{13}{2}\).

Step4: Calculate the perimeter of \(MODC\)

The perimeter of \(MODC\) is \(OM + MD+DC + OD=\frac{5}{2}+6 + 5+\frac{13}{2}=\frac{5 + 13}{2}+11=9 + 11=20\).

Answer:

D. 20