Question

1. a, b, and c are the midpoints of the sides of δdef.

a. find de, ef, and fd.
b. find the perimeter of δabc.
c. find the perimeter of δdef.
d. what is the ratio of the perimeters of δabc and δdef.

Explanation:

Step1: Recall Midline Theorem

The Midline Theorem (also known as the Midsegment Theorem) states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long. In \(\triangle DEF\), \(A\), \(B\), and \(C\) are midpoints, so \(AB\parallel EF\), \(AC\parallel DF\), and \(BC\parallel DE\), and \(AB = \frac{1}{2}EF\), \(AC=\frac{1}{2}DF\), \(BC = \frac{1}{2}DE\).

Step2: Solve part (a) - Find \(DE\), \(EF\), \(FD\)

• For \(DE\): Since \(BC\) is the midline and \(BC = 7\), by the Midline Theorem, \(DE=2\times BC = 2\times7 = 14\).
• For \(EF\): Since \(AB\) is the midline and \(AB = 5\), by the Midline Theorem, \(EF = 2\times AB=2\times5 = 10\).
• For \(FD\): Since \(AC\) is the midline and \(AC = 8\), by the Midline Theorem, \(FD=2\times AC = 2\times8 = 16\).

Step3: Solve part (b) - Perimeter of \(\triangle ABC\)

The perimeter of a triangle is the sum of its sides. For \(\triangle ABC\), sides are \(AB = 5\), \(BC = 7\), \(AC = 8\). So perimeter \(P_{ABC}=AB + BC+AC=5 + 7+8 = 20\).

Step4: Solve part (c) - Perimeter of \(\triangle DEF\)

The perimeter of \(\triangle DEF\) is the sum of its sides \(DE = 14\), \(EF = 10\), \(FD = 16\). So perimeter \(P_{DEF}=DE + EF+FD=14 + 10+16 = 40\).

Step5: Solve part (d) - Ratio of perimeters

The ratio of the perimeters of \(\triangle ABC\) and \(\triangle DEF\) is \(\frac{P_{ABC}}{P_{DEF}}=\frac{20}{40}=\frac{1}{2}\).

Answer:

s:
a. \(DE = 14\), \(EF = 10\), \(FD = 16\)
b. Perimeter of \(\triangle ABC\) is \(20\)
c. Perimeter of \(\triangle DEF\) is \(40\)
d. The ratio is \(\frac{1}{2}\)