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Identify the whole-number measure of segment AD¯¯¯¯¯¯¯¯

that would create △ABD
and △ACD
as shown.

An illustration shows a rhombus where the vertices are labeled as follows: A on the top vertex, B on the left, C on the right, and D at the bottom. A dashed line is drawn from A to D. The sides are labeled as follows: A C is labeled as 5, A B as 4, B D as 8, and C D as 15.

(1 point)
Responses

12
12

11
11

9
9

5

1 answer

To determine the length of segment \( AD \) that would create triangles \( \triangle ABD \) and \( \triangle ACD \) in the given rhombus configuration, we can utilize the triangle inequality theorem.

Since a rhombus has opposite sides that are equal, we know that:

  • \( AB = 4 \)
  • \( AC = 5 \)
  • \( BD = 8 \)
  • \( CD = 15 \)

We have to find \( AD \). Let's denote this length as \( AD = x \).

By applying the triangle inequality theorem to triangles \( \triangle ABD \) and \( \triangle ACD \):

For \( \triangle ABD \):

  1. \( AB + AD > BD \)

    • \( 4 + x > 8 \)
    • \( x > 4 \)

  2. \( AB + BD > AD \)

    • \( 4 + 8 > x \)
    • \( 12 > x \) or \( x < 12 \)

  3. \( AD + BD > AB \)

    • \( x + 8 > 4 \)
    • \( x > -4 \) (This is always true since \( x \) is a length)

Combining results for triangle \( \triangle ABD \):

  • \( 4 < x < 12 \)

For \( \triangle ACD \):

  1. \( AC + AD > CD \)

    • \( 5 + x > 15 \)
    • \( x > 10 \)

  2. \( AC + CD > AD \)

    • \( 5 + 15 > x \)
    • \( 20 > x \) or \( x < 20 \)

  3. \( AD + CD > AC \)

    • \( x + 15 > 5 \)
    • \( x > -10 \) (This is always true since \( x \) is a length)

Combining results for triangle \( \triangle ACD \):

  • \( 10 < x < 20 \)

Final Result:

Considering both sets of inequalities:

  • From \( \triangle ABD \): \( 4 < x < 12 \)
  • From \( \triangle ACD \): \( 10 < x < 20 \)

Combining these constraints gives us:

  • \( 10 < x < 12 \)

Thus, the only whole-number measure for \( AD \) in this range is \( 11 \).

So, the answer is \( 11 \).