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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

9
9

5
5

11
11

12

1 answer

To solve for the whole-number measure of segment \( AD \) that would create triangles \( \triangle ABD \) and \( \triangle ACD \), we can use the properties of a rhombus and the triangle inequality theorem.

In a rhombus, the opposite sides are equal, and the diagonals bisect each other at right angles. This means that:

  • \( AB = AD \) (sides of the rhombus)
  • \( AC = BC \)

From the information given:

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

Given that \( AD \) is also a side of the rhombus, we can denote its length as \( AD \).

We can form triangles from the sides given:

  1. For \( \triangle ABD \):

    • The sides are \( AB = 4 \), \( AD = x \), and \( BD = 8 \).
    • By the triangle inequality theorem:
      • \( AB + AD > BD \) → \( 4 + x > 8 \) → \( x > 4 \)
      • \( AB + BD > AD \) → \( 4 + 8 > x \) → \( 12 > x \)
      • \( BD + AD > AB \) → \( 8 + x > 4 \) → \( x > -4 \) (not a constraint)

  2. For \( \triangle ACD \):

    • The sides are \( AC = 5 \), \( AD = x \), and \( CD = 15 \).
    • Again by using the triangle inequality:
      • \( AC + AD > CD \) → \( 5 + x > 15 \) → \( x > 10 \)
      • \( AC + CD > AD \) → \( 5 + 15 > x \) → \( 20 > x \)
      • \( CD + AD > AC \) → \( 15 + x > 5 \) → \( x > -10 \) (not a constraint)

Now, we need to combine the inequalities obtained from both triangles:

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

The intersection of these two sets of inequalities gives us:

  • \( 10 < x < 12 \)

Since \( x \) must be a whole number, the only value that satisfies this inequality is \( x = 11 \).

Therefore, the whole-number measure of segment \( AD \) that would create triangles \( \triangle ABD \) and \( \triangle ACD \) is \( \boxed{11} \).