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

Question

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

5
5

12
12

11
11

9

Answer 1

To determine the length of segment \( AD \) such that triangles \( \triangle ABD \) and \( \triangle ACD \) can be formed, we can use the following relationships based on the properties of the rhombus and triangles.

In the rhombus, the lengths of the sides must be equal. Hence, we have:

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

We can define the length of segment \( AD \) as \( x \). Both triangles \( \triangle ABD \) and \( \triangle ACD \) will use \( AD \) in their calculations.

Using the triangle inequality for \( \triangle ABD \):

1. \( AB + AD > BD \) ⇒ \( 4 + x > 8 \) ⇒ \( x > 4 \)
2. \( AB + BD > AD \) ⇒ \( 4 + 8 > x \) ⇒ \( 12 > x \) ⇒ \( x < 12 \)
3. \( AD + BD > AB \) ⇒ \( x + 8 > 4 \) ⇒ \( x > -4 \) (this is always true)

From \( \triangle ABD \), we can conclude:

• \( 4 < x < 12 \)

Next, for \( \triangle ACD \):

1. \( AC + AD > CD \) ⇒ \( 5 + x > 15 \) ⇒ \( x > 10 \)
2. \( AC + CD > AD \) ⇒ \( 5 + 15 > x \) ⇒ \( 20 > x \) ⇒ \( x < 20 \)
3. \( AD + CD > AC \) ⇒ \( x + 15 > 5 \) ⇒ \( x > -10 \) (this is always true)

From \( \triangle ACD \), we can conclude:

• \( 10 < x < 20 \)

Combining the results from \( \triangle ABD \) and \( \triangle ACD \):

• The only measure that satisfies both conditions is \( 10 < x < 12 \).

Considering whole numbers, the potential candidates for \( x \) could be \( 11 \).

Thus, \( AD \) must be \( \boxed{11} \).