Question

1. in the figure shown, \\(c\\) is on the segment with endpoints \\(a\\) and \\(d\\). the distance between \\(a\\) and \\(b\\) is 2,000 km, between \\(a\\) and \\(c\\) is 1,600 km, between \\(a\\) and \\(d\\) is 2,500 km, and between \\(b\\) and \\(c\\) is 1,200 km. what is the distance, in kilometers, between \\(b\\) and \\(d\\)?

a. \\(300\sqrt{7}\\)
b. 900
c. 1,200
d. 1,500

Explanation:

Identify the triangle properties

We are given a triangle \(\triangle ABD\) where point \(C\) lies on the segment \(AD\). The given lengths are:

• \(AB = 2,000\text{ km}\)
• \(AC = 1,600\text{ km}\)
• \(AD = 2,500\text{ km}\)
• \(BC = 1,200\text{ km}\)

We need to find the distance \(BD\).

Analyze triangle ABC

Let's check if \(\triangle ABC\) is a right triangle. We apply the Pythagorean Theorem to the side lengths of \(\triangle ABC\):

$$ AC^2 + BC^2 = 1,600^2 + 1,200^2 = 2,560,000 + 1,440,000 = 4,000,000 $$

$$ AB^2 = 2,000^2 = 4,000,000 $$

Since \(AC^2 + BC^2 = AB^2\), by the converse of the Pythagorean Theorem, \(\triangle ABC\) is a right triangle with the right angle at vertex \(C\). Thus, \(\angle ACB = 90^\circ\).

Determine the geometry of triangle BCD

Since \(C\) lies on the straight line segment \(AD\), the angles \(\angle ACB\) and \(\angle DCB\) are supplementary:

$$ \angle DCB = 180^\circ - \angle ACB = 180^\circ - 90^\circ = 90^\circ $$

Therefore, \(\triangle BCD\) is also a right triangle with the right angle at vertex \(C\).

Calculate the length of CD

Using Right Triangle Geometry, we find the length of segment \(CD\):

$$ CD = AD - AC = 2,500 - 1,600 = 900\text{ km} $$

Solve for BD

We apply the Pythagorean Theorem to the right triangle \(\triangle BCD\):

$$ BD^2 = BC^2 + CD^2 $$

$$ BD^2 = 1,200^2 + 900^2 = 1,440,000 + 810,000 = 2,250,000 $$

$$ BD = \sqrt{2,250,000} = 1,500\text{ km} $$

Answer:

• A. \(300\sqrt{7}\)
• B. 900
• C. 1,200
• D. 1,500 (Correct answer)