Question

A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is

• A. 4units
• B. 8units
• C. 4√2units
• D. 2√2units

Answer 1

The Correct Option is A

Given: The area of triangle \( \triangle ABC = 32 \, \text{square units} \), and side \( BC \) lies on the vertical line \( x = 4 \), with length \( 8 \) units.

To find the shortest distance between point \( A \) and the origin \( (0, 0) \), we start by using the formula for the area of a triangle:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, base \( BC = 8 \) units and the area is given as 32, so we plug in:

\[ 32 = \frac{1}{2} \times 8 \times \text{height} \]

Solving for height:

\[ \text{height} = \frac{32 \times 2}{8} = 8 \, \text{units} \]

Since line \( BC \) lies on the vertical line \( x = 4 \), any point directly above or below it forming the height must also lie on a vertical line.

Thus, point \( A \) must lie on a line parallel to \( BC \), also at \( x = 4 \), so A’s x-coordinate is 4.

The shortest distance from point \( A \) to the origin \( (0, 0) \) is the horizontal distance, since both points lie on vertical lines:

\[ \text{Distance} = |4 - 0| = 4 \, \text{units} \]

Therefore, the shortest possible distance between A and the origin is:

\[ \boxed{4 \text{ units}} \]