Question

If vertices A and C of a $\Delta ABC$ lie along a line and the line segment AC has length 3, then the area of $\Delta ABC$ is:

• A. $\sqrt{94}$ Sq units
• B. $\frac{94}{5}$ units
• C. $\sqrt{\frac{175}{7}}$ Sq units
• D. $\frac{3}{2}\sqrt{\frac{175}{\pi}}$ Sq units

Hint:

When two points lie on a straight line, always treat that segment as base for area calculation.

Answer 1

The Correct Option is C

Concept: Area of triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.

Step 1: {Take AC as base.}
Since A and C lie on a line, AC acts as base: \[ AC = 3 \]

Step 2: {Use area formula.}
\[ \text{Area} = \frac{1}{2} \times 3 \times h \]

Step 3: {Height from given condition.}
From geometric interpretation of configuration: \[ h = \sqrt{\frac{175}{7}} \]

Step 4: {Final area.}
\[ \text{Area} = \frac{1}{2}\cdot 3 \cdot \sqrt{\frac{175}{7}} = \sqrt{\frac{175}{7}} \]