Question

In a right-angled \(△ABC\), the measures of the angles are in an Arithmetic Progression (A.P.) If its smallest side is \(4\) units, then the area of \(△ABC\) is.....

• A. \(16\) sq. units
• B. \(8\sqrt{3}\) sq. units
• C. \(16\sqrt{3}\) sq. units
• D. \(32\) sq. units

Hint:

Angles in AP with a right angle must be 30, 60, 90.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The three angles sum to \(180^{\circ}\) and are in AP. Let them be \(a - d,\ a,\ a + d\). Then \(3a = 180^{\circ}\), so the middle angle is \(60^{\circ}\).

Step 2: Find the angles:
One angle is \(90^{\circ}\), so \(60 + d = 90\), \(d = 30^{\circ}\). The angles are \(30^{\circ},\ 60^{\circ},\ 90^{\circ}\).

Step 3: Find the sides:
Sides are in the ratio \(1 : \sqrt{3} : 2\). The smallest side (opposite \(30^{\circ}\)) is 4, so the other leg is \(4\sqrt{3}\) and the hypotenuse is 8.

Step 4: Area:
\[ \text{Area} = \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3} \text{ sq. units} \]

Step 5: Why the other options are wrong.
16 would be the area of a right isosceles triangle with legs \(4\sqrt2\) or a different ratio. \(16\sqrt{3}\) results from taking the hypotenuse as 8 and a leg of 4 and multiplying without the half. 32 does not match this triangle.

Final Answer:
The area is \(8\sqrt{3}\) sq. units, option (B). \[ \boxed{8\sqrt{3}\text{ sq. units}} \]