Question

The sides of a right-angled triangle ae in an arithmetic progression. If the area of the triangle is 54, then the length of the longest side is

• A. 6
• B. 12
• C. 15
• D. 9
• E. 18

Answer 1

The Correct Option is C

Let the sides of the right-angled triangle be in arithmetic progression: \[ a - d, \quad a, \quad a + d \] where \( a > d > 0 \). Since it's a right-angled triangle, by the Pythagorean theorem: \[ (a - d)^2 + a^2 = (a + d)^2 \] Expanding: \[ a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2 \] Simplifying: \[ 2a^2 - 2ad = a^2 + 2ad \] \[ a^2 - 4ad = 0 \implies a(a - 4d) = 0 \] Since \( a > 0 \), we have \( a = 4d \). The area of the triangle is: \[ \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 = 54 \] \[ \frac{1}{2} \times (a - d) \times a = 54 \] Substitute \( a = 4d \): \[ \frac{1}{2} \times 3d \times 4d = 54 \implies 6d^2 = 54 \implies d^2 = 9 \implies d = 3 \] Thus: \[ a = 4d = 12 \] The sides are: \[ a - d = 9, \quad a = 12, \quad a + d = 15 \] The longest side (hypotenuse) is: \[ \boxed{15} \]

Answer 2

Let the sides of the right-angled triangle in arithmetic progression be \(a - d\), \(a\), and \(a + d\), where \(a + d\) is the hypotenuse.

By the Pythagorean theorem: \((a - d)^2 + a^2 = (a + d)^2\)

Expanding and simplifying: \(a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2\) which simplifies to \(a^2 - 4ad = 0\).

Factoring: \(a(a - 4d) = 0\). 

Since \(a\) cannot be zero, \(a = 4d\).

The area of the triangle is 54: \(\frac{1}{2} (a - d) a = 54\)

Substituting \(a = 4d\): \(\frac{1}{2} (4d - d) 4d = 54\), which simplifies to \(6d^2 = 54\).

Solving for \(d\): \(d^2 = 9\), so \(d = 3\) (since side lengths must be positive).

Finding \(a\): \(a = 4d = 4 \times 3 = 12\).

The longest side (hypotenuse) is \(a + d = 12 + 3 = 15\).

Therefore, the length of the longest side is 15.