Question

Let l, m, n be three consecutive natural numbers (l > m > n). If the angle A of DABC be given by

Sin A = $\frac { \left( \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } \right) \left( \mathrm { l } ^ { 2 } + \mathrm { m } ^ { 2 } + \mathrm { n } ^ { 2 } \right) } { ( \mathrm { al } + \mathrm { bm } + \mathrm { cn } ) ^ { 2 } }$ and the perimeter of the triangle is 12 unit then the area of the triangle ABC is

• A. 4$\sqrt { 3 }$ sq. units
• B. 6 sq. units
• C. $\frac { 15 } { 2 }$ sq. units
• D. None of these

Answer 1

Answer: (B)

6 sq. units

Answer 2

(a² + b² + c²) (l² + m² + n²) – (al + bm + cn)²

= (am – bl)² + (bn – cm)² + (cl – an)² ≥ 0

\ $\frac { \left( \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } \right) \left( \mathrm { l } ^ { 2 } + \mathrm { m } ^ { 2 } + \mathrm { n } ^ { 2 } \right) } { ( \mathrm { al } + \mathrm { bm } + \mathrm { cn } ) ^ { 2 } }$ ≥ 1

But sin A ≤ 1

̃ $\frac { \left( \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } \right) \left( \mathrm { l } ^ { 2 } + \mathrm { m } ^ { 2 } + \mathrm { n } ^ { 2 } \right) } { ( \mathrm { al } + \mathrm { bm } + \mathrm { cn } ) ^ { 2 } }$ = 1

̃ $\frac { \mathrm { a } } { \mathrm { l } } = \frac { \mathrm { b } } { \mathrm { m } } = \frac { \mathrm { c } } { \mathrm { n } } = \lambda$(say)

As l = m + 1, n = m – 1 ̃ a + b + c = l (3m)

̃ lm = 4 ̃ a = l(m + 1)

̃ 4 + l, b = 4, c = l(m – 1) = 4 – l

As sinA = 1 ̃ A = $\frac { \pi } { 2 }$, thus a² = b² + c²

̃ l = 1 ̃ a = 5, b = 4, c = 3

Area of DABC = $\frac { 1 } { 2 }$× 4 × 3 = 6