Question

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

= $\frac { \left( \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } \right) \left( \ell ^ { 2 } + \mathrm { m } ^ { 2 } + \mathrm { n } ^ { 2 } \right) } { ( \mathrm { a } \ell + \mathrm { bm } + \mathrm { cn } ) ^ { 2 } }$ and the perimeter of the

triangle is 12 unit then the area of the DABC 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 – cn)² + (cl – an)² ³ 0

\$\frac { \left( \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } \right) / \left( \ell ^ { 2 } + \mathrm { m } ^ { 2 } + \mathrm { n } ^ { 2 } \right) } { ( \mathrm { a } \ell + \mathrm { bm } + \mathrm { cn } ) ^ { 2 } }$ ³ 1

but sin A £ 1 ̃$\frac { \left( \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } \right) \left( \ell ^ { 2 } + \mathrm { m } ^ { 2 } + \mathrm { n } ^ { 2 } \right) } { ( \mathrm { a } \ell + \mathrm { bm } + \mathrm { cn } ) ^ { 2 } }$= 1

̃ $\frac { \mathrm { a } } { \ell }$ = = $\frac { \mathrm { c } } { \mathrm { n } }$ = l (say)

As $\ell$ = 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 sin A = 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