Question

In a triangle ABC, medians AD and CE are drawn. If AD = 5, ∠DAC =$\frac { \pi } { 4 }$, the area of triangle ABC is –

• A. 15/7
• B. 25/3
• C. 25/6
• D. None of these

Answer 1

Answer: (B)

25/3

Answer 2

Let O be the point of intersection of the medians of triangle ∆ABC shown in the figure. Then the area of ∆ABC is three times that of ∆AOC, because O being the centroid of ∆ABC.

It divides of median through B in the ratio 2 : 1 and the height of ∆AOC is one-third that of ∆ABC

Now, in ∆AOC, AO = 10/3 AD = 10/3.

Applying the sine rule to ∆AOC, we get

$\frac { \mathrm { OC } } { \sin \frac { \pi } { 8 } }$ = $\frac { \mathrm { AO } } { \sin \frac { \pi } { 4 } }$

⇒ OC =$\frac { 10 } { 3 }$ $\frac { \sin \left( \frac { \pi } { 8 } \right) } { \sin \left( \frac { \pi } { 4 } \right) }$

∴ Area of ∆AOC = $\frac { 1 } { 2 }$ . AO . OC. sin ∠AOC

= $\frac { 1 } { 2 } \cdot \frac { 10 } { 3 } \cdot \frac { 10 } { 3 } \cdot \frac { \sin ( \pi / 8 ) } { \sin ( \pi / 4 ) }$ .sin $\left( \frac { \pi } { 2 } + \frac { \pi } { 8 } \right)$

= $\frac { 50 } { 9 }$ .$\frac { \sin ( \pi / 8 ) \cos ( \pi / 8 ) } { \sin ( \pi / 4 ) }$=$\frac { 50 } { 18 }$= $\frac { 25 } { 9 }$

⇒ Area of ABC = $\frac { 3 \times 25 } { 9 }$ = $\frac { 25 } { 3 }$ .