Question

Triangle ABC is isosceles with AB = AC and BC = 65 cm. P is a point on BC such that the perpendicular distances from P to AB and AC are 24 cm and 36 cm, respectively. The area of triangle ABC in sq cm is-

• A. 1254
• B. 1950
• C. 2335
• D. 5070

Answer 1

Answer: (C)

2335

Answer 2

A = $\frac { 1 } { 2 }$b² sin2q = b²sinqcosq ....(i)

Now, cosecq = =

60x = (24)(65)

x = 26

\ sinq = $\frac { 12 } { 13 }$ and cosq = $\frac { 5 } { 13 }$

Again,

̃ b = $\frac { 65 } { 2 \cos \theta }$ = $\frac { ( 65 ) ( 13 ) } { ( 2 ) ( 5 ) }$ = $\frac { 13 ^ { 2 } } { 2 }$

From Eq.(i), we get

A = $\frac { 13 ^ { 4 } } { 4 } \cdot \frac { 12 } { 13 } \cdot \frac { 5 } { 13 }$ = (169)(15) = 2535