The base of an isosceles triangle is equal to 4 , the base angle is equal to 45°. A straigth line cuts the extension of the base at a point M at the angle θ where θ is acute and bisects the lateral side of the triangle which is nearest to M. The area 'A' of the quadrilateral which the straight line cuts off from given triangle is

Question

The base of an isosceles triangle is equal to 4 , the base angle is equal to 45°. A straigth line cuts the extension of the base at a point M at the angle θ where θ is acute and bisects the lateral side of the triangle which is nearest to M. The area 'A' of the quadrilateral which the straight line cuts off from given triangle is (A) (3+ tan θ/1+ tan θ) (B) (3+2 tan θ/1+ tan θ) (C) (3+ tan θ/1- tan θ) (D) (3+5 tan θ/1+ tan θ)

Tardigrade Admin · Accepted Answer

Correct answer is (d) (3+5 tan θ/1+ tan θ)

$\frac{3 + t a n θ}{1 + t a n θ}$

$\frac{3 + 2 t a n θ}{1 + t a n θ}$

$\frac{3 + t a n θ}{1 - t a n θ}$

$\frac{3 + 5 t a n θ}{1 + t a n θ}$

Correct answer is (d) $\frac{3 + 5 t a n θ}{1 + t a n θ}$

1+tanθ3+tanθ​

1+tanθ3+2tanθ​

1−tanθ3+tanθ​

1+tanθ3+5tanθ​

Correct answer is (d) 1+tanθ3+5tanθ​

$\frac{3 + t a n θ}{1 + t a n θ}$

$\frac{3 + 2 t a n θ}{1 + t a n θ}$

$\frac{3 + t a n θ}{1 - t a n θ}$

$\frac{3 + 5 t a n θ}{1 + t a n θ}$

Correct answer is (d) $\frac{3 + 5 t a n θ}{1 + t a n θ}$