The base of an isosceles triangle measures 4 units base angle is equal to `45^(@)`. A straight line cuts the extension of the base at a point M at the angle `theta` and bisects the lateral side of the triangle which is nearest to M.
The area of quadrilateral which the straight line cuts off from the given triangle is

Options

A. `(3+"tan" theta)/(1+"tan" theta)`
B. `(3+5"tan" theta)/(1+"tan" theta)`
C. `(3+"tan" theta)/(1-"tan" theta)`
D. `(3+2"tan" theta)/(1+"tan" theta)`

Answer

B

Explanation

For the given specification of the triangle, consider triangle ABC as shown in the figure.

Equation of line PM is
`y-1 = "tan " theta(x-1)`
For point Q, solving MP and AC,
`4-x-1 = "tan "theta(x-1)`
`rArr Q-=((3+"tan"theta)/(1+"tan" theta),(1+3 "tan" theta)/(1+"tan" theta))`
`"Area of "DeltaAPQ = (1)/(2)||{:(1,1,1),(2,2,1),((3+"tan"theta)/(1+"tan"theta), (1+3"tan"theta)/(1+"tan" theta), 1):}|| = (1-"tan"theta)/(1+"tan"theta)" " ("As " theta lt pi//4)`
Area of quadilateral BPQC,
`Delta = (1)/(2) xx 4 xx 2-((1-"tan"theta)/(1+"tan" theta))`
`"or "Delta = ((3+5"tan"theta)/(1+"tan" theta))`
`"or "Delta = 5-(2)/(1+"tan" theta)" where " theta in (0, (pi)/(4))`
`therefore Delta in (3,4)`
`(PQ)^(2) = ((3+"tan" theta)/(1+"tan" theta)-1)^(2) + ((1+3"tan" theta)/(1+"tan" theta)-1)^(2)`
`=(4(1+"tan"^(2) theta))/(1+"tan"^(2)theta + 2 "tan" theta)`
`=(4)/(1+(2 "tan" theta)/(1+"tan"^(2)theta))`
`=(4)/(1+"sin" 2theta)`
`"So, (PQ)^(2) in (2,4)`
`therefore PQ in (sqrt(2), 2)`