Consider a square ABCD with midpoints E, F, G, H of AB, BC,
CD and DA respectively. Let L denote the
line passing through F and H. Consider points P and Q, on L and inside ABCD,
such that the angles APD and BQC both equal 120°. What is the ratio of the area
of ABQCDP to the remaining area inside ABCD?

(1) 4√2/3 (2) 2+√3 (3) (10-3√3)/9 (4) 1+(1/√3) (5) 2√3-1

Solution follows here:

__Solution:__
Let the length of side of square be ‘x’

Angle ∠APD = ∠BQC = 120

^{0}
Compare the two triangles APH and HPD:

HP is the common side in both the triangles; AH = HD = x/2;

∠AHP
= ∠DHP
= 90

^{0}
=> so by S-A-S rule, triangles APH and HPD are congruent

=>
∠APH
= ∠HPD
= 120

^{0}/2 = 60^{0}^{}

^{}

tan(∠APH) = AH/HP => tan(60

^{0}) = (x/2)/HP => HP = x/2√3
Area of ΔAHP = ½ (AH) (HP) = ½ (x/2) (x/2√3) = x

^{2}/8√3
Similarly ΔHPD = ΔBQF = ΔQFC = x

^{2}√3/8
Area of ABQCDP = Area of square - (ΔAHP + ΔHPD + ΔBQF + ΔQFC)

= x

^{2}- 4(x^{2}/8√3) = x^{2}- x^{2}/2√3 = x^{2}(2√3-1)/2√3
Remaining area inside ABCD = ΔAHP + ΔHPD + ΔBQF + ΔQFC = x

^{2}/2√3
∴Ratio of the area of ABQCDP to the
remaining area inside ABCD

= {x

^{2}(2√3-1)/2√3} / {x^{2}/2√3} = 2√3-1**Answer (5)**

## No comments:

## Post a Comment