Geometry-15 (CAT-2008)

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 = 120⁰

Compare the two triangles APH and HPD:

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

∠AHP = ∠DHP = 90⁰ 

=> so by S-A-S rule, triangles APH and HPD are congruent

=> ∠APH = ∠HPD = 120⁰/2 = 60⁰

tan(∠APH) = AH/HP => tan(60⁰) = (x/2)/HP => HP = x/2√3

Area of ΔAHP = ½ (AH) (HP) = ½ (x/2) (x/2√3) = x²/8√3

Similarly ΔHPD = ΔBQF = ΔQFC = x²√3/8

Area of ABQCDP = Area of square - (ΔAHP + ΔHPD + ΔBQF + ΔQFC)

= x² - 4(x²/8√3) = x² - x²/2√3 = x² (2√3-1)/2√3

Remaining area inside ABCD = ΔAHP + ΔHPD + ΔBQF + ΔQFC = x²/2√3

∴Ratio of the area of ABQCDP to the remaining area inside ABCD

= {x² (2√3-1)/2√3} / {x²/2√3} = 2√3-1

Answer (5)