## Tuesday, 15 November 2011

### 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