Think along the lines
Geometry
Level
3
Triangle ABC is right angled at B. AB=4 BC=3 P is the midpoint of AB Q lies on AC and PQ is perpendicular to AC Find the value of (5BQ/4) times (5BQ/4)
The answer is 13.
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1 solution
we know that AP = PB = 2. Also, angle BAC = 37
and angle ACB = 53. In triangle APQ, angle A = 37. and hypotenuse = AP = 2. So, AQ = 2 cos37= 1.6 Hence AC = 3.4 Now drop a perpendicular on BC at D. In triangle QDC, angle D = 90. Angle C = 53. so, QD = QC sin53= 2.72 similarly, DC = QC cos53= 2.04 hence BD = 0.96 So, in triangle QBD angle D = 90` QD = 2.72 BD = 0.96 Applying Pythagoras theorem we get BQ^2 = 8.32 we want to know (5BQ/4) *(5BQ/4) i.e [25(BQ^2)] /16 = 13