P is a point on side AB of a rectangle ABCD, which divides AB ........

A. 105 sq. units
B. 150 sq. units
C. 175 sq. units
D. 210 sq. units
E. 270 sq. units

Solution

• AB and CD are two parallel chords of a circle
• O is the center of the circle
• AB = 12 and CD = 6
• Diameter of the circle = 6√5

• The distance between the two chords, AB and CD

• \(AE = \frac{AB}{2} = \frac{12}{2} = 6\)
• OA = radius of the circle =\(\frac{6√5}{2} = 3√5\)
• \(∠OEA = 90^o\)

• Implies, \((3√5)^2 = OE^2 + 6^2\)
• \(OE^2 = 45 – 36 = 9\)
• Thus, OE = √9 = 3

• \(OD^2 = OF^2 + FD^2\)

o \((3√5)^2 = OF^2 + (\frac{6}{2})^2\)
o \(OF^2 = 45 – 9 = 36\)

• Thus, OF = √36 = 6

Solution

• P is a point on side AB of rectangle ABCD

o P divides AB in the ratio 5: 3

• Q is a point on side CD of rectangle ABCD

o Q divides CD in the ratio 3: 1

• The area of triangle ABCD = 480 sq. units

• The area of quadrilateral ADQP

• This implies,

o Length of AP = \(\frac{5x}{(5 + 3)} = \frac{5x}{8}\)
o Length of PB = \(\frac{3x}{(5 + 3)} = \frac{3x}{8}\)
o Length of CQ = \(\frac{3x}{(1 + 3)} = \frac{3x}{4}\)
o Length of QD = \(\frac{x}{(1 + 3)} = \frac{x}{4}\)

o We are given that area of rectangle ABCD = 480 = AB * BC = x * h