ABDC is a trapezium with AB||CD. P and Q are midpoints of AC and BD respectively. If AB = 3 cm and CD = 5 cm, Find the ratio of the area of ABQP to the area of PQDC.

  1. 7 : 6
  2. 8 : 5
  3. 7 : 9
  4. 6 : 5

Answer (Detailed Solution Below)

Option 3 : 7 : 9
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Detailed Solution

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Given:

AB||CD, AB = 3 cm and CD = 5 cm

P and Q are midpoints of AC and BD ⇒ AP : PC = 1 : 1

F1 M.G 6.6.20 Pallavi D2

Formula used:

Area of trapezium = 1/2 × sum of lengths of parallel sides × height,

Where, height = distance between parallel sides

Calculation:

PQ = ½ [AB + CD]

⇒ ½ [3 + 5] = 4 cm

⇒ AP : PC = 1 : 1

⇒ height of trapezium ABQP = height of trapezium PQDC = h (assume)

⇒ area of ABQP : area of PQDC

⇒ [1/2 × (3 + 4) × h] : [1/2 × (4 + 5) × h]

⇒ 7 : 9

∴ The ratio is 7 : 9.

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