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In the figure shown below, ABC is a triangle with A = 45°. P and Q are the points on the sides AB and AC respectively such that PQ is parallel to BC. Also, the diagonals of the trapezium PQCB i.e. BQ and PC intersect at right angles. If PQ = 6 cm and BC = 14 cm, find the area (in cm2) of the trapezoid PQCB.
  • a)
    84
  • b)
    102
  • c)
    105
  • d)
    98
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
In the figure shown below, ABC is a triangle with A = 45°. P and Q ar...
We have ∠M=90∘
Area of the trapezium PQBC= Area of ΔPBQ+ Area of ΔCBQ
=1/2(BQ×PM)+1/2(BQ×CM)
=1/2BQ(PM+CM)
=1/2(BQ×CP)
We have ∠PQM≅∠CBM
…(∵PQ‖BC)
∠QPM≅∠BCM
Also, ∠QMP≅∠BMC
.. (Vertically opposite angles)
Hence ΔPMQ∼ΔCMB
(A-A-A Test of similarity)
Thus we have CP=10×&BQ=10y
Hence, area of the trapezium
We have,
∴ Area of trapezoid PQCB = 50 × xy=105cm2
Hence option 3.
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In the figure shown below, ABC is a triangle with A = 45°. P and Q are the points on the sides AB and AC respectively such that PQ is parallel to BC. Also, the diagonals of the trapezium PQCB i.e. BQ and PC intersect at right angles. If PQ = 6 cm and BC = 14 cm, find the area (in cm2) of the trapezoid PQCB.a)84b)102c)105d)98Correct answer is option 'C'. Can you explain this answer?
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