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In ∆ ABC the straight line parallel to the side BC meets AB and AC at the points P and Q respectively. If AP = QC, the length of AB is 12 units and the length of AQ is 2 units, then the length (in units) of CQ is
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- 4
- 6
- 8
- 10
- 4
Correct Option: A
On the basis of question we draw a figure of triangle ABC ,
Here , AB = 12 units and AQ = 2 units
∠P = ∠B
∠Q = ∠C
∴ By AA — similarity,
∆ APQ ~ ∆ ABC
| ∴ | = | ||
| AB | AC |
| ⇒ | = | ||
| AP | AQ |
| ⇒ | - 1 = | - 1 = | |||
| AP | AQ | AQ |
| ⇒ | - 1 = | ||
| AP | AQ |
| ⇒ | - 1 = | ||
| QC | 2 |
| ⇒ | - 1 = | ||
| y | 2 |
| ⇒ | = | ||
| y | 2 |
⇒ y² + 2y – 24 = 0
⇒ y² + 6y – 4y – 24 = 0
⇒ y (y + 6) – 4(y + 6) = 0
⇒ (y – 4) (y + 6) = 0
⇒ y = 4 because y ≠ –6
