In Δ 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 16 cm and the length of AQ is 4 cm, then the length (in cm) CQ is:

Question

Question

This question was previously asked in

SSC CHSL Exam 2023 Tier-I Official Paper (Held On: 02 Aug, 2023 Shift 2)

Answer 1

Given:

In Δ ABC , PQ is parallel to the side BC . AP = QC , Ab = 16cm and AQ = 4cm

Concept used:

In Δ ABC , PQ parallel to the side BC

So AP / PB = AQ /QC

Calculation:

F2 Savita SSC 15-2-24 D1

Let AP = QC = x , PB = 16 - x

As per the question,

\(\frac{x}{16-x}=\frac{4}{x}\)

⇒ x² + 4x - 64 = 0

⇒ x = \(-b \pm \sqrt{b^2-4ac} \over 2a\)

⇒ x = \(-4 \pm \sqrt{4^2-4(1)(-64)} \over 2(1)\)

⇒ x = \(-4 \pm \sqrt{16+256} \over 2\)

⇒ x = \(-4 \pm \sqrt{272} \over 2\)

⇒ x = \(-4 \pm 4\sqrt{17} \over 2\)

⇒ x = \(2\sqrt{17}-2 \)

∴ The correct option is 3.

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