Question : In $ riangle$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) of CQ is:
Option 1: $(2 \sqrt{21}+2)$
Option 2: $(2 \sqrt{18}-2)$
Option 3: $(2 \sqrt{17}-2)$
Option 4: $(2 \sqrt{19}+2)$

Question

Question : In $ riangle$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) of CQ is:

Option 1: $(2 \sqrt{21}+2)$

Option 2: $(2 \sqrt{18}-2)$

Option 3: $(2 \sqrt{17}-2)$

Option 4: $(2 \sqrt{19}+2)$

Team Careers360 · Accepted Answer

Correct Answer: $(2 \sqrt{17}-2)$

Solution :
In $ riangle $APQ and $ riangle $ABC,
$ \angle $P = $ \angle $B (corresponding angle)
$ \angle $Q = $ \angle $C (corresponding angle)
$ riangle $APQ $\sim$ $ riangle $ABC, by AA similarity.
Given that AP = QC, AB = 16 cm, and AQ = 4 cm.
Let the length of QC be $y$ cm.
We can use the property of similar triangles to set up the following proportion,
$⇒\frac{AP}{AB} = \frac{AQ}{AC}$
$⇒\frac{AB}{AP}= \frac{AC}{AQ}$
$⇒\frac{AB}{AP}-1= \frac{AC}{AQ}-1$
$⇒\frac{AB}{AP}-1= \frac{AC-AQ}{AQ}$
$⇒\frac{AB}{AP}-1= \frac{QC}{AQ}$
$⇒\frac{12}{QC}-1= \frac{QC}{4}$
$⇒\frac{12}{y}-1= \frac{y}{4}$
$⇒y^2+4y-64=0$
$⇒y=\frac{-4 \pm\sqrt{4^2-4 imes 1 imes (-64)}}{2 imes 1}$
$⇒y=\frac{-4 \pm\sqrt{272}}{2}$
$⇒y=\frac{-4 \pm 4\sqrt{17}}{2}$
$⇒y=(2 \sqrt{17}-2)$
Hence, the correct answer is $(2 \sqrt{17}-2)$.

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Question : 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 16 cm and the length of AQ is 4 cm, then the length (in cm) of CQ is:Option 1: (221+2)Option 2: (218−2)Option 3: (217−2)Option 4: (219+2)

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