SATHEE: Quantitative Aptitude Ques 1565

Quantitative Aptitude Ques 1565

Question: In the given figure below, $ \angle PQR=90{}^\circ $ and QL is a median, $ PQ=5cm $ and $ QR=12cm. $ Then, QL is equal to [CDS 2013]

Options:

A) $ 5cm $

B) $ 5.5cm $

C) $ 6cm $

D) $ 6.5cm $

Show Answer

Answer:

Correct Answer: D

Solution:

Given that, $ PQ=5cm, $ $ QR=12cm $ and QL is a median.

$ \therefore $ $ PL=LR=\frac{PR}{2} $ (i) In $ \Delta PQR, $ $ (PR^{2})={{(PQ)}^{2}}+{{(QR)}^{2}} $ [by Pythagoras theorem] $ ={{(5)}^{2}}+{{(12)}^{2}} $ $ =25+144=169={{(13)}^{2}} $

$ \Rightarrow $ $ PR^{2}={{(13)}^{2}} $
$ \Rightarrow $ $ PR=13 $ Now, by theorem, it L is the mid-point of the hypotenuse PR of a right angled $ \Delta PQR. $ Then, $ QL=\frac{1}{2}PR=\frac{1}{2}(13)=6.5cm $

Quantitative Aptitude Ques 1564

Quantitative Aptitude Ques 1566

Quantitative Aptitude Ques 1565

Question: In the given figure below, $ \angle PQR=90{}^\circ $ and QL is a median, $ PQ=5cm $ and $ QR=12cm. $ Then, QL is equal to [CDS 2013]

Options:

Answer:

Solution:

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