Passage

$$E$$ and $$F$$ are points on the sides $$PQ$$ and $$PR$$ respectively of a $$\triangle PQR$$. For the given case, state whether $$EF\parallel QR$$.
Mathematics

$$PQ=1.28cm, PR=2.56cm, PE=0.18cm$$ and $$PF=0.36cm$$


ANSWER

True


SOLUTION
Given PQ=1.28 cm,PR=2.56 cm,PE=0.18 cm and PF=0.36 cm where E and F are points on PQ and PR respectively.
Now, $$\dfrac{PE}{PQ} = \dfrac{0.18}{1.28} = \dfrac{9}{64}$$
$$\dfrac{PF}{PR} = \dfrac{0.36}{2.56} = \dfrac{9}{64}$$
thus, $$\dfrac{PE}{PQ} = \dfrac{PF}{PR}$$
By converse of equal Intercept theorem, $$EF \parallel QR$$
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Single Correct Medium Published on 09th 09, 2020
Mathematics

$$PE=4cm, QE=4.5cm, PF=8cm$$ and $$RF=9cm$$.If True enter 1 else if False enter 0.


ANSWER

1


SOLUTION
In figure
$$\cfrac {PE}{EQ}=\cfrac{4}{4.5}=\cfrac {8}{9}$$ and $$\cfrac{PF}{FR}=\cfrac{8}{9}$$
$$\Rightarrow$$ $$\cfrac {PE}{EQ}=\cfrac {PF}{FR}$$
From the converse of proportionality theorem, $$EF\parallel QR$$
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One Word Medium Published on 09th 09, 2020
Mathematics

$$PE=3.9cm, EQ=3cm, PF=3.6cm$$ and $$FR=2.4cm$$. If $$EF \parallel QR$$ enter 1 else enter 0


ANSWER


SOLUTION
In the figure, 
$$\cfrac {PE}{EQ}=\cfrac{3.9}{3}=1.3$$
$$\cfrac {PF}{FR}=\cfrac{3.6}{2.4}=\cfrac{3}{2}=1.5$$
$$\Rightarrow$$ $$\cfrac {PE}{EQ}\ne \cfrac{PF}{FR}$$
$$\Rightarrow$$ is not $$\parallel$$' $$QR$$
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One Word Medium Published on 09th 09, 2020
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