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 $PF=0.36cm$

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, $EF \parallel QR$

You're just one step away

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.

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, $EF\parallel QR$ You're just one step away

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

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 $\parallel$' $QR$ You're just one step away

One Word Medium Published on 09th 09, 2020
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