★ Triangles ★
1: E and F are the points on the side PQ and PR respectively of a ∆ PQR . For each of the following cases , state whether EF || QR :
• PE = 3.9 cm , EQ = 3cm , PF = 3.6 cm and FR = 2.4
• PE = 4 cm , QE = 4.5 cm , PF = 8 cm , RF = 9 cm
• PQ = 1.28 cm , PR = 2.56 cm , PE = 0.18 cm and PF = 0.36 cm
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Answers
Answer:
Using Basic proportionality theorem,
∴
EQ
PE
=
3
3.9
=
30
39
=
10
13
=1.3
FR
PF
=
2.4
3.6
=
24
36
=
2
3
=1.5
EQ
PE
=
FR
PF
So, EF is not parallel to QR.
Step-by-step explanation:
PE=4 cm, QE=4.5 cm, PF=8 cm, RF=9 cm
Using Basic proportionality theorem,
∴
QE
PE
=
4.5
4
=
45
40
=
9
8
RF
PF
=
9
8
QE
PE
=
RF
PF
So, EF is parallel to QR.
Answer:
Given:-
In APQR, E and F are two points on side PQ and PR respectively.
* Note:- Look into the attechment for the diagram
(i) Given:
• PE = 3.9 cm · EQ = 3 cm • PF = 3.6 cm • FR = 2,4 cm
Therefore, using BPT, we get:
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, PE/EQ PF/FR
Hence, EF is not parallel to QR.
(ii) Given:
• PE = 4 cm
• QE = 4.5 cm • PF = 8cm • RF = 9cm
* Using BPT :-
PE/QE = 4/4.5 = 40/45 = 8/9
And, PF/RF = 8/9
So, we get here,
PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) Given:
• PQ = 1.28 cm • PR = 2.56 cm
• PE = 0.18 cm • PF = 0.36 cm
EQ = PQ – P = 1.28 – 0.18 = 1.10 cm
FR = PR PF = 2.56 -0.36 = 2.20 cm
PE/EQ = 0.18/1.10 = 18/110 = 9/55 - (i)
PE/FR = 0.36/2.20 = 36/220 = 9/55 + (ii)
So, we get:
PE/EQ = PF/FR
Hence, EF is parallel to QR.
