Triangle
Exercise 6.2 Part 1
Question 1: In the given figures, DE || BC. Find EC in first figure and AD in second figure.
Solution: In the first figure;
Δ ADE ~ Δ ABC (Because DE || BC)
Hence;
`(AD)/(DC)=(AE)/(EC)`
Or, `(1.5)/(3)=(1)/(EC)`
Or, `EC=(3)/(1.5)=2 cm`
Similarly, in the second figure;
Δ ADE ~ Δ ABC (Because DE || BC)
Hence;
`(AD)/(DC)=(AE)/(EC)`
Or, `(AD)/(7.2)=(1.8)/(5.4)`
Or, `AD=(1.8xx7.2)/(5.4)=2.4 cm`
Question 2: E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR.
(a) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Solution: For EF || QR, the figure should fulfill following criterion;
`(PE)/(EQ)=(PF)/(FR)`
In this case;
`(PE)/(EQ)=(3.9)/(3)=1.3`
`(PF)/(FR)=(3.6)/(2.4)=3/2`
It is clear that;
`(PE)/(EQ)≠(PF)/(FR)`
Hence; EF and QR are not parallel.
(b) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Solution: In this case;
`(PE)/(EQ)=(4)/(4.5)=8/9`
`(PF)/(FR)=8/9`
It is clear that;
`(PE)/(EQ)=(PF)/(FR)`
Hence; EF || QR
(c) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution: In this case;
`(PE)/(EQ)=(0.18)/(1.28-0.18)=(0.18)/(1.10)=(9)/(55)`
`(PF)/(FR)=(0.36)/(2.56-0.36)=(0.36)/(2.20)=(9)/(55)`
It is clear that;
`(PE)/(EQ)=(PF)/(FR)`
Hence; EF || QR
Question 3: In the given figure, if LM || CB and LN || CD, prove that `(AM)/(AB)=(AN)/(AD)`
Solution: In Δ ABC and Δ AML;
Δ ABC ∼ Δ AML (because ML || BC)
Hence;
`(AM)/(AB)=(AL)/(AC)`Hence;
`(AN)/(AD)=(AL)/(AC)`From above two equations;
`(AM)/(AB)=(AN)/(AD)`Question 4: In the given figure, DE || AC and DF || AE. Prove that `(BF)/(FE)=(BE)/(EC)`
Solution: In Δ ABC and ΔDBE;
`(BE)/(EC)=(BD)/(BA)`
Because Δ ABC ∼ Δ DBE
Similarly, in Δ ABE and Δ DBF;
`(BF)/(FE)=(BD)/(BA)`
From above two equations, it is clear;
`(BF)/(FE)=(BE)/(EC)`