If D, E, F are the mid-points of sides BC, CA and AB respectively of ∆ABC, then the ratio of the areas of triangles DEF and ABC is
Options
1 : 4
1 : 2
2 : 3
4 : 5
Solution
GIVEN: In ΔABC, D, E and F are the midpoints of BC, CA, and AB respectively.
TO FIND: Ratio of the areas of ΔDEF and ΔABC
Since it is given that D and, E are the midpoints of BC, and AC respectively.
Therefore DE || AB, DE || FA……(1)
Again it is given that D and, F are the midpoints of BC, and, AB respectively.
Therefore, DF || CA, DF || AE……(2)
From (1) and (2) we get AFDE is a parallelogram.
Similarly we can prove that BDEF is a parallelogram.
Now, in ΔADE and ΔABC
`∠FDE=∠A\text{(opposite angles of}||^(gm)AFDE)`
`∠DEF = ∠ B=\text{(opposite angles of}||^(gm)BDEF)`
`⇒ Δ ABC ∼ Δ DEF (\text{AA similarity criterion})`
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
`{ar(Δ DEF)}/{ar(Δ ABC)}=((DE)/(AB))^2`
`{ar(Δ DEF)}/{ar(Δ ABC)}=((1/2(AB))/(AB))^2(Since DE =1/2AB)`
`{ar(Δ DEF)}/{ar(Δ ABC)}=(1/4)`
Hence the correct option is `a`
