Question

Question 5 D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

Class 10 - Math - Triangles Page 144

Answer 1

Given: D, E and F are the mid-points of the sides AB, BC and CA respectively of the ΔABC.

To Find: area(ΔDEF) and area(ΔABC)

Solution: In ΔABC, we have
F is the mid point of AB (Given)
E is the mid-point of AC (Given)
So, by the mid-point theorem, we have
FE || BC and FE = 1/2BC
⇒ FE || BC and FE || BD [BD = 1/2BC]
∴ BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common)
BD = FE (Opposite sides of parallelogram BDEF)
∴ ΔFBD ≅ ΔDEF
Similarly, we can prove that
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
If triangles are congruent,then they are equal in area.
So, area(ΔFBD) = area(ΔDEF) ...(i)
area(ΔAFE) = area(ΔDEF) ...(ii)
and, area(ΔEDC) = area(ΔDEF) ...(iii)
Now, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔAFE) + area(ΔEDC) ...(iv)
area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)
⇒ area(ΔDEF) = 1/4area(ΔABC) [From (i), (ii)and (iii)]
⇒ area(ΔDEF)/area(ΔABC) = 1/4
Hence, area(ΔDEF):area(ΔABC) = 1:4