D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC
Solution:
We know that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it (mid-point theorem).
According to the given question in ΔABC,
D, E, and F are respectively the mid-points of sides AB, BC, and CA.
So, by the mid-point theorem, we have,
FE || AB and FE = 1/2 AB ------------ (1)
⇒ FE || AB
Thus, FE || BD
Also, BD = 1/2 AB ---------- (2)
From equations (1) and (2),
FE = BD and FE || BD
As we know that if one pair of opposite sides of a quadrilateral are equal and parallel then it is a parallelogram.
Therefore, BDEF is a parallelogram.
Similarly, in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF are equal)
FD = FD (Common)
BD = FE (Opposite sides of parallelogram BDEF are equal)
Therefore, ΔFBD ≅ ΔDEF
Similarly, we can prove that ΔAFE ≅ ΔDEF and ΔEDC ≅ ΔDEF
As we know, if two triangles are congruent, then their areas must be equal.
Hence,
Area(ΔFBD) = Area(ΔDEF) …(1)
Area(ΔAFE) = Area(ΔDEF) …….(2)
and,
Area(ΔEDC) = Area(ΔDEF) ……(3)
Now,
Area(ΔABC) = Area(ΔFBD) + Area(ΔDEF) + Area(ΔAFE) + Area(ΔEDC) ……(4)
Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) [ Area(ΔFBD) = Area(ΔDEF) and Area(ΔEDC) = Area(ΔDEF) ]
From equation (1), (2) and (3),
⇒ Area(ΔDEF) = (1/4)Area(ΔABC)
⇒ Area(ΔDEF)/Area(ΔABC) = 1/4
Therefore,
Area(ΔDEF): Area(ΔABC) = 1 / 4 = 1:4
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC
NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.4 Question 5
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Then the ratio of the areas of ∆DEF and ∆ABC is 1:4
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