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# 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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