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In Fig. Below, Abc and Bde Are Two Equilateral Triangles Such that D is the Mid-point of Bc. Ae Intersects Bc in F. Prove that (1) Ar (δ Bde) = `1/2` Ar (δAbc) (2) Area ( δBde) `= 1/2 ` Ar (δBae) - Mathematics

In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. AE intersects BC in F. Prove that

(1)  ar (Δ BDE) = `1/2` ar (ΔABC) 

(2) Area ( ΔBDE) `= 1/2 ` ar (ΔBAE)

(3)  ar (BEF) = ar (ΔAFD)

(4) area (Δ ABC) = 2 area (ΔBEC)

(5) ar (ΔFED) `= 1/8` ar (ΔAFC) 

(6) ar (Δ BFE) = 2 ar (ΔEFD)

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Solution

Given that ,

ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x . Then `BD = x/2 = DE = BE`

(1)  We have 

  `ar  (ΔABC) =sqrt3/4 x^2`

  `ar (ΔABC) =sqrt3/4 (x/2)^2  = 1/4 xx sqrt3/4 x^2 `

⇒  `ar (ΔBDE) =sqrt3/4 (x/2)^2`

 (2) It is given that triangles ABC and BED are equilateral triangles 

∠ACB =∠DBE = 60°

 ⇒ BF ll AC  (Since alternative angles are equal)
Triangles BAF and BEC are on the same base
BE and between the same parallel BE and AC

  ∴  ar (ΔBAE) = area (ΔBEC)
 ⇒ ar (ΔBAE) = area (ΔBDE)

[ ∴ ED is a median of ΔEBC ; ar (ΔBEC) = 2ar (ΔBDE) ]

 ⇒ area (Δ BDE) =`1/2` ar (Δ BDE)

(3)  Since  ΔABC and  ΔBDE are equilateral triangles

    ∴ ∠ABC = 60°  and  ∠BDE = 60°

        ∠ ABC = ∠BDE 

       ⇒ AB ll DE      (Since alternative angles are equal)

Triangles BED and AED are on the same base ED and between the same parallels
AB and DE.

  ∴ ar  (ΔBED) =  area (ΔED)

⇒ ar (ΔBED) - area  (ΔEFD) = area (AED) -  area (ΔEFD)

 ⇒ ar (BEF) =  ar  (ΔAFD)

(4) Since ED is the median of  ΔBEC
      ∴ area (ΔBEC) = 2ar  (ΔBDE) 

⇒ `ar (ΔBEC) = 2 xx 1/4 ar (ΔABC)`           [form (1)]

⇒ ar(ΔBEC = area  (ΔABC)

⇒  area  (ΔABC)  = 2area (ΔBEC)

(5)  Let h be the height of vertex E, corresponding to the side BD on triangle BDE
Let H be the height of the vertex A corresponding to the side BC in triangle ABC
From part (i)

  ar (Δ BDE) = `1/4` ar (ΔABC)

⇒ `1/2 xx BD xx h = 1/4 ar ( Δ ABC)`

⇒ `BD xx h = 1/4 (1/2 xx BC xx H)`

⇒ `h = 1/2 H`     ............... (1)

From part …..(3)
Area (ΔBFE) = ar (ΔAFD)

 = `1/2 xx FD xx H`

  = `1/2 xx FD xxH`

  = `2(1/2 xx FD xx 2h)`

  = 2ar (Δ EFD)

(6)  area (ΔAFC) area ( AFD) + area ( ADC)

    ⇒ ar  (ΔBFE) ar + `1/2` ar (ΔABC)

[ using part (3); and AD is the median  ΔABC ]

   = ar (ΔBFE +`1/2 xx` 4ar (ΔBDE) using part (1)]

  =  ar (ΔBFE) =2ar (ΔFED)   ....... (3)

 Area ( ΔBDE) = ar (ΔBFE) + ar (ΔFED)

  ⇒ R ar (ΔFED) + ar (Δ  FED)

   ⇒  3 ar (ΔFED)

From (2), (3) and (4) we get
Area   (ΔAFC)  = 2area (ΔFED)  + 2 × 3ar (ΔFED)

= 8 ar (ΔFED)

Hence, area `(ΔFED) = 1/8`area (AFC)

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 29 | Page 48
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