How i solve the question of class 9^{th} ncert math of Areas of Parallelograms and Triangles chapter of exercise 9.4 of question no 5(1). I think it is very important question of class 9^{th} give me the tricky way for solving this question In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(i) ar (BDE) =1/4 ar (ABC)

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# In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(i) ar (BDE) =1/4 ar (ABC) Q.5 (1)

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Assume that G and H are the mid-points of the sides AB and AC respectively.

Join the mid-points with line-segment GH. Here, GH is parallel to third side.

, BC will be half of the length of BC by mid-point theorem.

âˆ´ GH =1/2 BC and GH || BD

âˆ´ GH = BD = DC and GH || BD (Since, D is the mid-point of BC)

Similarly,

GD = HC = HA

HD = AG = BG

, Î”ABC is divided into 4 equal equilateral triangles Î”BGD, Î”AGH, Î”DHC and Î”GHD

We can say that,

Î”BGD = Â¼ Î”ABC

Considering, Î”BDG and Î”BDE

BD = BD (Common base)

Since both triangles are equilateral triangle, we can say that,

BG = BE

DG = DE

, Î”BDG Î”BDE [By SSS congruency]

, area (Î”BDG) = area (Î”BDE)

ar (Î”BDE) = Â¼ ar (Î”ABC)

Hence proved