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Ex 9.4, 5 (Optional) - In Fig, ABC and BDE are 2 equilateral triangles

Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2
Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 4 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 5 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 6 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 7 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 8 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 9 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 10 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 11 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 12 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 13 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 14 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 15 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 16 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 17 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 18 Ex 9.4, 5 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 19

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Question 5 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) [Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.] Join EC and AD. Draw EL ⊥ BD. Given ∆ABC is an equilateral triangle, Let AB = BC = AC = 2a & ∠BAC = ∠ABC = ∠ACB = 60° Now, Since D is the mid point of BC ∴ BD = CD = 2𝑎/2 = a Also, since Δ BDE is equilateral Thus, BD = DE = BE = a ∠BDE = ∠BED = ∠DBE = 60° Now, In equilateral triangle Δ ABC D is mid-point, So, AD is median ∴ AD will also be the altitude In equilateral triangle, altitude and median are the same ∴ AD⊥ BC So, ∆ADC will be a right angle triangle By Pythagoras theorem AC2 = AD2 + DC2 (2a)2 = AD2 + (a)2 4a2 = AD2 + a2 4a2 − a2 = AD2 3a2 = AD2 √3𝑎2 = AD √3 a = AD AD = √3 a Similarly, For equilateral Δ BDE, We can find EL = √3/2 a Now, Finding area of Δ ABC and Δ BDE For ∆ ABC Base = BC = 2a Height = AD = √3 𝑎 ar (∆ABC) = 1/2 × Base × Height = 1/2 × 2a × √3 𝑎 = √3 𝑎2 For ∆ BDE Base = BD = a Height = EL = √3/2 𝑎 ar (∆BDE) = 1/2 × Base × Height = 1/2 × a × √3/2 𝑎 = √3/4 𝑎2 Now, (𝑎𝑟 (𝐵𝐷𝐸))/(𝑎𝑟 (𝐴𝐵𝐶)) = (√3/4 𝑎"2" )/(√3 𝑎"2" ) (𝑎𝑟 (𝐵𝐷𝐸))/(𝑎𝑟 (𝐴𝐵𝐶)) = (√3 𝑎"2" )/(√3 𝑎"2" ) × 1/4 (𝑎𝑟 (𝐵𝐷𝐸))/(𝑎𝑟 (𝐴𝐵𝐶)) = 1/4 ar (BDE) = 1/4 ar (ABC) Hence proved Question 5 Show that (ii) ar (BDE) = 1/2 ar (BAE) For lines AB and DE, with transversal BE ∠ABD = ∠BDE = 60° Since ∠ABD & ∠BDE are alternate angles, and they are equal So, lines must be parallel ∴ AB ∥ DE Similarly, we can prove BE ∥ AC Now, ∆BAE and ∆BCE are on the same base BE, and between same Parallels BE and AC. ∴ ar (BAE) = ar (BCE) But, In ∆BCE, D is the mid Point of BC ∴ ar (BDE) = ar (DCE) Thus, ar (BDE) = 1/2ar (BCE) (Median divides a triangle into two triangles of equal area) …(2) From (1) & 2 ar (BDE) = 1/2 ar (BAE) Hence proved Question 5 Show that (iii) ar (ABC) = 2 ar (BEC) In part (i), we proved that ar (BDE) = 1/4 ar (ABC) And, ar (BCE) = 2 ar (BDE) ar (BDE) = 1/2 ar (BCE) Comparing equations 1/2 ar (BCE) = 1/4 ar (ABC) 1/2 × 4 × ar (BCE) = ar (ABC) 2 ar (BCE) = ar (ABC) ar (ABC) = 2ar (BCE) Hence proved Question 5 Show that (iv) ar (BFE) = ar (AFD) ∆ BED and ∆AED are on the same base ED and between same parallel lines ED & AB ∴ ar (BED) = ar (AED) Subtracting ar (FED) both sides ar (BED) – ar (FED) = ar (AED) – ar (FED) ar (BFE) = ar (AFD). Hence proved Question 5 Show that (v) ar (BFE) = 2 ar (FED) We note that Δ BFE and Δ FED have the same height FE Let’s find their base - BF and FD To find BF, we have proved in part (iv) that ar (AFD) = ar (BFE) Let’s find their area’s using Base × Height formula For ∆ AFD, Base = FD Height = AD = √3 𝑎 ar (AFD) = 1/2 × Base × Height = 1/2 × FD × √3 𝑎 For ∆ BFE, Base = BF Height = EL = √3/2 𝑎 ar (BFE) = 1/2 × Base × Height = 1/2 × BF × √3/2 𝑎 Since ar (AFD) = ar (BFE) 1/2 × FD × √3 𝑎 = = 1/2 × BF × √3/2 𝑎 FD = 𝐵𝐹/2 2FD = BF BF = 2FD Now, BD = a BF + FD = a 2FD + FD = a 3FD = a FD = 𝑎/3 And, BF = 2FD = 2𝑎/3 Now, Finding area of ∆ FED Base = FD = 1/2 BF Height = EL = √3/2 𝑎 ar (FED) = 1/2 × Base × Height = 1/2 × 1/2 BF × √3/2 𝒂 Now, (𝑎𝑟 (𝐵𝐹𝐸))/(𝑎𝑟 (𝐹𝐸𝐷)) = (1/2 × 𝐵𝐹 × √3/2 " " 𝑎" " )/(1/2 × 1/2 𝐵𝐹 × √3/2 " " 𝑎) (𝑎𝑟 (𝐵𝐹𝐸))/(𝑎𝑟 (𝐹𝐸𝐷)) = 1/(1/2) (𝑎𝑟 (𝐵𝐹𝐸))/(𝑎𝑟 (𝐹𝐸𝐷)) = 2 ar (BFE) = 2 ar (FED) Hence proved Question 5 Show that (vi) ar (FED) = 1/8 ar (AFC) Finding Area of Δ FED Base = FD = 𝑎/3 Height = EL = √3/2 𝑎 Area of Δ FED = 1/2 × Base × Height = 1/2 × 𝑎/3 × √3/2 𝑎 = √3/12 𝑎^2 Finding Area of Δ AFC Base = FC = FD + DC = 𝑎/3 + 𝑎 = 4/3 𝑎 Height = AD = √3 𝑎 Area of Δ AFC = 1/2 × Base × Height = 1/2 × 4/3 𝑎 × √3 𝑎 = (2√3)/3 𝑎^2 Now (𝑎𝑟 (𝐹𝐸𝐷))/(𝑎𝑟 (𝐴𝐹𝐶)) = (√3/12 𝑎^2)/((2√3)/3 𝑎^2 ) (𝑎𝑟 (𝐹𝐸𝐷))/(𝑎𝑟 (𝐴𝐹𝐶)) =√3/12 𝑎^2×3/(2√3 𝑎^2 ) (𝑎𝑟 (𝐹𝐸𝐷))/(𝑎𝑟 (𝐴𝐹𝐶)) = 1/12×3/2 (𝑎𝑟 (𝐹𝐸𝐷))/(𝑎𝑟 (𝐴𝐹𝐶)) = 1/8 ar (FED) = 1/8 ar (AFC) Hence Proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.