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Important questions on Parallelograms and Triangles
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Important questions on Parallelograms and Triangles
Last updated at May 29, 2023 by Teachoo
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 β₯ BD. Given βABC is an equilateral triangle, Let AB = BC = AC = 2a & β BAC = β ABC = β ACB = 60Β° Now, Since D is the mid point of BC β΄ BD = CD = 2π/2 = a Also, since Ξ BDE is equilateral Thus, BD = DE = BE = a β BDE = β BED = β DBE = 60Β° Now, In equilateral triangle Ξ ABC D is mid-point, So, AD is median β΄ AD will also be the altitude In equilateral triangle, altitude and median are the same β΄ ADβ₯ BC So, βADC will be a right angle triangle By Pythagoras theorem AC2 = AD2 + DC2 (2a)2 = AD2 + (a)2 4a2 = AD2 + a2 4a2 β a2 = AD2 3a2 = AD2 β3π2 = AD β3 a = AD AD = β3 a Similarly, For equilateral Ξ BDE, We can find EL = β3/2 a Now, Finding area of Ξ ABC and Ξ BDE For β ABC Base = BC = 2a Height = AD = β3 π ar (βABC) = 1/2 Γ Base Γ Height = 1/2 Γ 2a Γ β3 π = β3 π2 For β BDE Base = BD = a Height = EL = β3/2 π ar (βBDE) = 1/2 Γ Base Γ Height = 1/2 Γ a Γ β3/2 π = β3/4 π2 Now, (ππ (π΅π·πΈ))/(ππ (π΄π΅πΆ)) = (β3/4 π"2" )/(β3 π"2" ) (ππ (π΅π·πΈ))/(ππ (π΄π΅πΆ)) = (β3 π"2" )/(β3 π"2" ) Γ 1/4 (ππ (π΅π·πΈ))/(ππ (π΄π΅πΆ)) = 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 β ABD = β BDE = 60Β° Since β ABD & β BDE are alternate angles, and they are equal So, lines must be parallel β΄ AB β₯ DE Similarly, we can prove BE β₯ AC Now, βBAE and βBCE are on the same base BE, and between same Parallels BE and AC. β΄ ar (BAE) = ar (BCE) But, In βBCE, D is the mid Point of BC β΄ ar (BDE) = ar (DCE) Thus, ar (BDE) = 1/2ar (BCE) (Median divides a triangle into two triangles of equal area) β¦(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 Γ 4 Γ ar (BCE) = ar (ABC) 2 ar (BCE) = ar (ABC) ar (ABC) = 2ar (BCE) Hence proved Question 5 Show that (iv) ar (BFE) = ar (AFD) β BED and βAED are on the same base ED and between same parallel lines ED & AB β΄ ar (BED) = ar (AED) Subtracting ar (FED) both sides ar (BED) β ar (FED) = ar (AED) β ar (FED) ar (BFE) = ar (AFD). Hence proved Question 5 Show that (v) ar (BFE) = 2 ar (FED) We note that Ξ BFE and Ξ FED have the same height FE Letβs find their base - BF and FD To find BF, we have proved in part (iv) that ar (AFD) = ar (BFE) Letβs find their areaβs using Base Γ Height formula For β AFD, Base = FD Height = AD = β3 π ar (AFD) = 1/2 Γ Base Γ Height = 1/2 Γ FD Γ β3 π For β BFE, Base = BF Height = EL = β3/2 π ar (BFE) = 1/2 Γ Base Γ Height = 1/2 Γ BF Γ β3/2 π Since ar (AFD) = ar (BFE) 1/2 Γ FD Γ β3 π = = 1/2 Γ BF Γ β3/2 π FD = π΅πΉ/2 2FD = BF BF = 2FD Now, BD = a BF + FD = a 2FD + FD = a 3FD = a FD = π/3 And, BF = 2FD = 2π/3 Now, Finding area of β FED Base = FD = 1/2 BF Height = EL = β3/2 π ar (FED) = 1/2 Γ Base Γ Height = 1/2 Γ 1/2 BF Γ β3/2 π Now, (ππ (π΅πΉπΈ))/(ππ (πΉπΈπ·)) = (1/2 Γ π΅πΉ Γ β3/2 " " π" " )/(1/2 Γ 1/2 π΅πΉ Γ β3/2 " " π) (ππ (π΅πΉπΈ))/(ππ (πΉπΈπ·)) = 1/(1/2) (ππ (π΅πΉπΈ))/(ππ (πΉπΈπ·)) = 2 ar (BFE) = 2 ar (FED) Hence proved Question 5 Show that (vi) ar (FED) = 1/8 ar (AFC) Finding Area of Ξ FED Base = FD = π/3 Height = EL = β3/2 π Area of Ξ FED = 1/2 Γ Base Γ Height = 1/2 Γ π/3 Γ β3/2 π = β3/12 π^2 Finding Area of Ξ AFC Base = FC = FD + DC = π/3 + π = 4/3 π Height = AD = β3 π Area of Ξ AFC = 1/2 Γ Base Γ Height = 1/2 Γ 4/3 π Γ β3 π = (2β3)/3 π^2 Now (ππ (πΉπΈπ·))/(ππ (π΄πΉπΆ)) = (β3/12 π^2)/((2β3)/3 π^2 ) (ππ (πΉπΈπ·))/(ππ (π΄πΉπΆ)) =β3/12 π^2Γ3/(2β3 π^2 ) (ππ (πΉπΈπ·))/(ππ (π΄πΉπΆ)) = 1/12Γ3/2 (ππ (πΉπΈπ·))/(ππ (π΄πΉπΆ)) = 1/8 ar (FED) = 1/8 ar (AFC) Hence Proved