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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Question 6 (Method 1) The vertices of a Ξ” ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that 𝐴𝐷/𝐴𝐡=𝐴𝐸/𝐴𝐢=1/4. Calculate the area of the Ξ” ADE and compare it with the area of Ξ” ABC. (Recall Theorem 6.2 and Theorem 6.6). Given 𝐴𝐷/𝐴𝐡=𝐴𝐸/𝐴𝐢=1/4 𝑨𝑫/𝑨𝑩=𝟏/πŸ’ 𝑨𝑬/𝑨π‘ͺ=𝟏/πŸ’ 4 AD = AB 4 AD = AD + BD 4 AD βˆ’ AD = BD 3 AD = BD 𝐴𝐷/𝐡𝐷=1/3 Thus point D divides AB in the ratio 1 : 3. Using section formula, coordinates of D are ((π‘š_1 π‘₯_2 + π‘š_2 π‘₯_1)/(π‘š_1 + π‘š_2 ),(π‘š_1 𝑦_2 + π‘š_2 𝑦_1)/(π‘š_1 + π‘š_2 )) 4 AE = AC 4 AE = AE + CE 4 AE βˆ’ AE = CE 3 AE = CE 𝐴𝐸/𝐢𝐸=1/3 Thus point E divides AC in the ratio 1 : 3. Using section formula, coordinates of E are ((π‘š_1 π‘₯_2 + π‘š_2 π‘₯_1)/(π‘š_1 + π‘š_2 ),(π‘š_1 𝑦_2 + π‘š_2 𝑦_1)/(π‘š_1 + π‘š_2 )) Put π‘š_1=1, π‘š_2=3 π‘₯_1=4, π‘₯_2=1 𝑦_1=6, 𝑦_2=5 Point D = ((1(1) + (3)(4))/(1 + 3), (1(5) + 3(6))/(1 + 3)) = ((1 + 12)/4,(5 + 18)/4) = (13/4,23/4) Put π‘š_1=1, π‘š_2=3 π‘₯_1=4, π‘₯_2=7 𝑦_1=6, 𝑦_2=2 Point E = ((1(7) + (3)(4))/(1 + 3), (1(2) + 3(6))/(1 + 3)) = ((7 + 12)/4,(2 + 18)/4) = (19/4,20/4) Now, finding Area of Ξ” ADE and Ξ” ABC Area Ξ” ADE Area of βˆ†ADE = 1/2 [π‘₯_1 (𝑦_2βˆ’π‘¦_3 )+π‘₯_(2 ) (𝑦_3 βˆ’π‘¦_1 )+π‘₯_(3 ) (𝑦_1 βˆ’π‘¦_2 )] Put π‘₯_1=4, π‘₯_2=13/4 π‘₯_3=19/4, 𝑦_1=6 𝑦_2=23/4, 𝑦_3=20/4 𝑦_2=23/4, 𝑦_3=20/4 Ar (βˆ†ADE) = 1/2 [4 (23/4βˆ’20/4)+13/4 (20/4βˆ’6)+19/4 (6 βˆ’23/4)] = 1/2 [4 (3/4)+13/4 ((20 βˆ’ 24)/4)+19/4 ((24 βˆ’ 23)/4)] = 1/2 [4 (3/4)+13/4 ((βˆ’4)/4)+19/4 (1/4)] = 1/2 [12/4βˆ’52/16+19/16] = 1/2 [(48 βˆ’ 52 + 19)/16] = πŸπŸ“/πŸ‘πŸ sq. units. Area Ξ” ABC Area of βˆ†ABC = 1/2 [π‘₯_1 (𝑦_2βˆ’π‘¦_3 )+π‘₯_(2 ) (𝑦_3 βˆ’π‘¦_1 )+π‘₯_(3 ) (𝑦_1 βˆ’π‘¦_2 )] Put π‘₯_1=4, π‘₯_2=1 π‘₯_3=7, 𝑦_1=6 𝑦_2=5, 𝑦_3=2 Ar (βˆ†ABC) = 1/2 [4 (5βˆ’2)+1 (2 βˆ’ 6)+7 (6 βˆ’5)] = 1/2 [4 (3)+1 (βˆ’4)+7 (1)] = 1/2 [12 βˆ’ 4+7] = 15/2 sq. units. Thus, (π‘Žπ‘Ÿ (βˆ†π΄π·πΈ))/(π‘Žπ‘Ÿ (βˆ†π΄π΅πΆ))=(15/32)/(15/2) = 15/32Γ—2/15 = 1/16 Thus, area of βˆ†ADE is πŸπŸ“/πŸ‘πŸ sq. units and (π‘Žπ‘Ÿ (βˆ†π΄π·πΈ))/(π‘Žπ‘Ÿ (βˆ†π΄π΅πΆ))=𝟏/πŸπŸ” Question 6 (Method 2) The vertices of a Ξ” ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that 𝐴𝐷/𝐴𝐡=𝐴𝐸/𝐴𝐢=1/4. Calculate the area of the Ξ” ADE and compare it with the area of Ξ” ABC. (Recall Theorem 6.2 and Theorem 6.6). Given 𝐴𝐷/𝐴𝐡=𝐴𝐸/𝐴𝐢=1/4 𝑨𝑫/𝑨𝑩=𝟏/πŸ’ 4 AD = AB 4 AD = AD + BD 4 AD βˆ’ AD = BD 3 AD = BD 𝐴𝐷/𝐡𝐷=1/3 𝑨𝑬/𝑨π‘ͺ=𝟏/πŸ’ 4 AE = AC 4 AE = AE + CE 4 AE βˆ’ AE = CE 3 AE = CE 𝐴𝐸/𝐢𝐸=1/3 Thus, 𝐴𝐷/𝐴𝐡=𝐴𝐸/𝐴𝐢=1/3 ∴ DE divides AB and AC in the ratio 1 : 3 Applying theorem 6.2, DE βˆ₯ BC From theorem 6.2 If a line divides any two sides of a triangle in the same ratio, then it is parallel to the third side. Now, In βˆ†ADE and βˆ†ABC ∠A = ∠A ∠ADE = ∠ABC ∴ βˆ†ADE ~ βˆ†ABC From theorem 6.6 The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Thus, (π‘Žπ‘Ÿ (βˆ†π΄π·πΈ))/(π‘Žπ‘Ÿ (βˆ†π΄π΅πΆ))=(𝐴𝐷/𝐴𝐡)^2 (Common angle) (Corresponding angles) (By AA similarity criterion) (π‘Žπ‘Ÿ (βˆ†π΄π·πΈ))/(π‘Žπ‘Ÿ (βˆ†π΄π΅πΆ))=(1/4)^2=1/16 π‘Žπ‘Ÿ (βˆ†π΄π·πΈ)=1/16 π‘Žπ‘Ÿ (βˆ†π΄π΅πΆ) Now, Finding Area Ξ” ABC π‘Žπ‘Ÿ (βˆ†π΄π΅πΆ)=1/2 [π‘₯_1 (𝑦_2βˆ’π‘¦_3 ) + π‘₯_2 (𝑦_3βˆ’π‘¦_1 ) + π‘₯_3 (𝑦_1βˆ’π‘¦_2 )] Put π‘₯_1=4, 𝑦_1=6 π‘₯_2=1, 𝑦_2=5 π‘₯_3=7, 𝑦_3=2 Ar (βˆ†ABC) = 1/2 [4 (5βˆ’2)+1 (2 βˆ’ 6)+7 (6 βˆ’5)] = 1/2 [4 (3)+1 (βˆ’4)+7 (1)] = 1/2 [12 βˆ’ 4+7] = 15/2 sq. units. Thus, From (1) ar (βˆ†ADE) = 1/16 ar (βˆ†ABC) = 1/16Γ—15/2=15/32 sq. units. Thus, area of βˆ†ADE is πŸπŸ“/πŸ‘πŸ sq. units and (π‘Žπ‘Ÿ (βˆ†π΄π·πΈ))/(π‘Žπ‘Ÿ (βˆ†π΄π΅πΆ))=𝟏/πŸπŸ”

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Davneet Singh

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.