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  1. Chapter 7 Class 10 Coordinate Geometry (Term 1)
  2. Serial order wise

Transcript

Ex 7.4, 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” Ex 7.4, 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.