Check sibling questions

Slide22.JPG

Slide23.JPG
Slide24.JPG Slide25.JPG Slide26.JPG Slide27.JPG Slide28.JPG Slide29.JPG Slide30.JPG Slide31.JPG Slide32.JPG Slide33.JPG

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


Transcript

Question 6 (Method 1) The vertices of a ฮ” 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 ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4. Calculate the area of the ฮ” ADE and compare it with the area of ฮ” ABC. (Recall Theorem 6.2 and Theorem 6.6). Given ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4 ๐‘จ๐‘ซ/๐‘จ๐‘ฉ=๐Ÿ/๐Ÿ’ ๐‘จ๐‘ฌ/๐‘จ๐‘ช=๐Ÿ/๐Ÿ’ 4 AD = AB 4 AD = AD + BD 4 AD โˆ’ AD = BD 3 AD = BD ๐ด๐ท/๐ต๐ท=1/3 Thus point D divides AB in the ratio 1 : 3. Using section formula, coordinates of D are ((๐‘š_1 ๐‘ฅ_2 + ๐‘š_2 ๐‘ฅ_1)/(๐‘š_1 + ๐‘š_2 ),(๐‘š_1 ๐‘ฆ_2 + ๐‘š_2 ๐‘ฆ_1)/(๐‘š_1 + ๐‘š_2 )) 4 AE = AC 4 AE = AE + CE 4 AE โˆ’ AE = CE 3 AE = CE ๐ด๐ธ/๐ถ๐ธ=1/3 Thus point E divides AC in the ratio 1 : 3. Using section formula, coordinates of E are ((๐‘š_1 ๐‘ฅ_2 + ๐‘š_2 ๐‘ฅ_1)/(๐‘š_1 + ๐‘š_2 ),(๐‘š_1 ๐‘ฆ_2 + ๐‘š_2 ๐‘ฆ_1)/(๐‘š_1 + ๐‘š_2 )) Put ๐‘š_1=1, ๐‘š_2=3 ๐‘ฅ_1=4, ๐‘ฅ_2=1 ๐‘ฆ_1=6, ๐‘ฆ_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 ๐‘š_1=1, ๐‘š_2=3 ๐‘ฅ_1=4, ๐‘ฅ_2=7 ๐‘ฆ_1=6, ๐‘ฆ_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 ฮ” ADE and ฮ” ABC Area ฮ” ADE Area of โˆ†ADE = 1/2 [๐‘ฅ_1 (๐‘ฆ_2โˆ’๐‘ฆ_3 )+๐‘ฅ_(2 ) (๐‘ฆ_3 โˆ’๐‘ฆ_1 )+๐‘ฅ_(3 ) (๐‘ฆ_1 โˆ’๐‘ฆ_2 )] Put ๐‘ฅ_1=4, ๐‘ฅ_2=13/4 ๐‘ฅ_3=19/4, ๐‘ฆ_1=6 ๐‘ฆ_2=23/4, ๐‘ฆ_3=20/4 ๐‘ฆ_2=23/4, ๐‘ฆ_3=20/4 Ar (โˆ†ADE) = 1/2 [4 (23/4โˆ’20/4)+13/4 (20/4โˆ’6)+19/4 (6 โˆ’23/4)] = 1/2 [4 (3/4)+13/4 ((20 โˆ’ 24)/4)+19/4 ((24 โˆ’ 23)/4)] = 1/2 [4 (3/4)+13/4 ((โˆ’4)/4)+19/4 (1/4)] = 1/2 [12/4โˆ’52/16+19/16] = 1/2 [(48 โˆ’ 52 + 19)/16] = ๐Ÿ๐Ÿ“/๐Ÿ‘๐Ÿ sq. units. Area ฮ” ABC Area of โˆ†ABC = 1/2 [๐‘ฅ_1 (๐‘ฆ_2โˆ’๐‘ฆ_3 )+๐‘ฅ_(2 ) (๐‘ฆ_3 โˆ’๐‘ฆ_1 )+๐‘ฅ_(3 ) (๐‘ฆ_1 โˆ’๐‘ฆ_2 )] Put ๐‘ฅ_1=4, ๐‘ฅ_2=1 ๐‘ฅ_3=7, ๐‘ฆ_1=6 ๐‘ฆ_2=5, ๐‘ฆ_3=2 Ar (โˆ†ABC) = 1/2 [4 (5โˆ’2)+1 (2 โˆ’ 6)+7 (6 โˆ’5)] = 1/2 [4 (3)+1 (โˆ’4)+7 (1)] = 1/2 [12 โˆ’ 4+7] = 15/2 sq. units. Thus, (๐‘Ž๐‘Ÿ (โˆ†๐ด๐ท๐ธ))/(๐‘Ž๐‘Ÿ (โˆ†๐ด๐ต๐ถ))=(15/32)/(15/2) = 15/32ร—2/15 = 1/16 Thus, area of โˆ†ADE is ๐Ÿ๐Ÿ“/๐Ÿ‘๐Ÿ sq. units and (๐‘Ž๐‘Ÿ (โˆ†๐ด๐ท๐ธ))/(๐‘Ž๐‘Ÿ (โˆ†๐ด๐ต๐ถ))=๐Ÿ/๐Ÿ๐Ÿ” Question 6 (Method 2) The vertices of a ฮ” 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 ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4. Calculate the area of the ฮ” ADE and compare it with the area of ฮ” ABC. (Recall Theorem 6.2 and Theorem 6.6). Given ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4 ๐‘จ๐‘ซ/๐‘จ๐‘ฉ=๐Ÿ/๐Ÿ’ 4 AD = AB 4 AD = AD + BD 4 AD โˆ’ AD = BD 3 AD = BD ๐ด๐ท/๐ต๐ท=1/3 ๐‘จ๐‘ฌ/๐‘จ๐‘ช=๐Ÿ/๐Ÿ’ 4 AE = AC 4 AE = AE + CE 4 AE โˆ’ AE = CE 3 AE = CE ๐ด๐ธ/๐ถ๐ธ=1/3 Thus, ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/3 โˆด DE divides AB and AC in the ratio 1 : 3 Applying theorem 6.2, DE โˆฅ 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 โˆ†ADE and โˆ†ABC โˆ A = โˆ A โˆ ADE = โˆ ABC โˆด โˆ†ADE ~ โˆ†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, (๐‘Ž๐‘Ÿ (โˆ†๐ด๐ท๐ธ))/(๐‘Ž๐‘Ÿ (โˆ†๐ด๐ต๐ถ))=(๐ด๐ท/๐ด๐ต)^2 (Common angle) (Corresponding angles) (By AA similarity criterion) (๐‘Ž๐‘Ÿ (โˆ†๐ด๐ท๐ธ))/(๐‘Ž๐‘Ÿ (โˆ†๐ด๐ต๐ถ))=(1/4)^2=1/16 ๐‘Ž๐‘Ÿ (โˆ†๐ด๐ท๐ธ)=1/16 ๐‘Ž๐‘Ÿ (โˆ†๐ด๐ต๐ถ) Now, Finding Area ฮ” ABC ๐‘Ž๐‘Ÿ (โˆ†๐ด๐ต๐ถ)=1/2 [๐‘ฅ_1 (๐‘ฆ_2โˆ’๐‘ฆ_3 ) + ๐‘ฅ_2 (๐‘ฆ_3โˆ’๐‘ฆ_1 ) + ๐‘ฅ_3 (๐‘ฆ_1โˆ’๐‘ฆ_2 )] Put ๐‘ฅ_1=4, ๐‘ฆ_1=6 ๐‘ฅ_2=1, ๐‘ฆ_2=5 ๐‘ฅ_3=7, ๐‘ฆ_3=2 Ar (โˆ†ABC) = 1/2 [4 (5โˆ’2)+1 (2 โˆ’ 6)+7 (6 โˆ’5)] = 1/2 [4 (3)+1 (โˆ’4)+7 (1)] = 1/2 [12 โˆ’ 4+7] = 15/2 sq. units. Thus, From (1) ar (โˆ†ADE) = 1/16 ar (โˆ†ABC) = 1/16ร—15/2=15/32 sq. units. Thus, area of โˆ†ADE is ๐Ÿ๐Ÿ“/๐Ÿ‘๐Ÿ sq. units and (๐‘Ž๐‘Ÿ (โˆ†๐ด๐ท๐ธ))/(๐‘Ž๐‘Ÿ (โˆ†๐ด๐ต๐ถ))=๐Ÿ/๐Ÿ๐Ÿ”

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.