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 Δ 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/𝐴𝐵=𝐴𝐸/𝐴𝐶=1/4. Calculate the area of the Δ ADE and compare it with the area of Δ ABC. (Recall Theorem 6.2 and Theorem 6.6). Given 𝐴𝐷/𝐴𝐵=𝐴𝐸/𝐴𝐶=1/4 𝑨𝑫/𝑨𝑩=𝟏/𝟒 𝑨𝑬/𝑨𝑪=𝟏/𝟒 4 AD = AB 4 AD = AD + BD 4 AD − AD = BD 3 AD = BD 𝐴𝐷/𝐵𝐷=1/3 Thus point D divides AB in the ratio 1 : 3. Using section formula, coordinates of D are ((𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ),(𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 )) 4 AE = AC 4 AE = AE + CE 4 AE − AE = CE 3 AE = CE 𝐴𝐸/𝐶𝐸=1/3 Thus point E divides AC in the ratio 1 : 3. Using section formula, coordinates of E are ((𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ),(𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 )) Put 𝑚_1=1, 𝑚_2=3 𝑥_1=4, 𝑥_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 𝑚_1=1, 𝑚_2=3 𝑥_1=4, 𝑥_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 Δ ADE and Δ ABC Area Δ ADE Area of ∆ADE = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+𝑥_(2 ) (𝑦_3 −𝑦_1 )+𝑥_(3 ) (𝑦_1 −𝑦_2 )] Put 𝑥_1=4, 𝑥_2=13/4 𝑥_3=19/4, 𝑦_1=6 𝑦_2=23/4, 𝑦_3=20/4 𝑦_2=23/4, 𝑦_3=20/4 Ar (∆ADE) = 1/2 [4 (23/4−20/4)+13/4 (20/4−6)+19/4 (6 −23/4)] = 1/2 [4 (3/4)+13/4 ((20 − 24)/4)+19/4 ((24 − 23)/4)] = 1/2 [4 (3/4)+13/4 ((−4)/4)+19/4 (1/4)] = 1/2 [12/4−52/16+19/16] = 1/2 [(48 − 52 + 19)/16] = 𝟏𝟓/𝟑𝟐 sq. units. Area Δ ABC Area of ∆ABC = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+𝑥_(2 ) (𝑦_3 −𝑦_1 )+𝑥_(3 ) (𝑦_1 −𝑦_2 )] Put 𝑥_1=4, 𝑥_2=1 𝑥_3=7, 𝑦_1=6 𝑦_2=5, 𝑦_3=2 Ar (∆ABC) = 1/2 [4 (5−2)+1 (2 − 6)+7 (6 −5)] = 1/2 [4 (3)+1 (−4)+7 (1)] = 1/2 [12 − 4+7] = 15/2 sq. units. Thus, (𝑎𝑟 (∆𝐴𝐷𝐸))/(𝑎𝑟 (∆𝐴𝐵𝐶))=(15/32)/(15/2) = 15/32×2/15 = 1/16 Thus, area of ∆ADE is 𝟏𝟓/𝟑𝟐 sq. units and (𝑎𝑟 (∆𝐴𝐷𝐸))/(𝑎𝑟 (∆𝐴𝐵𝐶))=𝟏/𝟏𝟔 Question 6 (Method 2) The vertices of a Δ 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/𝐴𝐵=𝐴𝐸/𝐴𝐶=1/4. Calculate the area of the Δ ADE and compare it with the area of Δ ABC. (Recall Theorem 6.2 and Theorem 6.6). Given 𝐴𝐷/𝐴𝐵=𝐴𝐸/𝐴𝐶=1/4 𝑨𝑫/𝑨𝑩=𝟏/𝟒 4 AD = AB 4 AD = AD + BD 4 AD − AD = BD 3 AD = BD 𝐴𝐷/𝐵𝐷=1/3 𝑨𝑬/𝑨𝑪=𝟏/𝟒 4 AE = AC 4 AE = AE + CE 4 AE − AE = CE 3 AE = CE 𝐴𝐸/𝐶𝐸=1/3 Thus, 𝐴𝐷/𝐴𝐵=𝐴𝐸/𝐴𝐶=1/3 ∴ DE divides AB and AC in the ratio 1 : 3 Applying theorem 6.2, DE ∥ 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 ∆ADE and ∆ABC ∠A = ∠A ∠ADE = ∠ABC ∴ ∆ADE ~ ∆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, (𝑎𝑟 (∆𝐴𝐷𝐸))/(𝑎𝑟 (∆𝐴𝐵𝐶))=(𝐴𝐷/𝐴𝐵)^2 (Common angle) (Corresponding angles) (By AA similarity criterion) (𝑎𝑟 (∆𝐴𝐷𝐸))/(𝑎𝑟 (∆𝐴𝐵𝐶))=(1/4)^2=1/16 𝑎𝑟 (∆𝐴𝐷𝐸)=1/16 𝑎𝑟 (∆𝐴𝐵𝐶) Now, Finding Area Δ ABC 𝑎𝑟 (∆𝐴𝐵𝐶)=1/2 [𝑥_1 (𝑦_2−𝑦_3 ) + 𝑥_2 (𝑦_3−𝑦_1 ) + 𝑥_3 (𝑦_1−𝑦_2 )] Put 𝑥_1=4, 𝑦_1=6 𝑥_2=1, 𝑦_2=5 𝑥_3=7, 𝑦_3=2 Ar (∆ABC) = 1/2 [4 (5−2)+1 (2 − 6)+7 (6 −5)] = 1/2 [4 (3)+1 (−4)+7 (1)] = 1/2 [12 − 4+7] = 15/2 sq. units. Thus, From (1) ar (∆ADE) = 1/16 ar (∆ABC) = 1/16×15/2=15/32 sq. units. Thus, area of ∆ADE is 𝟏𝟓/𝟑𝟐 sq. units and (𝑎𝑟 (∆𝐴𝐷𝐸))/(𝑎𝑟 (∆𝐴𝐵𝐶))=𝟏/𝟏𝟔

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.