Check sibling questions

Slide43.JPG

Slide44.JPG
Slide45.JPG
Slide46.JPG
Slide47.JPG

This video is only available for Teachoo black users

Slide48.JPG

This video is only available for Teachoo black users

Slide49.JPG

Slide50.JPG
Slide51.JPG
Slide52.JPG
Slide53.JPG

This video is only available for Teachoo black users

Slide54.JPG

This video is only available for Teachoo black users

Slide55.JPG

This video is only available for Teachoo black users


Transcript

Ex 12.2, 3 Radha made a picture of an aeroplane with coloured paper as shown in the given figure. Find the total area of the paper used. First labelling the figure Total Area = Area of I + Area of II + Area of III + Area of IV + Area of V Area of I This is ∆ABC with sides 5 cm, 5 cm and 1 cm Finding area using Heron’s Formula Area of triangle = √(𝐬(𝐬−𝐚)(𝐬−𝐛)(𝐬 −𝐜)) Here, a = 5 cm, b = 5 cm , c = 1 cm s = (𝒂 + 𝒃 + 𝒄)/𝟐 Area of Δ ABK = √(𝟓.𝟓(𝟓.𝟓 −𝟓)(𝟓.𝟓 −𝟓)(𝟓.𝟓 −𝟏))cm2 = √((5.5)×(0.5)×(0.5)×(4.5)) cm2 = √((55)×(5)×(5)×(45)×𝟏/𝟏𝟎𝟒) cm2 = √((𝟓×𝟏𝟏)×(5)×(5)×(𝟗×𝟓)×1/104) cm2 = √((5×5)×(5×5)×(9)×11×1/104) cm2 = √((25)×(25)×(9)×11×10^(−4) ) cm2 = √((25)2×(3)2×11×10^(−4) ) cm2 = √((25)2)×√((3)2) ×√11×√(10^(−4) ) cm2 = 25 × 3 × √𝟏𝟏 × 〖𝟏𝟎〗^(−𝟐)cm2 = 75/100 √11 cm2 = 0.75√11 cm2 = (0.75 × 3.31) cm2 = 2.4825 cm2 (approximately) Area of II This is a rectangle BCDE with Length = 6.5 cm Breadth = 1 cm Finding area of rectangle Area = Length × Breadth = (6.5 × 1) cm2 = 6.5 cm2 Area of III This is a trapezium with parallel sides 1 cm and 2 cm Now, Area of Trapezium = 𝟏/𝟐 × Sum of parallel sides × Height Since, we need height for finding Area, so, first we will find height of trapezium Finding Height of Trapezium Since height is the perpendicular distance from one parallel side to another Let DM ⊥ FG and EN ⊥ FG Here, MN = DE = 1 cm FM = NG Finding FM FM + MN + NG = FG FM + MN + FM = FG 2FM + MN = FG 2FM + 1 cm = 2 cm 2FM = 2 cm – 1 cm 2FM = 1 cm FM = 1/2 cm FM = 0.5 cm Since ΔDFM is a right angled triangle, By pythagoras theorem DF2 = DM2 + FM2 DM2 = DF2 − FM2 DM = √("DF2 − FM2" ) Putting DF = 1 cm and FM = 0.5 cm DM = (√(12 −(0.5)2)) cm = √(1 −0.25) cm = √0.75 cm = √(75/100) cm = √(75×10^(−2) ) cm = √(3×25×10^(−2) ) cm = √(3×5^2×10^(−2) ) cm = √3× √52 ×√(10^(−2) )cm = √3×5 × 〖(10)〗^(−1) = 1.732 × 5 ×1/10 = 8.66 ×1/10 = 0.866 cm Thus, height of trapezium is 0.866 cm Finding Area of Trapezium Area of Trapezium = 𝟏/𝟐 × Sum of parallel sides × Height = 1/2 × (1 + 2) × 0.866 = 1/2 × (3) × 0.866 = 1.299 cm2 Area of IV and Area of V By Symmetry Area of triangle (IV) = Area of triangle in (V) So, we will find Area of triangle (IV) Triangle IV is a right angled triangle with height 6 cm and base 1.5 cm Now, Area of Triangle = 1/2 × Base × Height = 1/2×1.5×6 = 4.5 cm2 Thus, Area of triangle (IV) = Area of triangle in (V) = 4.5 cm2 Now, Total area of the paper used = Area of I + Area of II + Area of III + Area of IV + Area of V = 2.4825 + 6.5 + 1.299 + 4.5 + 4.5 = 19.287 cm2 = 19.3 cm2 (Approx.)

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.