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


Transcript

Question 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 βˆ†ABC with sides 5 cm, 5 cm and 1 cm Finding area using Heron’s Formula Area of triangle = √(𝐬(π¬βˆ’πš)(π¬βˆ’π›)(𝐬 βˆ’πœ)) Here, a = 5 cm, b = 5 cm , c = 1 cm s = (𝒂 + 𝒃 + 𝒄)/𝟐 Area of Ξ” ABK = √(πŸ“.πŸ“(πŸ“.πŸ“ βˆ’πŸ“)(πŸ“.πŸ“ βˆ’πŸ“)(πŸ“.πŸ“ βˆ’πŸ))cm2 = √((5.5)Γ—(0.5)Γ—(0.5)Γ—(4.5)) cm2 = √((55)Γ—(5)Γ—(5)Γ—(45)Γ—πŸ/πŸπŸŽπŸ’) cm2 = √((πŸ“Γ—πŸπŸ)Γ—(5)Γ—(5)Γ—(πŸ—Γ—πŸ“)Γ—1/104) cm2 = √((5Γ—5)Γ—(5Γ—5)Γ—(9)Γ—11Γ—1/104) cm2 = √((25)Γ—(25)Γ—(9)Γ—11Γ—10^(βˆ’4) ) cm2 = √((25)2Γ—(3)2Γ—11Γ—10^(βˆ’4) ) cm2 = √((25)2)Γ—βˆš((3)2) Γ—βˆš11Γ—βˆš(10^(βˆ’4) ) cm2 = 25 Γ— 3 Γ— √𝟏𝟏 Γ— γ€–πŸπŸŽγ€—^(βˆ’πŸ)cm2 = 75/100 √11 cm2 = 0.75√11 cm2 = (0.75 Γ— 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 Γ— Breadth = (6.5 Γ— 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 Γ— Sum of parallel sides Γ— 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 βŠ₯ FG and EN βŠ₯ 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 Ξ”DFM is a right angled triangle, By pythagoras theorem DF2 = DM2 + FM2 DM2 = DF2 βˆ’ FM2 DM = √("DF2 βˆ’ FM2" ) Putting DF = 1 cm and FM = 0.5 cm DM = (√(12 βˆ’(0.5)2)) cm = √(1 βˆ’0.25) cm = √0.75 cm = √(75/100) cm = √(75Γ—10^(βˆ’2) ) cm = √(3Γ—25Γ—10^(βˆ’2) ) cm = √(3Γ—5^2Γ—10^(βˆ’2) ) cm = √3Γ— √52 Γ—βˆš(10^(βˆ’2) )cm = √3Γ—5 Γ— γ€–(10)γ€—^(βˆ’1) = 1.732 Γ— 5 Γ—1/10 = 8.66 Γ—1/10 = 0.866 cm Thus, height of trapezium is 0.866 cm Finding Area of Trapezium Area of Trapezium = 𝟏/𝟐 Γ— Sum of parallel sides Γ— Height = 1/2 Γ— (1 + 2) Γ— 0.866 = 1/2 Γ— (3) Γ— 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 Γ— Base Γ— Height = 1/2Γ—1.5Γ—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.)

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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.