Ex 6.4, 7 - Prove that area of an equilateral triangle - Area of similar triangles

Advertisement

Ex 6.4, 7 - Chapter 6 Class 10 Triangles - Part 2

Advertisement

Ex 6.4, 7 - Chapter 6 Class 10 Triangles - Part 3

Ex 6.4, 7 - Chapter 6 Class 10 Triangles - Part 4 Ex 6.4, 7 - Chapter 6 Class 10 Triangles - Part 5 Ex 6.4, 7 - Chapter 6 Class 10 Triangles - Part 6

  1. Chapter 6 Class 10 Triangles (Term 1)
  2. Serial order wise

Transcript

Ex 6.4, 7 (Introduction) Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. Concept 1 Two equilateral triangle are always similar In βˆ† 𝐴𝐡𝐢 π‘Žπ‘›π‘‘ βˆ† 𝐷𝐸𝐹 𝐷𝐸/𝐴𝐡=12/6 𝐸𝐹/𝐡𝐢=12/6 𝐷𝐹/𝐴𝐢=12/6 Hence by SSS similarity βˆ† 𝐴𝐡𝐢 ~ βˆ† 𝐷𝐸𝐹 Concept 2 Diagonal of a square is √2 of its side Since all sides of square are equal, Let AB = BC = CD = AD = x cm All angles of square are 90Β° So, ∠ C = 90Β° According to Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (BD)2 = BC2 + DC2 (BD)2 = BC2 + DC2 BD2 = x2 + x2 BD2 = 2x2 BD = √2 x Hence, π·π‘–π‘Žπ‘”π‘œπ‘›π‘Žπ‘™/𝑠𝑖𝑑𝑒 = 𝐡𝐷/𝐷𝐢 = (√2 π‘₯)/π‘₯ = √2/1 Diagonal = √2 side Now we can apply these concepts in Ex 6.4, 7 Ex 6.4, 7 Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. Given: Square ABCD with Diagonal BD βˆ† 𝐡𝐢𝐸 which is described on base BC βˆ† 𝐡𝐷𝐹 which is descried on base BD Both βˆ† 𝐡𝐢𝐸 and βˆ† 𝐡𝐷𝐹 equilateral To prove: (π‘Žπ‘Ÿ βˆ† 𝐡𝐢𝐸)/(π‘Žπ‘Ÿ βˆ† 𝐹𝐷𝐡)=1/2 Proof: Both Ξ” 𝐡𝐢𝐸 and βˆ† 𝐡𝐷𝐹 equilateral Both βˆ† 𝐡𝐢𝐸 and βˆ† 𝐡𝐷𝐹 equilateral In βˆ† 𝐡𝐷𝐹 and βˆ† 𝐡𝐢𝐸 𝐷𝐹/𝐢𝐸=𝐹𝐡/𝐸𝐡=𝐷𝐡/𝐢𝐡 Hence by SSS similarity βˆ† 𝐹𝐡𝐷 ~ βˆ† 𝐡𝐢𝐸 We know that in similar triangles, Ratio of area of triangle is equal to ratio of square of corresponding sides (π‘Žπ‘Ÿ βˆ† 𝐹𝐡𝐷)/(π‘Žπ‘Ÿ βˆ† 𝐡𝐢𝐸)=( 𝐷𝐡/𝐡𝐢 )^2 But DB = √2 𝐡𝐢 as DB is the diagonal of square ABCD Hence (π‘Žπ‘Ÿ βˆ† 𝐹𝐡𝐷)/(π‘Žπ‘Ÿ βˆ† 𝐡𝐢𝐸)=( (√2 𝐡𝐢)/𝐡𝐢 )^2 (π‘Žπ‘Ÿ βˆ† 𝐹𝐡𝐷)/(π‘Žπ‘Ÿ βˆ† 𝐡𝐢𝐸)=(√2)2 (π‘Žπ‘Ÿ βˆ† 𝐹𝐡𝐷)/(π‘Žπ‘Ÿ βˆ† 𝐡𝐢𝐸)=2 (π‘Žπ‘Ÿ βˆ† 𝐡𝐢𝐸)/(π‘Žπ‘Ÿ βˆ† 𝐹𝐡𝐷)=1/2 Hence proved

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.