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

Ex 6.4, 7 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.4, 7 - Chapter 6 Class 10 Triangles - Part 3

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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

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.