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Ex 6.4

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Chapter 6 Class 10 Triangles

Serial order wise

Last updated at March 29, 2023 by Teachoo

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 β π΄π΅πΆ πππ β π·πΈπΉ π·πΈ/π΄π΅=12/6 πΈπΉ/π΅πΆ=12/6 π·πΉ/π΄πΆ=12/6 Hence by SSS similarity β π΄π΅πΆ ~ β π·πΈπΉ Concept 2 Diagonal of a square is β2 of its side Since all sides of square are equal, Let AB = BC = CD = AD = x cm All angles of square are 90Β° So, β 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 = β2 x Hence, π·πππππππ/π πππ = π΅π·/π·πΆ = (β2 π₯)/π₯ = β2/1 Diagonal = β2 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 β π΅πΆπΈ which is described on base BC β π΅π·πΉ which is descried on base BD Both β π΅πΆπΈ and β π΅π·πΉ equilateral To prove: (ππ β π΅πΆπΈ)/(ππ β πΉπ·π΅)=1/2 Proof: Both Ξ π΅πΆπΈ and β π΅π·πΉ equilateral Both β π΅πΆπΈ and β π΅π·πΉ equilateral In β π΅π·πΉ and β π΅πΆπΈ π·πΉ/πΆπΈ=πΉπ΅/πΈπ΅=π·π΅/πΆπ΅ Hence by SSS similarity β πΉπ΅π· ~ β π΅πΆπΈ We know that in similar triangles, Ratio of area of triangle is equal to ratio of square of corresponding sides (ππ β πΉπ΅π·)/(ππ β π΅πΆπΈ)=( π·π΅/π΅πΆ )^2 But DB = β2 π΅πΆ as DB is the diagonal of square ABCD Hence (ππ β πΉπ΅π·)/(ππ β π΅πΆπΈ)=( (β2 π΅πΆ)/π΅πΆ )^2 (ππ β πΉπ΅π·)/(ππ β π΅πΆπΈ)=(β2)2 (ππ β πΉπ΅π·)/(ππ β π΅πΆπΈ)=2 (ππ β π΅πΆπΈ)/(ππ β πΉπ΅π·)=1/2 Hence proved