» kajal: nice side but what hapen yr down load dear
» Arjun: very very hot and tiet ur lund ahaaaaaa ha
» David: SirLet the base of right angled triglane be b' , altitude a and hypotenuse c'Then by Pythagorean theorem c = (a^2 + b^2)^1/2The right triglanes have perimeter P = a + b + c = a + b + (a^2 + b^2)^1/2and area A = a b / 2. ab - 2a - 2b = 2 (a^2 + b^2)^1/2 or (ab - 2a - 2b)^2 = 4(a^2 + b^2). Simplifying the sides of the first right triglane are 6,8,10AREA = PERIMETER = 24 In the Same way sides of second right angled triglane are5,12,13 AREA = PERIMETER = 30. The others have perimeter P = a + b + c THEN, area A = [s(s-a)(s-b)(s-c)] ^1/2 S= P / 2. On solving a=b=c=100 reveals the sides of the first triglane are 5,5,8, Area = 12 , Perimeter = 18 The sides of the second triglane are 3,25,26 .practically it is a small angles triglane Area = 36 , Perimeter = 541)For right angled triglaneArea =Perimeter = 242)For Second right angled triglaneArea =Perimeter = 30For other triglanes1)Area=12 and Perimeter = 182)Area=36,Perimeter = 54
» Immi: Gr8 wrk bhai, plz ass prade k vdo upld krna. Take care

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