['Answer : [[[216]]]\n', 'Answer_type : [answer_types([[string(Perimeter = ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)),string(If latex(AB = 18) cm and latex(BC = 90) cm are two adjacent sides of a rectangle latex(ABCD), find the perimeter of latex(ABCD). )]\n', 'Hint : [string(Perimeter = 2 x Summation of two adjacent sides.)]\n', 'Solution : [,string(Perimeter of latex(ABCD) = 2 x Summation of two adjacent sides latex( = 2 \\\\times (AB + BC) = 2 \\\\times (18 + 90) = 216 cm)),]\n', 'Config : [name(x,X),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(a,A)]\n', 'Qid : 401\n', 'Concepts_used : [["word_problem_find_rec_perimeter",0]]\n', 'Config_type : ["rectangle_sides_value"]\n', 'Question_type : ["cts_not_to_be_shown_individually"]\n', '\n', '----------------------\n', 'Answer : [[[216]]]\n', 'Answer_type : [answer_types([[string(Perimeter = ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)), string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter, find the perimeter of latex(ABCD). )]\n', 'Hint : [string(Perimeter = 2 x Summation of adjacent sides.),string(Area = Product of two adjacent sides.),string(If latex(ax^2 + bx + c = 0) then latex( x = \\\\frac{-b \\\\pm \\\\sqrt {b^2 - 4ac} }{2a }).),string(Put the value of latex(x) in the equation of sides.),string(Perimeter = 2 x Summation of two adjacent sides.)]\n', 'Solution : [string(Let latex(AB = x) then),string(latex(\\\\because BC) is 5 times side latex(AB)),string(latex(\\\\therefore BC = 5x)),string(Area is 1404 more than the perimeter),string(latex(AB \\\\times BC = 2 \\\\times (AB + BC) + 1404)),string(latex((x) \\\\times (5x) = 2 \\\\times (x + 5x) + 1404)),string(latex(5x^2 = 12x + 1404)),,string(latex(\\\\therefore 5x^2 - 12x - 1404 = 0)),string(Comparing latex(5x^2 - 12x - 1404 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = 5, b = -12, c = -1404)),string(latex(\\\\therefore b^2 - 4ac = (-12)^2 - 4 \\\\times 5 \\\\times -1404 = 28224)),string(Now, latex(x = \\\\frac{-(-12) \\\\pm \\\\sqrt {28224} }{2 \\\\times 5}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-12) - 168}{2 \\\\times 5} = -15.6) or latex(x = \\\\frac{-(-12) + 168}{2 \\\\times 5} = 18)),string(latex(\\\\therefore) latex(x = -15.6) or latex(x = 18)),string(When latex(x = -15.6)),string(latex(AB = x = -15.6 \\\\leq 0 ) Hence, this side is not possible.),string(latex(BC = 5x = 5 \\\\times -15.6 = -78 \\\\leq 0 ) Hence, this side is not possible.),string(When latex(x = 18)),string(latex(AB = x = 18) cm),string(latex(BC = 5x = 5 \\\\times 18 = 90) cm),string(Perimeter of latex(ABCD) = 2 x Summation of two adjacent sides latex( = 2 \\\\times (AB + BC) = 2 \\\\times (18 + 90) = 216 cm)),]\n', 'Config : [name(X,"X"),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(A,"A")]\n', 'Qid : -1\n', 'Concepts_used : [["word_problem_construct_equation_rec_peri_area_relation",753],["word_problem_quad_formula",886],["word_problem_find_rec_side",864],["word_problem_find_rec_perimeter",401]]\n', 'Config_type : ["rectangle_sides_relation_with_perimeter_area_relation_find_perimeter"]\n', 'Question_type : ["area_based"]\n', '\n', '----------------------\n', 'Answer : [[[1620]]]\n', 'Answer_type : [answer_types([[string(Area of ABCD = ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)),string(If latex(AB = 18) cm and latex(BC = 90) cm are two adjacent sides of a rectangle latex(ABCD), find the area of latex(ABCD). )]\n', 'Hint : [string(Area = Product of two adjacent sides.)]\n', 'Solution : [,string(Area of latex(ABCD) = Product of two adjacent sides = latex(AB \\\\times BC = 18 \\\\times 90 = 1620) latex( cm^2)),]\n', 'Config : [name(x,X),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(a,A)]\n', 'Qid : 509\n', 'Concepts_used : [["word_problem_find_rec_area",0]]\n', 'Config_type : ["rectangle_sides_value"]\n', 'Question_type : ["cts_not_to_be_shown_individually"]\n', '\n', '----------------------\n', 'Answer : [[[1620]]]\n', 'Answer_type : [answer_types([[string(Area of ABCD = ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)), string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter, find the area of latex(ABCD). )]\n', 'Hint : [string(Perimeter = 2 x Summation of adjacent sides.),string(Area = Product of two adjacent sides.),string(If latex(ax^2 + bx + c = 0) then latex( x = \\\\frac{-b \\\\pm \\\\sqrt {b^2 - 4ac} }{2a }).),string(Put the value of latex(x) in the equation of sides.),string(Area = Product of two adjacent sides.)]\n', 'Solution : [string(Let latex(AB = x) then),string(latex(\\\\because BC) is 5 times side latex(AB)),string(latex(\\\\therefore BC = 5x)),string(Area is 1404 more than the perimeter),string(latex(AB \\\\times BC = 2 \\\\times (AB + BC) + 1404)),string(latex((x) \\\\times (5x) = 2 \\\\times (x + 5x) + 1404)),string(latex(5x^2 = 12x + 1404)),,string(latex(\\\\therefore 5x^2 - 12x - 1404 = 0)),string(Comparing latex(5x^2 - 12x - 1404 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = 5, b = -12, c = -1404)),string(latex(\\\\therefore b^2 - 4ac = (-12)^2 - 4 \\\\times 5 \\\\times -1404 = 28224)),string(Now, latex(x = \\\\frac{-(-12) \\\\pm \\\\sqrt {28224} }{2 \\\\times 5}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-12) - 168}{2 \\\\times 5} = -15.6) or latex(x = \\\\frac{-(-12) + 168}{2 \\\\times 5} = 18)),string(latex(\\\\therefore) latex(x = -15.6) or latex(x = 18)),string(When latex(x = -15.6)),string(latex(AB = x = -15.6 \\\\leq 0 ) Hence, this side is not possible.),string(latex(BC = 5x = 5 \\\\times -15.6 = -78 \\\\leq 0 ) Hence, this side is not possible.),string(When latex(x = 18)),string(latex(AB = x = 18) cm),string(latex(BC = 5x = 5 \\\\times 18 = 90) cm),string(Area of latex(ABCD) = Product of two adjacent sides = latex(AB \\\\times BC = 18 \\\\times 90 = 1620) latex( cm^2)),]\n', 'Config : [name(X,"X"),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(A,"A")]\n', 'Qid : -1\n', 'Concepts_used : [["word_problem_construct_equation_rec_peri_area_relation",753],["word_problem_quad_formula",886],["word_problem_find_rec_side",864],["word_problem_find_rec_area",509]]\n', 'Config_type : ["rectangle_sides_relation_with_perimeter_area_relation_find_area"]\n', 'Question_type : ["area_based"]\n', '\n', '----------------------\n', 'Answer : [[[18],[90]],[[18],[90]]]\n', 'Answer_type : [answer_types([,[string(AB = ),type(s_textbox)],[string(and)],[string(BC = ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)),string(If latex(AB = (x))cm and latex(BC = (5x))cm are two adjacent sides of a rectangle latex(ABCD) and if values of latex(x) are latex(-15.6) and latex(18), find the value of sides latex(AB) and latex(BC). )]\n', 'Hint : [string(Put the value of latex(x) in the equation of sides.)]\n', 'Solution : [,string(When latex(x = -15.6)),string(latex(AB = x = -15.6 \\\\leq 0 ) Hence, this side is not possible.),string(latex(BC = 5x = 5 \\\\times -15.6 = -78 \\\\leq 0 ) Hence, this side is not possible.),string(When latex(x = 18)),string(latex(AB = x = 18) cm),string(latex(BC = 5x = 5 \\\\times 18 = 90) cm),]\n', 'Config : [name(x,X),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(a,A)]\n', 'Qid : 864\n', 'Concepts_used : [["word_problem_find_rec_side",0]]\n', 'Config_type : ["rectangle_var_sides_with_eqn"]\n', 'Question_type : ["cts_not_to_be_shown_individually"]\n', '\n', '----------------------\n', 'Answer : [[[18],[90]],[[18],[90]]]\n', 'Answer_type : [answer_types([,[string(AB = ),type(s_textbox)],[string(and)],[string(BC = ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)), string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter, find the value of sides latex(AB) and latex(BC). )]\n', 'Hint : [string(Perimeter = 2 x Summation of adjacent sides.),string(Area = Product of two adjacent sides.),string(If latex(ax^2 + bx + c = 0) then latex( x = \\\\frac{-b \\\\pm \\\\sqrt {b^2 - 4ac} }{2a }).),string(Put the value of latex(x) in the equation of sides.)]\n', 'Solution : [string(Let latex(AB = x) then),string(latex(\\\\because BC) is 5 times side latex(AB)),string(latex(\\\\therefore BC = 5x)),string(Area is 1404 more than the perimeter),string(latex(AB \\\\times BC = 2 \\\\times (AB + BC) + 1404)),string(latex((x) \\\\times (5x) = 2 \\\\times (x + 5x) + 1404)),string(latex(5x^2 = 12x + 1404)),,string(latex(\\\\therefore 5x^2 - 12x - 1404 = 0)),string(Comparing latex(5x^2 - 12x - 1404 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = 5, b = -12, c = -1404)),string(latex(\\\\therefore b^2 - 4ac = (-12)^2 - 4 \\\\times 5 \\\\times -1404 = 28224)),string(Now, latex(x = \\\\frac{-(-12) \\\\pm \\\\sqrt {28224} }{2 \\\\times 5}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-12) - 168}{2 \\\\times 5} = -15.6) or latex(x = \\\\frac{-(-12) + 168}{2 \\\\times 5} = 18)),string(latex(\\\\therefore) latex(x = -15.6) or latex(x = 18)),string(When latex(x = -15.6)),string(latex(AB = x = -15.6 \\\\leq 0 ) Hence, this side is not possible.),string(latex(BC = 5x = 5 \\\\times -15.6 = -78 \\\\leq 0 ) Hence, this side is not possible.),string(When latex(x = 18)),string(latex(AB = x = 18) cm),string(latex(BC = 5x = 5 \\\\times 18 = 90) cm),]\n', 'Config : [name(X,"X"),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(A,"A")]\n', 'Qid : -1\n', 'Concepts_used : [["word_problem_construct_equation_rec_peri_area_relation",753],["word_problem_quad_formula",886],["word_problem_find_rec_side",864]]\n', 'Config_type : []\n', 'Question_type : []\n', '\n', '----------------------\n', 'Answer : [[[-15.6],[18]],[[18],[-15.6]]]\n', 'Answer_type : [answer_types([[latex(x = ~),type(s_textbox)],[string(or)],[latex( x = ~ ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)),string(If latex(5x^2 - 12x - 1404 = 0), form a quadratic equation in latex(x) and solve it. )]\n', 'Hint : [string(If latex(ax^2 + bx + c = 0) then latex( x = \\\\frac{-b \\\\pm \\\\sqrt {b^2 - 4ac} }{2a }).)]\n', 'Solution : [,string(Comparing latex(5x^2 - 12x - 1404 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = 5, b = -12, c = -1404)),string(latex(\\\\therefore b^2 - 4ac = (-12)^2 - 4 \\\\times 5 \\\\times -1404 = 28224)),string(Now, latex(x = \\\\frac{-(-12) \\\\pm \\\\sqrt {28224} }{2 \\\\times 5}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-12) - 168}{2 \\\\times 5} = -15.6) or latex(x = \\\\frac{-(-12) + 168}{2 \\\\times 5} = 18)),string(latex(\\\\therefore) latex(x = -15.6) or latex(x = 18)),]\n', 'Config : [name(x,X),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(a,A)]\n', 'Qid : 886\n', 'Concepts_used : [["word_problem_quad_formula",0]]\n', 'Config_type : ["generic_quadratic_eqn_with_rhs_zero_word_problem"]\n', 'Question_type : ["cts_not_to_be_shown_individually"]\n', '\n', '----------------------\n', 'Answer : [[[-15.6],[18]],[[18],[-15.6]]]\n', 'Answer_type : [answer_types([[latex(x = ~),type(s_textbox)],[string(or)],[latex( x = ~ ),type(s_textbox)]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)), string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter, form a quadratic equation in latex(x) and solve it. )]\n', 'Hint : [string(Perimeter = 2 x Summation of adjacent sides.),string(Area = Product of two adjacent sides.),string(If latex(ax^2 + bx + c = 0) then latex( x = \\\\frac{-b \\\\pm \\\\sqrt {b^2 - 4ac} }{2a }).)]\n', 'Solution : [string(Let latex(AB = x) then),string(latex(\\\\because BC) is 5 times side latex(AB)),string(latex(\\\\therefore BC = 5x)),string(Area is 1404 more than the perimeter),string(latex(AB \\\\times BC = 2 \\\\times (AB + BC) + 1404)),string(latex((x) \\\\times (5x) = 2 \\\\times (x + 5x) + 1404)),string(latex(5x^2 = 12x + 1404)),,string(latex(\\\\therefore 5x^2 - 12x - 1404 = 0)),string(Comparing latex(5x^2 - 12x - 1404 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = 5, b = -12, c = -1404)),string(latex(\\\\therefore b^2 - 4ac = (-12)^2 - 4 \\\\times 5 \\\\times -1404 = 28224)),string(Now, latex(x = \\\\frac{-(-12) \\\\pm \\\\sqrt {28224} }{2 \\\\times 5}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-12) - 168}{2 \\\\times 5} = -15.6) or latex(x = \\\\frac{-(-12) + 168}{2 \\\\times 5} = 18)),string(latex(\\\\therefore) latex(x = -15.6) or latex(x = 18)),]\n', 'Config : [name(X,"X"),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(A,"A")]\n', 'Qid : -1\n', 'Concepts_used : [["word_problem_construct_equation_rec_peri_area_relation",753],["word_problem_quad_formula",886]]\n', 'Config_type : []\n', 'Question_type : []\n', '\n', '----------------------\n', 'Answer : [[5,-12,-1404]]\n', 'Answer_type : [answer_types([[string(A),type(continuous(input(string(x),type(s_textbox)),latex(x^2 + ), input(string(y),type(s_textbox)),latex( x + ),input(string(z),type(s_textbox)),latex( = 0)))]])]\n', 'Question : [rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A)),string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter, form a quadratic equation in latex(x).)]\n', 'Hint : [string(Perimeter = 2 x Summation of adjacent sides.),string(Area = Product of two adjacent sides.)]\n', 'Solution : [,string(Let latex(AB = x) then),string(latex(\\\\because BC) is 5 times side latex(AB)),string(latex(\\\\therefore BC = 5x)),string(Area is 1404 more than the perimeter),string(latex(AB \\\\times BC = 2 \\\\times (AB + BC) + 1404)),string(latex((x) \\\\times (5x) = 2 \\\\times (x + 5x) + 1404)),string(latex(5x^2 = 12x + 1404)),,string(latex(\\\\therefore 5x^2 - 12x - 1404 = 0)),]\n', 'Config : [name(x,X),rectangle_side_with_relation("ABCD","AB","BC",x,5,"times"),poly_perimeter_area_with_relation("ABCD",1404,"more_than"),config_for_parser(["rectangle(dimension(6,13),label_1(string()),label_2(string()),label_3(string()),label_4(string()),name(B,C,D,A))"]),config_text(["string(latex(ABCD) is a rectangle in which the side latex(BC) is 5 times the side latex(AB). If the area is 1404 more than the perimeter,)"]),name(a,A)]\n', 'Qid : 753\n', 'Concepts_used : [["word_problem_construct_equation_rec_peri_area_relation",0]]\n', 'Config_type : ["rectangle_sides_relation_with_perimeter_area_relation"]\n', 'Question_type : ["area_based"]\n', '\n', '----------------------\n', '\n']