{'answer': ' [[[30]]]', 'answer_type': ' [answer_types([[string(Perimeter = ),type(s_textbox)]])]', 'question': ' [right_angle_triangle(dimension(5,12,13),perp_label(string(6x - 115)),base_label(string(2x - 28)),hypt_label(string()),name(Q,R,P)), string(In latex(\\\\triangle {PQR}), latex(PQ = (6x - 115))cm, latex(QR = (2x - 28))cm and latex(\\\\angle {PQR} = 90^{\\\\circ}). If the area of the triangle is latex(30 cm^2), find the perimeter of latex(\\\\triangle {PQR}). )]', 'hint': ' [string(latex(\\\\frac{1}{2}) x Base x Height = Area.),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 of triangle = Summation of all sides.)]', 'solution': ' [string(latex(\\\\because \\\\angle{PQR} = 90^{\\\\circ})),string(latex(\\\\therefore \\\\triangle{PQR}) is a right angle triangle),string(latex(\\\\frac{1}{2}) x Base x Height = Area),string(PQ x QR = 2 x Area),string(latex(\\\\therefore (6x - 115) \\\\times (2x - 28) = 2 \\\\times 30)),string(latex(6\\\\times 2 \\\\times x^2 + (6 \\\\times -28 - 115 \\\\times 2) \\\\times x + (-115 \\\\times -28 - 2 \\\\times 30) = 0)),string(latex(2(6x^2 - 199x + 1580) = 0)),string(latex(\\\\therefore 6x^2 - 199x + 1580 = 0)),string(Comparing latex(12x^2 - 398x + 3160 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = 12, b = -398, c = 3160)),string(latex(\\\\therefore b^2 - 4ac = (-398)^2 - 4 \\\\times 12 \\\\times 3160 = 6724)),string(Now, latex(x = \\\\frac{-(-398) \\\\pm \\\\sqrt {6724} }{2 \\\\times 12}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-398) - 82}{2 \\\\times 12} = 13.166666666666666) or latex(x = \\\\frac{-(-398) + 82}{2 \\\\times 12} = 20)),string(latex(\\\\therefore) latex(x \\\\approx 13.17) or latex(x = 20)),string(When latex(x = 13.17)),string(latex(PQ = 6x - 115 = 6 \\\\times 13.17 - 115 = -35.98 \\\\leq 0 ) Hence, this side is not possible.),string(latex(QR = 2x - 28 = 2 \\\\times 13.17 - 28 = -1.6 \\\\leq 0 ) Hence, this side is not possible.),string(When latex(x = 20)),string(latex(PQ = 6x - 115 = 6 \\\\times 20 - 115 = 5) cm),string(latex(QR = 2x - 28 = 2 \\\\times 20 - 28 = 12) cm),string(Hypotenuse = latex(\\\\sqrt{{\\\\text{Base}}^2 + {\\\\text{Height}}^2})),string(Hypotenuse = latex(\\\\sqrt{{5}^2 + {12}^2} = \\\\sqrt{169} = 13)),string(latex(\\\\therefore) Perimeter of latex(\\\\triangle{ PQR}) = Summation of all sides latex( = 5 + 12 + 13 = 30 cm)),]', 'config': ' [name(X,"X"),rt_triangle_base_height_with_var("PQR","PQ","QR",x,6,-115,2,-28),poly_area("PQR",30),config_for_parser(["right_angle_triangle(dimension(5,12,13),perp_label(string(6x - 115)),base_label(string(2x - 28)),hypt_label(string()),name(Q,R,P))"]),config_text(["string(In latex(\\\\\\\\triangle {PQR}), latex(PQ = (6x - 115))cm, latex(QR = (2x - 28))cm and latex(\\\\\\\\angle {PQR} = 90^{\\\\\\\\circ}). If the area of the triangle is latex(30 cm^2),)"]),name(P,"P")]', 'qid': '-1', 'concepts_used': ' [["word_problem_construct_equation_rt_tri_var_base_height",243],["word_problem_quad_formula",867],["word_problem_find_rt_tri_base_height",655],["word_problem_find_rt_tri_perimeter",822]]', 'config_type': <ConfigType: Rt Triangle Var Base Height With Area Find Perimeter>, 'question_type': <QuestionType: Area Based>, 'metadata': <QuestionMeta: Quadratic Equation1:Word Problems706:2356>, 'topic': <Topic: Quadratic Equation1>, 'sub_topic': <SubTopic: Word Problems>, 'concept_list': ['word_problem_construct_equation_rt_tri_var_base_height', 'word_problem_quad_formula', 'word_problem_find_rt_tri_base_height', 'word_problem_find_rt_tri_perimeter'], 'sub_question_list': [558316, 558317, 558318, 558319], 'is_complex': True}, Request body - {'isOldForm': False, 'must_get_html_from_browser': True, 'subparts': [], 'configString': ' [name(X,"X"),rt_triangle_base_height_with_var("PQR","PQ","QR",x,6,-115,2,-28),poly_area("PQR",30),config_for_parser(["right_angle_triangle(dimension(5,12,13),perp_label(string(6x - 115)),base_label(string(2x - 28)),hypt_label(string()),name(Q,R,P))"]),config_text(["string(In latex(\\\\\\\\triangle {PQR}), latex(PQ = (6x - 115))cm, latex(QR = (2x - 28))cm and latex(\\\\\\\\angle {PQR} = 90^{\\\\\\\\circ}). If the area of the triangle is latex(30 cm^2),)"]),name(P,"P")]', 'questionString': ' [right_angle_triangle(dimension(5,12,13),perp_label(string(6x - 115)),base_label(string(2x - 28)),hypt_label(string()),name(Q,R,P)), string(In latex(\\\\triangle {PQR}), latex(PQ = (6x - 115))cm, latex(QR = (2x - 28))cm and latex(\\\\angle {PQR} = 90^{\\\\circ}). If the area of the triangle is latex(30 cm^2), find the perimeter of latex(\\\\triangle {PQR}). )]', 'hintString': ' [string(latex(\\\\frac{1}{2}) x Base x Height = Area.),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 of triangle = Summation of all sides.)]', 'answerTypeString': ' [answer_types([[string(Perimeter = ),type(s_textbox)]])]', 'solutionString': ' [string(latex(\\\\because \\\\angle{PQR} = 90^{\\\\circ})),string(latex(\\\\therefore \\\\triangle{PQR}) is a right angle triangle),string(latex(\\\\frac{1}{2}) x Base x Height = Area),string(PQ x QR = 2 x Area),string(latex(\\\\therefore (6x - 115) \\\\times (2x - 28) = 2 \\\\times 30)),string(latex(6\\\\times 2 \\\\times x^2 + (6 \\\\times -28 - 115 \\\\times 2) \\\\times x + (-115 \\\\times -28 - 2 \\\\times 30) = 0)),string(latex(2(6x^2 - 199x + 1580) = 0)),string(latex(\\\\therefore 6x^2 - 199x + 1580 = 0)),string(Comparing latex(12x^2 - 398x + 3160 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = 12, b = -398, c = 3160)),string(latex(\\\\therefore b^2 - 4ac = (-398)^2 - 4 \\\\times 12 \\\\times 3160 = 6724)),string(Now, latex(x = \\\\frac{-(-398) \\\\pm \\\\sqrt {6724} }{2 \\\\times 12}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-398) - 82}{2 \\\\times 12} = 13.166666666666666) or latex(x = \\\\frac{-(-398) + 82}{2 \\\\times 12} = 20)),string(latex(\\\\therefore) latex(x \\\\approx 13.17) or latex(x = 20)),string(When latex(x = 13.17)),string(latex(PQ = 6x - 115 = 6 \\\\times 13.17 - 115 = -35.98 \\\\leq 0 ) Hence, this side is not possible.),string(latex(QR = 2x - 28 = 2 \\\\times 13.17 - 28 = -1.6 \\\\leq 0 ) Hence, this side is not possible.),string(When latex(x = 20)),string(latex(PQ = 6x - 115 = 6 \\\\times 20 - 115 = 5) cm),string(latex(QR = 2x - 28 = 2 \\\\times 20 - 28 = 12) cm),string(Hypotenuse = latex(\\\\sqrt{{\\\\text{Base}}^2 + {\\\\text{Height}}^2})),string(Hypotenuse = latex(\\\\sqrt{{5}^2 + {12}^2} = \\\\sqrt{169} = 13)),string(latex(\\\\therefore) Perimeter of latex(\\\\triangle{ PQR}) = Summation of all sides latex( = 5 + 12 + 13 = 30 cm)),]'}