{'answer': ' [[[19],[900]],[[19],[900]]]', 'answer_type': ' [answer_types([[string(AB = ),type(s_textbox)],[string(and)],[string(BC = ),type(s_textbox)],])]', 'question': ' [right_angle_triangle(dimension(6,8,10),perp_label(string(x + 15)),base_label(string(3x + 168)),hypt_label(string()),name(Q,R,P)),right_angle_triangle(dimension(5,12,13),perp_label(string(10x - 21)),base_label(string(3x + 888)),hypt_label(string()),name(B,C,A)), string(In latex(\\\\triangle {PQR}), latex(PQ = (x + 15))cm, latex(QR = (3x + 168))cm, latex(\\\\angle {PQR} = 90^{\\\\circ}) and in latex(\\\\triangle {ABC}), latex(AB = (10x - 21))cm, latex(BC = (3x + 888))cm, latex(\\\\angle {ABC} = 90^{\\\\circ}). If ratio of the area of triangles latex(PQR) and latex(ABC) is latex(\\\\frac{1}{5}), find the value of sides latex(AB) and latex(BC). )]', 'hint': ' [string(Area = latex(\\\\frac{1}{2}) x Base x Height.),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.)]', 'solution': ' [,string(latex(\\\\because \\\\angle{PQR} = 90^{\\\\circ}) and latex(\\\\angle{ABC} = 90^{\\\\circ})) ,string(latex(\\\\therefore \\\\triangle{PQR}) and latex(\\\\triangle{ABC}) are right angle triangles),string(latex(PQ = x + 15), latex(QR = 3x + 168) are base and height of latex(\\\\triangle {PQR}) and latex(AB = 10x - 21), latex(BC = 3x + 888) are base and height of latex(\\\\triangle {ABC}) and their ratio of area is latex(\\\\frac{1}{5})),string(latex(\\\\therefore \\\\frac{1}{2} \\\\times (x + 15) \\\\times (3x + 168) = \\\\frac{1}{5}\\\\times \\\\frac{1}{2} \\\\times (10x - 21) \\\\times (3x + 888))),string(latex(1 \\\\times 3 \\\\times x^2 + (1 \\\\times 168  + 15 \\\\times 3) \\\\times x + (15 \\\\times 168) = \\\\frac{1}{5} \\\\times 10\\\\times 3 \\\\times x^2 + (10 \\\\times 888  - 21 \\\\times 3) \\\\times x + (-21 \\\\times 888))),string(latex(5 \\\\times 1 \\\\times 3 \\\\times x^2 + 5 \\\\times (1 \\\\times 168  + 15 \\\\times 3) \\\\times x + 5 \\\\times (15 \\\\times 168) = 1 \\\\times 10\\\\times 3 \\\\times x^2 + 1 \\\\times (10 \\\\times 888  - 21 \\\\times 3) \\\\times x + 1 \\\\times (-21 \\\\times 888))),string(latex((15 - 30)x^2 + (1065 - 8817)x + (12600 + 18648) = 0)),string(latex(\\\\therefore -15x^2 - 7752x + 31248 = 0)),,string(Comparing latex(-15x^2 - 7752x + 31248 = 0) with latex(ax^2 + bx + c = 0)),string(Thus, latex(a = -15, b = -7752, c = 31248)),string(latex(\\\\therefore b^2 - 4ac = (-7752)^2 - 4 \\\\times -15 \\\\times 31248 = 61968384)),string(Now, latex(x = \\\\frac{-(-7752) \\\\pm \\\\sqrt {61968384} }{2 \\\\times -15}) latex(\\\\text{~~~~}) (Quadratic formula)),string(latex(x = \\\\frac{-(-7752) - 7872}{2 \\\\times -15} = 4) or latex(x = \\\\frac{-(-7752) + 7872}{2 \\\\times -15} = -520.8)),string(latex(\\\\therefore) latex(x = 4) or latex(x = -520.8)),,string(When latex(x = 4)),string(latex(AB = 10x - 21 = 10 \\\\times 4 - 21 = 19) cm),string(latex(BC = 3x + 888 = 3 \\\\times 4 + 888 = 900) cm),string(When latex(x = -520.8)),string(latex(AB = 10x - 21 = 10 \\\\times -520.8 - 21 = -5229 \\\\leq 0 ) Hence, this side is not possible.),string(latex(BC = 3x + 888 = 3 \\\\times -520.8 + 888 = -674.4 \\\\leq 0 ) Hence, this side is not possible.),]', 'config': ' [name(A,"A"),name(X,"X"),rt_triangle_base_height_with_var("PQR","PQ","QR",x,1,15,3,168),rt_triangle_base_height_with_var("ABC","AB","BC",x,10,-21,3,888),poly_area_ratio("PQR","ABC",1,5),config_for_parser(["right_angle_triangle(dimension(6,8,10),perp_label(string(x + 15)),base_label(string(3x + 168)),hypt_label(string()),name(Q,R,P)),right_angle_triangle(dimension(5,12,13),perp_label(string(10x - 21)),base_label(string(3x + 888)),hypt_label(string()),name(B,C,A))"]),config_text(["string(In latex(\\\\\\\\triangle {PQR}), latex(PQ = (x + 15))cm, latex(QR = (3x + 168))cm, latex(\\\\\\\\angle {PQR} = 90^{\\\\\\\\circ}) and in latex(\\\\\\\\triangle {ABC}), latex(AB = (10x - 21))cm, latex(BC = (3x + 888))cm, latex(\\\\\\\\angle {ABC} = 90^{\\\\\\\\circ}). If ratio of the area of triangles latex(PQR) and latex(ABC) is latex(\\\\\\\\frac{1}{5}),)"]),name(P,"P")]', 'qid': '-1', 'concepts_used': ' [["word_problem_quad_formula",816],["word_problem_construct_equation_rt_tri_area_ratio",146],["word_problem_find_rt_tri_base_height",462]]', 'config_type': ' []', 'question_type': ' []'}