the hypotenuse of a right triangle is 52 in. one leg of the triangle is 8 in. more than twice the length of the other. what is the perimeter of the triangle?

20 in
26 in
120 in
138 in

Question

the hypotenuse of a right triangle is 52 in. one leg of the triangle is 8 in. more than twice the length of the other. what is the perimeter of the triangle?

20 in
26 in
120 in
138 in

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Solving Quadratic Equations", "Quadratic Equation Formulation" ], "new_concepts": [ "Pythagorean Theorem Application" ], "current_concepts": [ "Solving Quadratic Equations", "Quadratic Equation Formulation", "Pythagorean Theorem Application" ] } ## Set up the equations using the Pythagorean theorem $$ \begin{aligned} &\text{Let the shorter leg be } x.\ &\text{The other leg is } 2x + 8.\ &\text{The hypotenuse is } 52.\ &x^2 + (2x + 8)^2 = 52^2 \end{aligned} $$ ## Solve the quadratic equation for x $$ \begin{aligned} &x^2 + 4x^2 + 32x + 64 = 2704\ &5x^2 + 32x - 2640 = 0\ &(5x + 132)(x - 20) = 0\ &x = 20 \quad (x > 0) \end{aligned} $$ ## Calculate the perimeter of the triangle $$ \begin{aligned} &\text{Shorter leg: } x = 20\text{ in}\ &\text{Longer leg: } 2(20) + 8 = 48\text{ in}\ &\text{Hypotenuse: } 52\text{ in}\ &\text{Perimeter: } 20 + 48 + 52 = 120\text{ in} \end{aligned} $$

(A) 20 in

(B) 26 in

(C) 120 in

(D) 138 in

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Quadratic Equation Formulation" ] }

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QUESTION IMAGE

Question

Explanation:

Set up the equations using the Pythagorean theorem

Solve the quadratic equation for x

Calculate the perimeter of the triangle

Answer: