{"id":"cb_question_bank:9648fa12-82e1-4e58-9d8e-e7ff78aab75b","source_id":"cb_question_bank","source_item_id":"9648fa12-82e1-4e58-9d8e-e7ff78aab75b","external_id":"8c6f39d2-0804-4eb5-8c97-b801dcd8a2e8","ibn":null,"question_id":"6ab30ce3","program":"SAT","module":"math","domain_code":"S","domain":"Geometry and Trigonometry","skill_code":"S.C.","skill":"Right triangles and trigonometry","difficulty":"H","score_band":7,"answer_type":"mcq","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"Triangle upper A upper B upper C is similar to triangle upper D upper E upper F , where upper A corresponds to upper D and upper C corresponds to upper F . Angles upper C and upper F are right angles. If tangent left parenthesis upper A right parenthesis equals StartRoot 3 EndRoot and upper D upper F equals 125 , what is the length of line segment upper D upper E ?","stimulus":"","stimulus_html":null,"stem_html":"<p style=\"text-align: left;\">Triangle <math alttext=\"upper A upper B upper C\"><mrow>\n\t<mi>A</mi>\n\t<mi>B</mi>\n\t<mi>C</mi>\n</mrow>\n</math> is similar to triangle <math alttext=\"upper D upper E upper F\"><mrow>\n\t<mi>D</mi>\n\t<mi>E</mi>\n\t<mi>F</mi>\n</mrow>\n</math>, where <math alttext=\"upper A\"><mi>A</mi>\n</math> corresponds to <math alttext=\"upper D\"><mi>D</mi>\n</math> and <math alttext=\"upper C\"><mi>C</mi>\n</math> corresponds to <math alttext=\"upper F\"><mi>F</mi>\n</math>. Angles <math alttext=\"upper C\"><mi>C</mi>\n</math> and <math alttext=\"upper F\"><mi>F</mi>\n</math> are right angles. If <math alttext=\"tangent left parenthesis upper A right parenthesis equals StartRoot 3 EndRoot\"><mi>tan</mi><mfenced><mi>A</mi></mfenced><mo>=</mo><msqrt><mn>3</mn></msqrt></math> and <math alttext=\"upper D upper F equals 125\"><mi>D</mi><mi>F</mi><mo>=</mo><mrow><mn>125</mn></mrow></math>, what is the length of&nbsp;<math alttext=\"line segment upper D upper E\"><mover><mrow><mi>D</mi><mi>E</mi></mrow><mo>¯</mo></mover></math>?</p>","rationale_html":"<p style=\"text-align: left;\">Choice D is correct. Corresponding angles in similar triangles have equal measures. It's given that triangle <math alttext=\"upper A upper B upper C\"><mi>A</mi><mi>B</mi><mi>C</mi></math> is similar to triangle <math alttext=\"upper D upper E upper F\"><mi>D</mi><mi>E</mi><mi>F</mi></math>, where <math alttext=\"upper A\"><mi>A</mi>\n</math> corresponds to <math alttext=\"upper D\"><mi>D</mi>\n</math>, so the measure of angle <math alttext=\"upper A\"><mi>A</mi>\n</math> is equal to the measure of angle <math alttext=\"upper D\"><mi>D</mi>\n</math>. Therefore, if <math alttext=\"tangent left parenthesis upper A right parenthesis equals StartRoot 3 EndRoot\"><mi>tan</mi><mfenced><mi>A</mi></mfenced><mo>=</mo><msqrt><mn>3</mn></msqrt></math>,&nbsp;then <math alttext=\"tangent left parenthesis upper D right parenthesis equals StartRoot 3 EndRoot\"><mi>tan</mi><mfenced><mi>D</mi></mfenced><mo>=</mo><msqrt><mn>3</mn></msqrt></math>. It's given that angles <math alttext=\"upper C\"><mi>C</mi>\n</math> and <math alttext=\"upper F\"><mi>F</mi>\n</math> are right angles, so triangles <math alttext=\"upper A upper B upper C\"><mi>A</mi><mi>B</mi><mi>C</mi></math> and <math alttext=\"upper D upper E upper F\"><mi>D</mi><mi>E</mi><mi>F</mi></math> are right triangles. The adjacent side of an acute angle in a right triangle is the side closest to the angle that is not the hypotenuse. It follows that the adjacent side of angle <math alttext=\"upper D\"><mi>D</mi>\n</math> is side <math alttext=\"upper D upper F\"><mrow>\n\t<mi>D</mi>\n\t<mi>F</mi>\n</mrow>\n</math>. The opposite side of an acute angle in a right triangle is the side across from the acute angle. It follows that the opposite side of angle <math alttext=\"upper D\"><mi>D</mi>\n</math> is side <math alttext=\"upper E upper F\"><mrow>\n\t<mi>E</mi>\n\t<mi>F</mi>\n</mrow>\n</math>. The tangent of an acute angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side. Therefore, <math alttext=\"tangent left parenthesis upper D right parenthesis equals StartFraction upper E upper F Over upper D upper F EndFraction\"><mi>tan</mi><mfenced><mi>D</mi></mfenced><mo>=</mo><mfrac><mrow><mi>E</mi><mi>F</mi></mrow><mrow><mi>D</mi><mi>F</mi></mrow></mfrac></math>. If <math alttext=\"upper D upper F equals 125\"><mi>D</mi><mi>F</mi><mo>=</mo><mn>125</mn></math>, the length of side <math alttext=\"upper E upper F\"><mrow>\n\t<mi>E</mi>\n\t<mi>F</mi>\n</mrow>\n</math> can be found by substituting <math alttext=\"StartRoot 3 EndRoot\"><msqrt><mn>3</mn></msqrt></math> for&nbsp;<math alttext=\"tangent left parenthesis upper D right parenthesis\"><mi>tan</mi><mfenced><mi>D</mi></mfenced></math> and <math alttext=\"125\"><mn>125</mn>\n</math> for <math alttext=\"upper D upper F\"><mrow>\n\t<mi>D</mi>\n\t<mi>F</mi>\n</mrow>\n</math> in the equation <math alttext=\"tangent left parenthesis upper D right parenthesis equals StartFraction upper E upper F Over upper D upper F EndFraction\"><mi>tan</mi><mfenced><mi>D</mi></mfenced><mo>=</mo><mfrac><mrow><mi>E</mi><mi>F</mi></mrow><mrow><mi>D</mi><mi>F</mi></mrow></mfrac></math>, which yields <math alttext=\"StartRoot 3 EndRoot equals StartFraction upper E upper F Over 125 EndFraction\"><msqrt><mn>3</mn></msqrt><mo>=</mo><mfrac><mrow><mi>E</mi><mi>F</mi></mrow><mn>125</mn></mfrac></math>. Multiplying both sides of this equation by <math alttext=\"125\"><mn>125</mn>\n</math> yields <math alttext=\"125 StartRoot 3 EndRoot equals upper E upper F\"><mn>125</mn><msqrt><mn>3</mn></msqrt><mo>=</mo><mi>E</mi><mi>F</mi></math>. Since the length of side <math alttext=\"upper E upper F\"><mrow>\n\t<mi>E</mi>\n\t<mi>F</mi>\n</mrow>\n</math> is <math alttext=\"StartRoot 3 EndRoot\"><msqrt><mn>3</mn></msqrt></math> times the length of side <math alttext=\"upper D upper F\"><mrow>\n\t<mi>D</mi>\n\t<mi>F</mi>\n</mrow>\n</math>, it follows that&nbsp;triangle&nbsp;<math alttext=\"upper D upper E upper F\"><mi>D</mi><mi>E</mi><mi>F</mi></math> is a special right triangle with angle measures <math alttext=\"30 degree\"><mn>30</mn><mo>°</mo></math>, <math alttext=\"60 degree\"><mn>60</mn><mo>°</mo></math>, and <math alttext=\"90 degree\"><mn>90</mn><mo>°</mo></math>. Therefore, the length of the hypotenuse, <math alttext=\"line segment upper D upper E\"><mover><mrow><mi>D</mi><mi>E</mi></mrow><mo>¯</mo></mover></math>, is <math alttext=\"2\"><mn>2</mn>\n</math> times the length of side <math alttext=\"upper D upper F\"><mrow>\n\t<mi>D</mi>\n\t<mi>F</mi>\n</mrow>\n</math>, or <math alttext=\"upper D upper E equals 2 left parenthesis upper D upper F right parenthesis\"><mi>D</mi><mi>E</mi><mo>=</mo><mn>2</mn><mfenced><mrow><mi>D</mi><mi>F</mi></mrow></mfenced></math>. Substituting <math alttext=\"125\"><mn>125</mn>\n</math> for <math alttext=\"upper D upper F\"><mrow>\n\t<mi>D</mi>\n\t<mi>F</mi>\n</mrow>\n</math> in this equation yields <math alttext=\"upper D upper E equals 2 left parenthesis 125 right parenthesis\"><mi>D</mi><mi>E</mi><mo>=</mo><mn>2</mn><mfenced><mn>125</mn></mfenced></math>, or <math alttext=\"upper D upper E equals 250\"><mi>D</mi><mi>E</mi><mo>=</mo><mn>250</mn></math>. Thus, if&nbsp;<math alttext=\"tangent left parenthesis upper A right parenthesis equals StartRoot 3 EndRoot\"><mi>tan</mi><mfenced><mi>A</mi></mfenced><mo>=</mo><msqrt><mn>3</mn></msqrt></math> and&nbsp;<math alttext=\"upper D upper F equals 125\"><mi>D</mi><mi>F</mi><mo>=</mo><mn>125</mn></math>, the length of&nbsp;<math alttext=\"line segment upper D upper E\"><mover><mrow><mi>D</mi><mi>E</mi></mrow><mo>¯</mo></mover></math> is <math alttext=\"250\"><mn>250</mn>\n</math>.</p>\n<p style=\"text-align: left;\">Choice A is incorrect and may result from conceptual or calculation errors.</p>\n<p style=\"text-align: left;\">Choice B is incorrect and may result from conceptual or calculation errors.</p>\n<p style=\"text-align: left;\">Choice C is incorrect. This is the length of <math alttext=\"line segment upper E upper F\"><mover><mrow><mi>E</mi><mi>F</mi></mrow><mo>¯</mo></mover></math>, not <math alttext=\"line segment upper D upper E\"><mover><mrow><mi>D</mi><mi>E</mi></mrow><mo>¯</mo></mover></math>.</p>","correct_answer":["D"],"has_media":false,"has_mathml":true,"has_table":false,"source_created_at":"2023-08-02T20:25:59.840000+00:00","source_updated_at":"2023-08-02T20:25:59.840000+00:00","options":[{"label":"A","content_html":"<p><math alttext=\"125 StartFraction StartRoot 3 EndRoot Over 3 EndFraction\"><mrow><mn>125</mn></mrow><mfrac><msqrt><mn>3</mn></msqrt><mn>3</mn></mfrac></math></p>","ord":0},{"label":"B","content_html":"<p><math alttext=\"125 StartFraction StartRoot 3 EndRoot Over 2 EndFraction\"><mrow><mn>125</mn></mrow><mfrac><msqrt><mn>3</mn></msqrt><mn>2</mn></mfrac></math></p>","ord":1},{"label":"C","content_html":"<p><math alttext=\"125 StartRoot 3 EndRoot\"><mrow><mn>125</mn></mrow><msqrt><mn>3</mn></msqrt></math></p>","ord":2},{"label":"D","content_html":"<p><math alttext=\"250\"><mn>250</mn>\n</math></p>","ord":3}],"attribution":{"source_id":"cb_question_bank","source_name":"College Board digital SAT question bank (community dump)","rights_holder":"College Board","canonical_url":"https://satsuitequestionbank.collegeboard.org/","retrieved_from":"https://raw.githubusercontent.com/mdn522/sat-question-bank/main/data/cb-digital-questions.json","attribution":"SAT practice questions © College Board, from the official SAT Suite Question Bank. Retrieved via the community dump at github.com/mdn522/sat-question-bank.","disclaimer":"SAT® and PSAT/NMSQT® are trademarks registered by the College Board, which is not affiliated with, does not sponsor, and does not endorse this project.","license":"College Board copyright; source repository carries no licence. Local use only.","redistributable":false,"item_count":2017}}