Di\ufb03culty Medium NCTM Standard Geometry Standard for Grades 912 explore

# Diﬃculty medium nctm standard geometry standard for

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Diﬃculty: Medium NCTM Standard: Geometry Standard for Grades 9–12: explore relationships (including congruence and similarity) among classes of two- and three-dimensional geometric objects, make and test conjectures about them, and solve problems involving them.
The number n satisfies the equation 80 + 125 = n . What is the value of n ? (A) 205 (B) 305 (C) 405 (D) 500 (E) 1000 2010 AMC 10 B, Problem #62010 AMC 12 B, Problem #4“Note that80 = 45and125 = 55.” Solution Answer (C): Simplifying gives 80 + 125 = 4 5 + 5 5 = 9 5 = 405 . Hence n = 405 . Diﬃculty: Medium-easy NCTM Standard: Number and Operations Standard for Grades 9–12: understand numbers, ways of representing numbers, relationships among numbers, and number systems.
The square of x is 16 larger than the square of y , and x is 2 larger than y . What is the product xy ? (A)4(B)8(C)15(D)24(E)32 2010 AMC 10 B, Problem #7“Note thatx2y2= (xy)(x+y) = 16.” Solution Answer (C): Because x 2 y 2 = ( x y )( x + y ) = 16 , and x y = 2 , we have x + y = 16 2 = 8 . Solving the simultaneous equations x y = 2 and x + y = 8 yields x = 5 and y = 3 . Thus the product xy = 5 · 3 = 15 . Diﬃculty: Medium-easy NCTM Standard: Algebra Standard for Grades 9–12: write equivalent forms of equations, inequalities, and systems of equations and solve them with ﬂuency-mentally or with paper and pencil in simple cases and using technology in all cases.
The area of triangle EBD is one third of the area of 3 4 5 triangle ABC . Segment DE is perpendicular to segment AB . What is BD ? A B C D E 3 4 5 2010 AMC 10 B, Problem #8“The area ofABCis12·3·4 = 6.” Solution Answer (D): The area of ABC is 1 2 · 3 · 4 = 6 , so the area of EBD is 1 3 · 6 = 2 . Note that triangles ABC and EBD are right triangles with an angle in common, so they are similar. Therefore BD and DE are in the ratio 4 to 3. Let BD = x and DE = 3 4 x . Then the area of EBD can be expressed as 1 2 · x · 3 4 x = 3 8 x 2 . Because EBD has area 2, solving yields BD = 4 3 3 . OR Because EBD and ABC are similar triangles, their areas are in the ratio of the squares of their corresponding linear parts. Therefore ( BD 4 ) 2 = 1 3 and BD = 4 3 3 . Diﬃculty: Medium NCTM Standard: Geometry Standard for Grades 9–12: analyze properties and determine attributes of two- and three-dimensional objects.
If a rectangle’s width is increased by 3, the rectangle becomes a square and the area of the rectangle is increased by 54. What is the area of the original rectangle? 2010 AMC 10 B, Problem #9“Letxandybe the width and length of the originalrectangle, respectively. Form an equation to show thedifference in areas.” Solution Answer (E): Let x and y be the width and length of the original rectangle, respectively. Increasing the width by 3 is equivalent to attaching a rectangle with dimensions 3 and y , as shown. The extra area is 54, so 3 y = 54 and y = 18 . Then x = 15 and the area of the original rectangle is 15 · 18 = 270 . x y 3 Diﬃculty: Medium NCTM Standard: Geometry Standard for Grades 9–12: analyze properties and determine attributes of two- and three-dimensional objects.
On the way to the fair, Paul, Sue, and Ellen compare how much money they have. Paul has 20 percent more than Sue and 25 percent more than Ellen. Ellen has x percent less than Sue. What is x ?

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• Winter '13
• Kramer
• Platonic solid, United States of America Mathematical Olympiad