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

Difficulty medium nctm standard geometry standard for

This preview shows page 5 - 11 out of 25 pages.

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.
Image of page 5
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.
Image of page 6
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.
Image of page 7
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.
Image of page 8
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.
Image of page 9
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 ?
Image of page 10
Image of page 11

You've reached the end of your free preview.

Want to read all 25 pages?

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

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture