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AMC 10 Geometry Math Olympiad USA Math Olympiad

Lengths of Rectangle Problem | AMC-10A, 2009 | Problem 14

Try this beautiful problem from Geometry based on lengths of the rectangle from AMC-10A, 2009. You may use sequential hints to solve the problem.

Try this beautiful problem from geometry based on Lengths of Rectangle Problem.

Lengths of Rectangle Problem – AMC-10A, 2009- Problem 14


Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?

length of rectangle problem
  • \(3\)
  • \(\sqrt 10\)
  • \(2+\sqrt 2\)
  • \(2\sqrt 3\)
  • \(4\)

Key Concepts


Geometry

Square

Rectangle

Check the Answer


Answer: \(3\)

AMC-10A (2009) Problem 14

Pre College Mathematics

Try with Hints


length of rectangle

Given that The area of the outer square is $4$ times that of the inner square.therefore we can say that Therefore the side of the outer square is $\sqrt 4 = 2$ times that of the inner square.Can you find out length of the longer side of each rectangle?

Can you now finish the problem ……….

breaking the figure

Let the side length of the outer square is \(4x\) then the side length of the inner square be \(2x\).Hence the side length of the red region is \(2x\) .As the rectangles are congruent ,therefore side length of green shaded region and the side length of blue shaded region will be x

understanding the figure

Therefore the length of the longer side of each rectangle be \(3x\) and length of the shoter side will be \(x\)

finding the ratio of length of rectangles

Therefore the ratio of the length of the longer side of each rectangle to the length of its shorter side will be \(\frac{3x}{x}=3\)

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