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AMC 8 Geometry Math Olympiad USA Math Olympiad

Ratio of the area of Square and Pentagon | AMC 8, 2013

Try this beautiful problem from Geometry: Ratio of the area between Square and Pentagon.

Ratio of the area between Square and Pentagon – AMC-8, 2013 – Problem 24


Squares ABCD ,EFGH and GHIJ are equal in area .Points C and D are the mid points of the side IH and HE ,respectively.what is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares?

ratio of area of square and pentagon

  • $\frac{1}{4}$
  • $\frac{1}{3}$
  • $\frac{3}{8}$

Key Concepts


Geometry

Area of square

Area of Triangle

Check the Answer


Answer:$\frac{1}{3}$

AMC-8(2013) Problem 24

Pre College Mathematics

Try with Hints


extend  IJ until it hits the extension of  AB .

Can you now finish the problem ……….

find the area of the pentagon

can you finish the problem……..

solution figure

First let L=2 (where L is the side length of the squares) for simplicity. We can extend  IJ until it hits the extension of  AB . Call this point  X.

Then clearly length of AX=3 unit & length of XJ = 4 unit .

Therefore area of \(\triangle AXJ= (\frac{1}{2} \times AX \times XJ)=(\frac{1}{2} \times 4 \times 3)=6\) sq.unit

And area of Rectangle BXIC= \(( 1 \times 2)\)=2 sq.unit

Therefore the of the pentagon ABCIJ=6-2=4 sq.unit

The combined area of three given squares be \( (3 \times 2^2)\)=12 sq.unit

Now, the ratio of the shaded area(pentagon) to the combined area of the three squares is \(\frac{4}{12}=\frac{1}{3}\)

.

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of star and circle | AMC-8, 2012|problem 24

Try this beautiful problem from Geometry: Ratio of the area of the star figure to the area of the original circle

Area of the star and circle – AMC-8, 2012 – Problem 24


A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?

Area of star and circle
  • $\frac{1}{\pi}$
  • $\frac{4-\pi}{\pi}$
  • $\frac{\pi – 1}{\pi}$

Key Concepts


Geometry

Circle

Arc

Check the Answer


Answer:$\frac{4-\pi}{\pi}$

AMC-8 (2012) Problem 24

Pre College Mathematics

Try with Hints


Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.

Can you now finish the problem ……….

find the area of the star figure

can you finish the problem……..

Star and circle

Draw a square around the star figure. Then the length of one side of the square be 4(as the diameter of the circle is 4)

Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.

circle

The area of the above circle is \(\pi (2)^2 =4\pi\)

circle and square

and the area of the outer square is \((4)^2=16\)

star

Thus, the area of the star figure is \(16-4\pi\)

Therefore \(\frac{(the \quad area \quad of \quad the \quad star \quad figure)}{(the \quad area \quad of \quad the \quad original \quad circle )}=\frac{16-4\pi}{4\pi}\)

= \(\frac{4-\pi}{\pi}\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of cube’s cross section |Ratio | AMC 8, 2018 – Problem 24

Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

Area of cube’s cross section – AMC-8, 2018 – Problem 24


In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?

area of cube's cross section

  • $\frac{5}{4}$
  • $\frac{3}{2}$
  • $\frac{4}{3}$

Key Concepts


Geometry

Area

Pythagorean theorem

Check the Answer


Answer:$\frac{3}{2}$

AMC-8(2018) Problem 24

Pre College Mathematics

Try with Hints


EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Area of rhombus is half product of its diagonals….

can you finish the problem……..

area of cube's cross section

Let Side length of a cube be x.

then by the pythagorean  theorem$ EC=X \sqrt {3}$

$JI =X \sqrt {2}$

Now the area of the rhombus is half product of its diagonals

therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$

This shows that $R= \frac{\sqrt6}{2}$

i.e$ R^2=\frac{3}{2}$

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