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?
$\frac{1}{4}$
$\frac{1}{3}$
$\frac{3}{8}$
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........
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 \(\frac{4}{12}=\frac{1}{3}\)
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?
$\frac{1}{\pi}$
$\frac{4-\pi}{\pi}$
$\frac{\pi - 1}{\pi}$
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........
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.
The area of the above circle is \(\pi (2)^2 =4\pi\)
and the area of the outer square is \((4)^2=16\)
Thus, the area of the star figure is \(16-4\pi\)
Therefore \(\frac{(the \quad area \quad of \quad the \quad star \quad figure)}{(the \quad area \quad of \quad the \quad original \quad circle )}=\frac{16-4\pi}{4\pi}\)
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 $R^2$ ?
$\frac{5}{4}$
$\frac{3}{2}$
$\frac{4}{3}$
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........
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 $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$
