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 \(\frac{4}{12}=\frac{1}{3}\)
Area of Triangle and Square | AMC 8, 2012 | Problem 25
Try this beautiful problem from AMC-8-2012 (Geometry) based on area of a Triangle and square and congruency.
Area of a Triangle- AMC 8, 2012 - Problem 25
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?
\(\frac{1}{5}\)
\(\frac{2}{5}\)
\(\frac{1}{2}\)
Key Concepts
Geometry
Square
Triangle
Check the Answer
Answer:\(\frac{1}{2}\)
AMC-8, 2014 problem 25
Challenges and Thrills of Pre College Mathematics
Try with Hints
Find the area of four triangles
Can you now finish the problem ..........
Four triangles are congruent
can you finish the problem........
Total area of the big square i.e ABCD is 5 sq.unit
and total area of the small square i.e EFGH is 4 sq.unit
So Toal area of the \((\triangle AEH + \triangle BEF + \triangle GCF + \triangle DGH)\)=\((5-4)=1\) sq.unit
Now clearly Four triangles\(( i.e \triangle AEH , \triangle BEF , \triangle GCF , \triangle DGH )\) are congruent.
Total area of the four congruent triangles formed by the squares is 5-4=1 sq.unit
So area of the one triangle is \(\frac{1}{4}\) sq.unit
Now "a" be the height and "b" be the base of one triangle
The area of one triangle be \((\frac{1}{2} \times base \times height )\)=\(\frac{1}{4}\)
i.e \((\frac{1}{2} \times b \times a)\)= \(\frac{1}{4}\)
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 \(\frac{(the \quad area \quad of \quad the \quad star \quad figure)}{(the \quad area \quad of \quad the \quad original \quad circle )}=\frac{16-4\pi}{4\pi}\)
Try this beautiful problem from Geometry based on hexagon and Triangle.
Area of Triangle | AMC-8, 2015 |Problem 21
In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK $ is equilateral and FE=BC. What is the area of $\triangle KBC$?
9
12
32
Key Concepts
Geometry
Triangle
hexagon
Check the Answer
Answer:$12$
AMC-8, 2015 problem 21
Pre College Mathematics
Try with Hints
Clearly FE=BC
Can you now finish the problem ..........
$\triangle KBC$ is a Right Triangle
can you finish the problem........
Clearly ,since FE is a side of square with area 32
Therefore FE=$\sqrt 32$=$4\sqrt2$
Now since FE=BC,We have BC=$4\sqrt2$
Now JB is a side of a square with area 18
so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$
Lastly $\triangle KBC$ is a right triangle ,we see that
Try this beautiful problem from Geometry based on Area of a Triangle Using similarity
Area of Triangle - AMC-8, 2018 - Problem 20
In $\triangle ABC $ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.
What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?
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 $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$
Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.
AMC-8(2017) - Geometry (Problem 22)
In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?
$\frac{7}{6}$
$\frac{10}{3}$
$\frac{9}{8}$
Key Concepts
Geometry
congruency
similarity
Check the Answer
Answer:$\frac{10}{3}$
AMC-8(2017)
Pre College Mathematics
Try with Hints
Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.
Can you now finish the problem ..........
Now the $\triangle ODB $and $\triangle OCB$ are congruent
can you finish the problem........
Let x be the radius of the semi circle
Now the $\triangle ODB$ and $\triangle OCB$ we have
OD=OC
OB=OB
$\angle ODB$=$\angle OCB$= 90 degree`
so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)
BD=BC=5
And also $\triangle ODA$ and $\triangle BCA$ are similar....
