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AMC 10 Geometry Math Olympiad USA Math Olympiad

Side of Square | AMC 10A, 2013 | Problem 3

Try this beautiful problem from Geometry: Side of Square.

Sides of Square – AMC-10A, 2013- Problem 3


Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?

,

 i

Side of Square - Problem
  • $4$
  • $5$
  • $6$
  • $7$
  • \(8\)

Key Concepts


Geometry

Square

Triangle

Check the Answer


Answer: $8$

AMC-10A (2013) Problem 3

Pre College Mathematics

Try with Hints


Side of Square

Given that Square $ABCD$ has side length $10$ and area of $\triangle ABE$ is $40$.we have to find out length of \(BE\) where \(E\) is the point on \(BC\). we know area of the \(\triangle ABE=\frac{1}{2} AB.BE=40\)

Can you find out the side length of \(BE\)?

Can you now finish the problem ……….

Side of Square

\(\triangle ABE=\frac{1}{2} AB.BE=40\)

\(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

\(\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40\)

\(\Rightarrow BE=8\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Problem on Cube | AMC 10A, 2008 | Problem 21

Try this beautiful problem from Geometry: Problem on Cube.

Problem on Cube – AMC-10A, 2008- Problem 21


A cube with side length 1 is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $A B C D ?$

,

 i

  • $\frac{\sqrt{6}}{2}$
  • $\frac{5}{4}$
  • $\sqrt{2}$
  • $\frac{5}{8}$
  • $\frac{3}{4}$

Key Concepts


Geometry

Square

Pythagoras

Check the Answer


Answer: $\frac{\sqrt{6}}{2}$

AMC-10A (2008) Problem 21

Pre College Mathematics

Try with Hints


Problem on Cube - figure

The above diagram is a cube and given that side length $1$ and \(B\) and \(D\) are the mid points .we have to find out area of the \(ABCD\).Now since $A B=A D=C B=C D=\sqrt{\frac{1}{2}^{2}}+1^{2},$ it follows that $A B C D$ is a rhombus. can you find out area of the rhombus?

Can you now finish the problem ……….

Problem on Cube - figure

The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^{2}+1^{2}}=\sqrt{2} \cdot A C$ is a space diagonal, with length $\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$

can you finish the problem……..

Therefore area $A=\frac{1}{2} \times \sqrt{2} \times \sqrt{3}=\frac{\sqrt{6}}{2}$

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Television Problem | AMC 10A, 2008 | Problem 14

Try this beautiful Television Problem from AMC – 10A, 2008.

Television Problem – AMC-10A, 2008- Problem 14


Older television screens have an aspect ratio of 4: 3 . That is, the ratio of the width to the height is 4: 3 . The aspect ratio of many movies is not $4: 3,$ so they are sometimes shown on a television screen by “letterboxing” – darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of 2: 1 and is shown on an older television screen with a 27 -inch diagonal. What is the height, in inches, of each darkened strip?

,

 i

Television Problem
  • $2$
  • $2.25$
  • $2.5$
  • $2.7$
  • $3$

Key Concepts


Geometry

Square

Pythagoras

Check the Answer


Answer: $2.7$

AMC-10A (2008) Problem 14

Pre College Mathematics

Try with Hints


Television Problem

The above diagram is a diagram of Television set whose aspect ratio of $4: 3$.Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$-inch diagonal. Then we have to find the height, in inches, of each darkened strip.

we assume that the width and height of the screen be $4x$ and $3x$ respectively, and let the width and height of the movie be $2y$ and $y$ respectively. If we can find out the value of \(x\) and \(y\) then  the height of each strip can be calculate eassily

Can you now finish the problem ……….

Television Problem

By the Pythagorean Theorem, the diagonal is $\sqrt{(3 x)^{2}+(4 x)^{2}}=5 x=27 .$ So $x=\frac{27}{5}$

Now the movie and the screen have the same width, $2 y=4 x \Rightarrow y=2 x$

can you finish the problem……..

Thus, the height of each strip is $\frac{3 x-y}{2}=\frac{3 x-2 x}{2}=\frac{x}{2}=\frac{27}{10}=2.7$

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AMC-8 Geometry Math Olympiad USA Math Olympiad

The area of trapezoid | AMC 8, 2003 | Problem 21

Try this beautiful problem from Geometry: The area of trapezoid

The area of trapezoid – AMC-8, 2003- Problem 21


The area of trapezoid ABCD is 164 \(cm^2\). The altitude is  8 cm, AB is 10 cm, and CD is 17 cm. What is BC in centimeters?

The area of trapezoid - problem figure

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 i

  • $8$
  • $ 10 $
  • $15$

Key Concepts


Geometry

trapezoid

Triangle

Check the Answer


Answer: $ 10 $

AMC-8 (2003) Problem 21

Pre College Mathematics

Try with Hints


Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Can you now finish the problem ……….

finding the area of trapezoid

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

can you finish the problem……..

finding the area of trapezoid

Given that the area of the trapezoid is 164 sq.unit

Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

Using Pythagorean rules on the triangle ABD,we have…

\((AD)^2 + (BD)^2 =(AB)^2\)

\( \Rightarrow (AD)^2 + (8)^2 =(10)^2\)

\( \Rightarrow (AD)^2 =(10)^2 – (8)^2 \)

\( \Rightarrow (AD)^2 = 36\)

\( \Rightarrow (AD) =6\)

Using Pythagorean rules on the triangle CED,we have…

\((CE)^2 + (DE)^2 =(DC)^2\)

\( \Rightarrow (CE)^2 + (8)^2 =(17)^2\)

\( \Rightarrow (CE)^2 =(17)^2 – (8)^2 \)

\( \Rightarrow (CE)^2 = 225\)

\( \Rightarrow (CE) =15\)

Let BC= DE=x

Therefore area of the trapezoid=\(\frac{1}{2} \times (AD+BC) \times 8\)=164

\(\Rightarrow \frac{1}{2} \times (6+x+15) \times 8\) =164

\(\Rightarrow x=10\)

Therefore BC=10 cm

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AMC-8 Math Olympiad USA Math Olympiad

Largest area Problem | AMC 8, 2003 | Problem 22

Try this beautiful problem from Geometry based Largest area.

Largest area – AMC-8, 2003 – Problem 22


The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

Problem figure
  • $A$
  • $B$
  • $C$

Key Concepts


Geometry

Circle

Square

Check the Answer


Answer:$C$

AMC-8 (2003) Problem 22

Pre College Mathematics

Try with Hints


To find out the largest area at first we have to find out the radius of the circles . all the circles are inscribed ito the squares .now there is a relation between the radius and the side length of the squares….

Can you now finish the problem ……….

area of circle =\(\pi r^2\)

can you finish the problem……..

Largest  area Problem

In A:

Total area of the square =\(2^2=4\)

Now the radius of the inscribed be 1(as the diameter of circle = side length of the side =2)

Area of the inscribed circle is \(\pi (1)^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In B:

Largest  area Problem - figure

Total area of the square =\(2^2=4\)

There are 4 circle and radius of one circle be \(\frac{1}{2}\)

Total area pf 4 circles be \(4 \times \pi \times (\frac{1}{2})^2=\pi\)

Therefore the shaded area =\(4- \pi\)

In C:

finding the largest area

Total area of the square =\(2^2=4\)

Now the length of the diameter = length of the diagonal of the square=2

Therefore radius of the circle=\(\pi\) and lengthe of the side of the square=\(\sqrt 2\)

Thertefore area of the shaded region=Area of the square-Area of the circle=\(\pi (1)^2-(\sqrt 2)^2\)=\(\pi – 2\)

Therefore the answer is  C

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of the Octagon | AMC-10A, 2005 | Problem 20

Try this beautiful problem from Geometry based on the Area of the Octagon from AMC 10A, 2005, Problem 2005

Area of the octagon – AMC-10A, 2005- Problem 20


An equiangular octagon has four sides of length 1 and four sides of length \(\frac{\sqrt{2}}{2}\), arranged so that no two consecutive sides have the same length. What is the area of the octagon?

  • \(\frac{4+5\sqrt 2}{2}\)
  • \(\frac{7}{2}\)
  • \(7\)

Key Concepts


Geometry

Triangle

Octagon

Check the Answer


Answer: \(\frac{7}{2}\)

AMC-10A (2005) Problem 20

Pre College Mathematics

Try with Hints


Octagon figure

We have to find out the equiangular octagon whose four sides of length 1 and four sides of length \(\frac{\sqrt{2}}{2}\),

we join \(AD\),\(HE\),\(BG\) and \(CF\).We assume that side lengths of \(AB=CD=EF=GH=1\) and side lengths of \(AH=BC=DE=GF=\frac{\sqrt{2}}{2}\)( As no two consecutive sides have the same length). Now

Can you now finish the problem ……….

finding the Area of Octagon


There are 5 squares with side lengths \(\frac{\sqrt{2}}{2}\) and 4 Triangles of side lengths \(1\)

Now area of \(5\) squares=\( 5 \times (\frac{\sqrt{2}}{2})^2\)=\(\frac{5}{2}\) and area of each Triangle is half of the area of a square.so the area of \(4\) Triangles=\(4 \times \frac{1}{2} \times \frac{1}{2}\)=\(1\)

can you finish the problem……..

Therefore the Total area of the required octagon=Total area of Five squares + Total areas of Four Triangles=\(\frac{5}{2} +1\)=\(\frac{7}{2}\)

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AMC 10 Geometry Math Olympiad USA Math Olympiad

Rectangle Pattern | AMC-10A, 2016 | Problem 10

Try this beautiful problem from Geometry based on Rectangle Pattern from AMC 10A, 2016, Problem 10.

Rectangle Pattern- AMC-10A, 2016- Problem 10


A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?

Rectangle Pattern Problem
  • \(1\)
  • \(2\)
  • \(4\)
  • \(6\)
  • \(8\)

Key Concepts


Geometry

Rectangle

square

Check the Answer


Answer: \(2\)

AMC-10A (2016) Problem 10

Pre College Mathematics

Try with Hints


Rectangle Pattern Problem figure

Given that length of the inner rectangle be $x$. Therefore the area of that rectangle is $x \cdot 1=x$
The second largest rectangle has dimensions of $x+2$ and 3 , Therefore area $3 x+6$. Now area of the second shaded rectangle= $3 x+6-x=2 x+6$

can you finish the problem……..

Rectangle Pattern Problem figure

Now the dimension of the largest rectangle is $x+4$ and 5 , and the area= $5 x+20$. The area of the largest shaded region is the largest rectangle- the second largest rectangle, which is $(5 x+20)-(3 x+6)=2 x+14$

can you finish the problem……..

Now The problem states that $x, 2 x+6,2 x+14$ is an arithmetic progression,i.e the common difference will be same . So we can say $(2 x+6)-(x)=(2 x+14)-(2 x+6) \Longrightarrow x+6=8 \Longrightarrow x=2$

Therefore the side length =\(2\)

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AMC 10 Math Olympiad USA Math Olympiad

Area of the Inner Square | AMC-10A, 2005 | Problem 8

Try this beautiful problem from Geometry: Area of the inner square

Area of Inner Square – AMC-10A, 2005- Problem 8


In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$. What is the area of the inner square $EFGH$?

Area of the Inner Square - Problem Figure
  • \(25\)
  • \(32\)
  • \(36\)
  • \(42\)
  • \(40\)

Key Concepts


Geometry

Square

similarity

Check the Answer


Answer: \(36\)

AMC-10A (2005) Problem 8

Pre College Mathematics

Try with Hints


Area of the Inner Square - Shaded Figure

We have to find out the area of the region \(EFGH\) Which is a square shape .so if we can find out one of it’s side length then we can easily find out the area of \(EFGH\). Now given that \(BE=1\) i.e \(BE=CF=DG=AH=1\) and side length of the square \(ABCD=\sqrt {50}\).Therefore \((AB)^2=(\sqrt {50})^2=50\).so using this information can you find out the length of \(EH\)?

Can you find out the required area…..?

Explanatory Shading of the figure

Since \(EFGH\) is a square,therefore \(ABH\) is a Right -angle Triangle.

Therefore,\((AH)^2+(BH)^2=(AB)^2\)

\(\Rightarrow (AH)^2+(HE+EB)^2=(AB)^2\)

\(\Rightarrow (1)^2+(HE+1)^2=50\)

\(\Rightarrow (HE+1)^2=49\)

\(\Rightarrow (HE+1)=7\)

\(\Rightarrow HE=6\)

Therefore area of the inner square (red shaded region) =\({6}^2=36\)

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AMC 10 Math Olympiad USA Math Olympiad

Pentagon & Square Pattern | AMC-10A, 2001 | Problem 18

Try this beautiful problem from Geometry based on Pentagon and square Pattern.

Pentagon & Square Pattern – AMC-10A, 2001- Problem 18


The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to

pentagon and square pattern
  • \(50\)
  • \(58\)
  • \(60\)
  • \(56\)
  • \(64\)

Key Concepts


Geometry

Pentagon

square

Check the Answer


Answer: \(56\)

AMC-10A (2001) Problem 18

Pre College Mathematics

Try with Hints


pentagon and square pattern

The given square is the above square.we have to find out The percent of the plane that is enclosed by the pentagons.Notice that there are \(9\) tiles in the square box.so if we can find out the area of pentagon and small square in single tile,then we can find out the total area of the pentagon in the total big square……

can you finish the problem……..

Shaded pattern

Now consider a single tile from the big square,Let us take the side of the small square is $a$.There are four squares which is in the area \(4a^2\) and there are five pentagons which are in areas \(5a^2\).Then the area of the single tile is $9a^2$

Therefore we can say that exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane also

can you finish the problem……..

pattern

Therefore for the whole square, expressed as a percentage,it becomes $55.\overline{5}\%$, and the closest integer to this value is \(56\)

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AMC 10 Math Olympiad USA Math Olympiad

Area of Hexagon Problem | AMC-10A, 2014 | Problem 13

Try this beautiful problem from Geometry based on Area of Hexagon Problem

Area of Hexagon Problem – AMC-10A, 2014- Problem 13


Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?

area of hexagon

  • \(3+{\sqrt 5}\)
  • \(4+{\sqrt 3}\)
  • \(3+{\sqrt 3}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{9}\)

Key Concepts


Geometry

Triangle

square

Check the Answer


Answer: \(3+{\sqrt 3}\)

AMC-10A (2014) Problem 13

Pre College Mathematics

Try with Hints


shaded hexagon

Given that \(\triangle ABC\) is an Equilateral Triangle with side length \(1\) and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle.Now we have to find out the area of hexagon $DEFGHI$.Now area of the the Hexagon $DEFGHI$=Area of \(\triangle ABC\)+Area of three squares( i.e $ABDE$, $BCHI$, $CAFG$)+Area of Triangles(( \(\triangle AEF,\triangle DBI,\triangle HCG\))

Since the side length of Equilateral Triangle \(\triangle ABC\) is given then we can find out the area of the \(\triangle ABC\) and area of the squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle(as side length of one square =side length of the equilateral \(\triangle ABC\).Now we have to find out the area of other three Triangles( \(\triangle AEF,\triangle DBI,\triangle HCG\))

can you finish the problem……..

Area of the \(\triangle ABC\)(Red shaded Region)=\(\frac{\sqrt 3}{4}\) (as side lengtjh is 1)

Area of 3 squares =\(3\times {1}^2=3\)

solution figure of hexagon problem

Now we have to find out the area of the \(\triangle GCH\).At first draw a perpendicular \(CL\) on \(HG\). As \(\triangle GCH\) is an isosceles triangle (as \(HC=CG=1\)),Therefore \(HL=GL\)

Now in the \(\triangle CGL\),

\(\angle GCL=60^{\circ}\) (as \(\angle GCH=360^{\circ}-\angle ACB -\angle ACG-\angle BCH \) \(\Rightarrow \angle GCH=360^{\circ}-60^{\circ}-90^{\circ}-90^{\circ}=120^{\circ}\))

So \(\angle GCL=60^{\circ}\)

So \(\angle CGL=30^{\circ}\)

\(\frac{CL}{CG}\)=Sin \(30^{\circ}\)

\(\Rightarrow CL=\frac{1}{2}\) (as CG=1)

And ,

\(\frac{GL}{CG}\)=Sin \(60^{\circ}\)

\(\Rightarrow GL=\frac{\sqrt 3}{2}\) (as CG=1)

So \(GH=\sqrt 3\)

Therefore area of the \(\triangle CGH=\frac{1}{2}\times \sqrt 3 \times{1}{2}=\frac{\sqrt 3}{4}\)

Therefore area of three Triangles ( \(\triangle AEF,\triangle DBI,\triangle HCG\))=\(3\times \frac{\sqrt 3}{4}\)

can you finish the problem……..

Shaded area of hexagon

Therefore area of the the Hexagon $DEFGHI$=Area of \(\triangle ABC\)+Area of three squares( i.e $ABDE$, $BCHI$, $CAFG$)+Area of Triangles(( \(\triangle AEF,\triangle DBI,\triangle HCG\))=(\(\frac{\sqrt 3}{4}+3+3\times \frac{\sqrt 3}{4}\))=\(3+{\sqrt 3}\)

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