Categories
AMC 8 Math Olympiad USA Math Olympiad

Area of the Region Problem | AMC-10A, 2007 | Problem 24

Try this beautiful problem from Geometry: Area of the region

Problem on Area of the Region – AMC-10A, 2007- Problem 24


Circle centered at \(A\) and \(B\) each have radius \(2\), as shown. Point \(O\) is the midpoint of \(\overline{AB}\), and \(OA = 2\sqrt {2}\). Segments \(OC\) and \(OD\) are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region \(ECODF\)?

  • \(\pi\)
  • \(7\sqrt 3 -\pi\)
  • \(8\sqrt 2 -4-\pi\)

Key Concepts


Geometry

Triangle

similarity

Check the Answer


Answer: \(8\sqrt 2 -4-\pi\)

AMC-10A (2007) Problem 24

Pre College Mathematics

Try with Hints


Area of the region problem

We have to find out the area of the region \(ECODF\) i.e of gray shaded region.this is not any standard geometrical figure (such as circle,triangle…etc).so we can not find out the value easily.Now if we join \(AC\),\(AE\),\(BD\),\(BF\).Then \(ABFE\) is a rectangle.then we can find out the required area by [ area of rectangle \(ABEF\)- (area of arc \(AEC\)+area of \(\triangle ACO\)+area of \(\triangle BDO\)+ area of arc \(BFD\))]

Shaded region

Can you find out the required area…..?

Given that Circle centered at \(A\) and \(B\) each have radius \(2\) and Point \(O\) is the midpoint of \(\overline{AB}\), and \(OA = 2\sqrt {2}\)

Shaded area of the region

Area of \(ABEF\)=\(2 \times 2 \times 2\sqrt 2\)=\(8\sqrt 2\)

Now \(\triangle{ACO}\) is a right triangle. We know \(AO=2\sqrt{2}\)and \(AC=2\), so \(\triangle{ACO}\) is isosceles, a \(45\)-\(45\) right triangle.\(\overline{CO}\) with length \(2\). The area of \(\triangle{ACO}=\frac{1}{2} \times base \times height=2\). By symmetry, \(\triangle{ACO}\cong\triangle{BDO}\), and so the area of \(\triangle{BDO}\) is also \(2\).now the \(\angle CAO\) = \(\angle DBO\)=\(45^{\circ}\). therefore \(\frac{360}{45}=8\)

So the area of arc \(AEC \) and arc \(BFD\)=\(\frac{1}{8} \times\) area of the circle=\(\frac{\pi 2^2}{8}\)=\(\frac{\pi}{2}\)

can you finish the problem……..

Therefore the required area by [ area of rectangle \(ABEF\)- (area of arc \(AEC\)+area of \(\triangle ACO\)+area of \(\triangle BDO\)+ area of arc \(BFD\))]=\(8\sqrt 2-(\frac{\pi}{2}+2+2+\frac{\pi}{2}\))=\(8\sqrt 2 -4-\pi\)

Subscribe to Cheenta at Youtube


Categories
AMC 8 Math Olympiad USA Math Olympiad

Circular Cylinder Problem | AMC-10A, 2001 | Problem 21

Try this beautiful problem from Geometry based on Circular Cylinder.

Circular Cylinder Problem – AMC-10A, 2001- Problem 21


A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

  • \(\frac{30}{23}\)
  • \(\frac{30}{11}\)
  • \(\frac{15}{11}\)
  • \(\frac{17}{11}\)
  • \(\frac{3}{2}\)

Key Concepts


Geometry

Cylinder

cone

Check the Answer


Answer: \(\frac{30}{11}\)

AMC-10A (2001) Problem 21

Pre College Mathematics

Try with Hints


Circular Cylinder Problem

Given that the diameter equal to its height is inscribed in a right circular cone.Let the diameter and the height of the right circular cone be \(2r\).And also The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide.we have to find out the radius of the cylinder.Now if we can show that \(\triangle AFE \sim \triangle AGC\), then we can find out the value of \(r\)

Can you now finish the problem ……….

Circular cylinder problem

Given that \(Bc=10\),\(AG=12\),\(HL=FG=2r\). Therefore \(AF=12-2r\),\(FE=r\),\(GC=5\)

Now the \(\triangle AFE \sim \triangle AGC\), Can you find out the radius from from this similarity property…….?

can you finish the problem……..

Since \(\triangle AFE \sim \triangle AGC\), we can write \(\frac{AF}{FE}=\frac{AG}{GC}\)

\(\Rightarrow \frac{12-2r}{r}=\frac{12}{5}\)

\(\Rightarrow r=\frac{30}{11}\)

Therefore the radius of the cylinder is \(\frac{30}{11}\)

Subscribe to Cheenta at Youtube


Categories
AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of Triangle Problem | AMC-10A, 2009 | Problem 10

Try this beautiful problem from Geometry: Area of triangle

Area of the Triangle- AMC-10A, 2009- Problem 10


Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\triangle ABC$?

Problem to find the area of triangle
  • \(8\)
  • \(7\sqrt 3\)
  • \(8\sqrt 3\)

Key Concepts


Geometry

Triangle

similarity

Check the Answer


Answer: \(7\sqrt 3\)

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


Right- angled triangle

We have to find out the area of \(\triangle ABC\).now the given that \(BD\) perpendicular on \(AC\).now area of \(\triangle ABC\) =\(\frac{1}{2} \times base \times height\). but we don’t know the value of \(AB\) & \(BC\).

Given \(AC=AD+DC=3+4=7\) and \(BD\) is perpendicular on \(AC\).So if you find out the value of \(BD\) then you can find out the area .can you find out the length of \(BD\)?

Can you now finish the problem ……….

 area of triangle problem

If we proof that \(\triangle ABD \sim \triangle BDC\), then we can find out the value of \(BD\)

Let \(\angle C =x\) \(\Rightarrow DBA=(90-X)\) and \(\angle BAD=(90-x)\),so \(\angle ABD=x\) (as sum of the angles of a triangle is 180)

In Triangle \(\triangle ABD\) & \(\triangle BDC\) we have…

\(\angle BDA=\angle BDC=90\)

\(\angle ABD=\angle BCD=x\)

\(\angle BAD=\angle DBC=(90-x)\)

So we can say that \(\triangle ABD \sim \triangle BDC\)

Therefore \(\frac{BD}{AD}=\frac{CD}{BD}\) \(\Rightarrow (BD)^2=AD .CD \Rightarrow BD=\sqrt{3.4}=2\sqrt 3\)

can you finish the problem……..

Therefore area of the \(\triangle ABC =\frac {1}{2} \times AC \times BD=\frac {1}{2} \times 7 \times 2\sqrt 3=7 \sqrt 3\) sq.unit

Subscribe to Cheenta at Youtube


Categories
AMC 8 Math Olympiad USA Math Olympiad

Measuring the length in Triangle | AMC-10B, 2011 | Problem 9

Try this beautiful problem from Geometry and solve it by measuring the length in triangle.

Measuring the length in Triangle- AMC-10B, 2011- Problem 9


The area of $\triangle$$EBD$ is one third of the area of $\triangle$$ABC$. Segment $DE$ is perpendicular to segment $AB$. What is $BD$?

Measuring the length in Triangle- Problem
  • \(8\sqrt 3\)
  • \(\frac {4\sqrt3}{3}\)
  • \(6\sqrt 3\)

Key Concepts


Geometry

Triangle

similarity

Check the Answer


Answer: \(\frac{ 4\sqrt 3}{3}\)

AMC-10B (2011) Problem 9

Pre College Mathematics

Try with Hints


Find BD

We have to find out the length of \(BD\). The given informations are “The area of $\triangle$$EBD$ is one third of the area of $\triangle$$ABC$. Segment $DE$ is perpendicular to segment $AB$”

If you notice very carefully about the side lengths of the \(\triangle ABC\) then \(AC=3,BC=4,AB=5\) i.e \((AC)^2+(AB)^2=(3)^2+(4)^2=25=(AB)^2\)……..So from the pythagorean theorm we can say that \(\angle ACB=90^{\circ} \)

Therefore area of \(\triangle ACB=\frac{1}{2} \times 3 \times 4=6\)

so area of the \(\triangle BDE=\frac{1}{3} \times 6=2\)

Now the \(\triangle BDE\) and \(\triangle ABC\) If we can show that two triangles are similar then we will get the value of \(BD\).Can you prove \(\triangle BDE \sim \triangle ABC\) ?

Can you now finish the problem ……….

Finding the measurement

In \(\triangle BDE\) & \(\triangle ACB\) we have…..

\(\angle B=X\) \(\Rightarrow \angle BED=(90-x)\) and \(\angle CAB=(90-X)\) (AS \(\angle ACB=90\) & sum of the angles of a triangle is 180)

Therefore \(\triangle BDE \sim \triangle BCD\)

can you finish the problem……..

The value of BD:

Now \(\triangle BDE \sim \triangle BCD\) \(\Rightarrow \frac{(BD)^2}{(BC)^2}=\frac{Area of \triangle BDE}{Area of triangle ACB}\) =\(\frac{(BD)^2}{16}=\frac{2}{6}\)

So \(BD=\frac{ 4\sqrt 3}{3}\)

Subscribe to Cheenta at Youtube


Categories
AMC 8 Geometry Math Olympiad

Area of a Triangle -AMC 8, 2018 – Problem 20

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 $\triangle ABC$?

  • $\frac{2}{3}$
  • $\frac{4}{9}$
  • $\frac{3}{5}$

Key Concepts


Geometry

Area

similarity

Check the Answer


Answer:$\frac{4}{9}$

AMC-8, 2018 problem 20

Pre College Mathematics

Try with Hints


$\triangle ADE$ $\sim$ $\triangle ABC$

Can you now finish the problem ……….

$\triangle BEF$ $\sim$ $\triangle ABC$

can you finish the problem……..

Area of triangle- figure

Since $\triangle ADE$$\sim$ $\triangle ABC$

$\frac{ \text {area of} \triangle ADE}{ \text {area of} \triangle ABC}$=$\frac{AE^2}{AB^2}$

i.e $\frac{\text{area of} \triangle ADE}{\text{area of} \triangle ABC}$ =$\frac{(1)^2}{(3)^2}$=$\frac{1}{9}$

Again $\triangle BEF$ $\sim$ $\triangle ABC$

Therefore $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$=$\frac{BE^2}{AB^2}$

i.e $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$ =$\frac{(2)^2}{(3)^2}$=$\frac{4}{9}$

Therefore Area of quad. CDEF =$\frac {4}{9}$ of area $\triangle ABC$

i.e The ratio of the area of quad.CDEF to the area of $\triangle ABC$ is $\frac{4}{9}$

Subscribe to Cheenta at Youtube


Categories
AMC 8 Geometry Math Olympiad

Radius of a Semi Circle -AMC 8, 2017 – Problem 22

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?

radius of a 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……..

Radius of a semi circle

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….

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$

Subscribe to Cheenta at Youtube