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
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 \(BFD\))]
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 \(OA = 2\sqrt {2}\)
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 \(BFD\))]=\(8\sqrt 2-(\frac{\pi}{2}+2+2+\frac{\pi}{2}\))=\(8\sqrt 2 -4-\pi\)
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
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 ..........
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 \(\frac{30}{11}\)
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 $\triangle ABC$?
\(8\)
\(7\sqrt 3\)
\(8\sqrt 3\)
Key Concepts
Geometry
Triangle
similarity
Check the Answer
Answer: \(7\sqrt 3\)
AMC-10A (2009) Problem 10
Pre College Mathematics
Try with Hints
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 ..........
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 \(\triangle ABD \sim \triangle BDC\)
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 $BD$?
\(8\sqrt 3\)
\(\frac {4\sqrt3}{3}\)
\(6\sqrt 3\)
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
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 ..........
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 \(\triangle BDE \sim \triangle BCD\) \(\Rightarrow \frac{(BD)^2}{(BC)^2}=\frac{Area of \triangle BDE}{Area of triangle ACB}\) =\(\frac{(BD)^2}{16}=\frac{2}{6}\)
Try this beautiful problem from the PRMO, 2018 based on Centroids and Area.
Centroids and Area - PRMO 2018
Let ABC be an acute angled triangle and let H be its orthocentre. Let \(G_1\),\(G_2\) and \(G_3\) be the centroids of the triangles HBC, HCA, HAB. If area of triangle \(G_1G_2G_3\) =7 units, find area of triangle ABC.
is 107
is 63
is 840
cannot be determined from the given information
Key Concepts
Orthocentre
Centroids
Similarity
Check the Answer
Answer: is 63.
PRMO, 2018, Question 21
Geometry Vol I to IV by Hall and Stevens
Try with Hints
AB=2DE in triangle \(HG_1G_2\) and triangle \(HDE\) \(\frac{AG_1}{HD}=\frac{G_1G_2}{DE}=\frac{2}{3}\) then \(G_1G_2=\frac{2DE}{3}=\frac{2AB}{3 \times 2}=\frac{AB}{3}\)
triangle \(G_1G_2G_3\) is similar triangle ABC then \(\frac{AreaatriangleABC}{Area G_1G_2G_3}=\frac{AB^{2}}{G_1G_2^{2}}=9\)
then area triangle ABC=\(9 \times area triangle G_1G_2G_3\)=(9)(7)=63.
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 $\triangle ABC$?
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....
