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## 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 $ECODF$?

• $\pi$
• $7\sqrt 3 -\pi$
• $8\sqrt 2 -4-\pi$

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 $BFD$)]=$8\sqrt 2-(\frac{\pi}{2}+2+2+\frac{\pi}{2}$)=$8\sqrt 2 -4-\pi$

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## 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.

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

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 $\frac{30}{11}$

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## 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 $\triangle ABC$?

• $8$
• $7\sqrt 3$
• $8\sqrt 3$

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 $\triangle ABD \sim \triangle BDC$

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

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## 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 $BD$?

• $8\sqrt 3$
• $\frac {4\sqrt3}{3}$
• $6\sqrt 3$

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

So $BD=\frac{ 4\sqrt 3}{3}$

Categories

## 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 $\triangle ABC$?

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

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……..

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 $\frac{4}{9}$

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## 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?

• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$

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….

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

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