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

# 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 $$\frac{30}{11}$$

# 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

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

# Centroids and Area | PRMO 2018 | Question 21

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

Orthocentre

Centroids

Similarity

## Check the Answer

PRMO, 2018, Question 21

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

AB=2DE in triangle $$HG_1G_2$$ and triangle $$HDE$$ $$\frac{AG_1}{HD}=\frac{G_1G_2}{DE}=\frac{2}{3}$$ then $$G_1G_2=\frac{2DE}{3}=\frac{2AB}{3 \times 2}=\frac{AB}{3}$$

Second Hint

triangle $$G_1G_2G_3$$ is similar triangle ABC then $$\frac{AreaatriangleABC}{Area G_1G_2G_3}=\frac{AB^{2}}{G_1G_2^{2}}=9$$

Final Step

then area triangle ABC=$$9 \times area triangle G_1G_2G_3$$=(9)(7)=63.

# 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 $\frac{4}{9}$

# 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 =$\frac{10}{3}$