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## Ratio of the area of Square and Pentagon | AMC 8, 2013

Try this beautiful problem from Geometry: Ratio of the area between Square and Pentagon.

## Ratio of the area between Square and Pentagon – AMC-8, 2013 – Problem 24

Squares ABCD ,EFGH and GHIJ are equal in area .Points C and D are the mid points of the side IH and HE ,respectively.what is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares?

• $\frac{1}{4}$
• $\frac{1}{3}$
• $\frac{3}{8}$

### Key Concepts

Geometry

Area of square

Area of Triangle

Answer:$\frac{1}{3}$

AMC-8(2013) Problem 24

Pre College Mathematics

## Try with Hints

extend  IJ until it hits the extension of  AB .

Can you now finish the problem ……….

find the area of the pentagon

can you finish the problem……..

First let L=2 (where L is the side length of the squares) for simplicity. We can extend  IJ until it hits the extension of  AB . Call this point  X.

Then clearly length of AX=3 unit & length of XJ = 4 unit .

Therefore area of $\triangle AXJ= (\frac{1}{2} \times AX \times XJ)=(\frac{1}{2} \times 4 \times 3)=6$ sq.unit

And area of Rectangle BXIC= $( 1 \times 2)$=2 sq.unit

Therefore the of the pentagon ABCIJ=6-2=4 sq.unit

The combined area of three given squares be $(3 \times 2^2)$=12 sq.unit

Now, the ratio of the shaded area(pentagon) to the combined area of the three squares is $\frac{4}{12}=\frac{1}{3}$

.

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## Area of Triangle and Square | AMC 8, 2012 | Problem 25

Try this beautiful problem from AMC-8-2012 (Geometry) based on area of a Triangle and square and congruency.

## Area of a Triangle- AMC 8, 2012 – Problem 25

A square with area 4 is inscribed in a square with area 5,  with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?

• $\frac{1}{5}$
• $\frac{2}{5}$
• $\frac{1}{2}$

### Key Concepts

Geometry

Square

Triangle

Answer:$\frac{1}{2}$

AMC-8, 2014 problem 25

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Find the area of four triangles

Can you now finish the problem ……….

Four triangles are congruent

can you finish the problem……..

Total area of the big square i.e ABCD is 5 sq.unit

and total area of the small square i.e EFGH is 4 sq.unit

So Toal area of the $(\triangle AEH + \triangle BEF + \triangle GCF + \triangle DGH)=(5-4)=1$ sq.unit

Now clearly Four triangles$( i.e \triangle AEH , \triangle BEF , \triangle GCF , \triangle DGH )$ are congruent.

Total area of the four congruent triangles formed by the squares is 5-4=1 sq.unit

So area of the one triangle is $\frac{1}{4}$ sq.unit

Now “a” be the height and “b” be the base of one triangle

The area of one triangle be $(\frac{1}{2} \times base \times height )$=$\frac{1}{4}$

i.e $(\frac{1}{2} \times b \times a)$= $\frac{1}{4}$

i.e $ab$= $\frac{1}{2}$

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## Area of star and circle | AMC-8, 2012|problem 24

Try this beautiful problem from Geometry: Ratio of the area of the star figure to the area of the original circle

## Area of the star and circle – AMC-8, 2012 – Problem 24

A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?

• $\frac{1}{\pi}$
• $\frac{4-\pi}{\pi}$
• $\frac{\pi – 1}{\pi}$

### Key Concepts

Geometry

Circle

Arc

Answer:$\frac{4-\pi}{\pi}$

AMC-8 (2012) Problem 24

Pre College Mathematics

## Try with Hints

Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.

Can you now finish the problem ……….

find the area of the star figure

can you finish the problem……..

Draw a square around the star figure. Then the length of one side of the square be 4(as the diameter of the circle is 4)

Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.

The area of the above circle is $\pi (2)^2 =4\pi$

and the area of the outer square is $(4)^2=16$

Thus, the area of the star figure is $16-4\pi$

Therefore $\frac{(the \quad area \quad of \quad the \quad star \quad figure)}{(the \quad area \quad of \quad the \quad original \quad circle )}=\frac{16-4\pi}{4\pi}$

= $\frac{4-\pi}{\pi}$

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## Hexagon and Triangle |AMC 8- 2015 -|Problem 21

Try this beautiful problem from Geometry based on hexagon and Triangle.

## Area of Triangle | AMC-8, 2015 |Problem 21

In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK$ is equilateral and FE=BC. What is the area of $\triangle KBC$?

• 9
• 12
• 32

### Key Concepts

Geometry

Triangle

hexagon

Answer:$12$

AMC-8, 2015 problem 21

Pre College Mathematics

## Try with Hints

Clearly FE=BC

Can you now finish the problem ……….

$\triangle KBC$ is a Right Triangle

can you finish the problem……..

Clearly ,since FE is a side of square with area 32

Therefore FE=$\sqrt 32$=$4\sqrt2$

Now since FE=BC,We have BC=$4\sqrt2$

Now JB is a side of a square with area 18

so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$

Lastly $\triangle KBC$ is a right triangle ,we see that

$\angle JBA + \angle ABC +\angle CBK +\angle KBJ$ =$360^\circ$

i.e$90^\circ + 120^\circ +\angle CBK + 60^\circ=360^\circ$

i.e $\angle CBK=90^\circ$

So $\triangle KBC$ is a right triangle with legs $3\sqrt 2$ and $4\sqrt2$

Now its area is $3\sqrt2 \times 4\sqrt 2 \times \frac {1}{2}$=$\frac{24}{2}$=12

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## 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

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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## Area of cube’s cross section |Ratio | AMC 8, 2018 – Problem 24

Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

## Area of cube’s cross section – AMC-8, 2018 – Problem 24

In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?

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

### Key Concepts

Geometry

Area

Pythagorean theorem

Answer:$\frac{3}{2}$

AMC-8(2018) Problem 24

Pre College Mathematics

## Try with Hints

EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Area of rhombus is half product of its diagonals….

can you finish the problem……..

Let Side length of a cube be x.

then by the pythagorean  theorem$EC=X \sqrt {3}$

$JI =X \sqrt {2}$

Now the area of the rhombus is half product of its diagonals

therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$

This shows that $R= \frac{\sqrt6}{2}$

i.e$R^2=\frac{3}{2}$

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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?

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

### Key Concepts

Geometry

congruency

similarity

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 =$\frac{10}{3}$