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## Area of square and circle | AMC 8, 2011|Problem 25

Try this beautiful problem from Geometry based on Ratio of the area of square and circle.

## Area of the star and circle – AMC-8, 2011 – Problem 25

A circle with radius 1 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle’s shaded area to the area between the two squares?

• $\frac{3}{2}$
• $\frac{1}{2}$
• $1$

### Key Concepts

Geometry

Circle

Square

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

AMC-8 (2011) Problem 25

Pre College Mathematics

## Try with Hints

Join the diagonals of the smaller square (i.e GEHF)

Can you now finish the problem ……….

The circle’s shaded area is the area of the smaller square(i.e. GEHF) subtracted from the area of the circle

and The area between the two squares is Area of the square ABCD – Area of the square EFGH

can you finish the problem……..

Given that the Radius of the circle with centre O is 1.Therefore The area of the circle is $\pi (1)^2$=$\pi$ sq.unit

The diameter of the circle is 2 i.e $EF=BC=2$ unit

The area of the big square i.e $ABCD=2^2=4$ sq.unit

$OE=OH=1$ i.e $EH=\sqrt{(1^2+1^2)}=\sqrt 2$

Therefore the area of the smaller square is $(\sqrt 2)^2=2$

The circle’s shaded area is the area of the smaller square(i.e. GEHF) subtracted from the area of the circle =$\pi$ – 2

The area between the two squares is Area of the square ABCD – Area of the square EFGH=4-2=2 sq.unit

The ratio of the circle’s shaded area to the area between the two squares is $\frac{\pi – 2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \frac{1}{2}$

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## Rolling ball Problem | Semicircle |AMC 8- 2013 -|Problem 25

Try this beautiful problem from Geometry based on a rolling ball on a semicircular track.

## A Ball rolling Problem from AMC-8, 2013

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3  semicircular arcs whose radii are $R_1=100$ inches ,$R_2=60$ inches ,and $R_3=80$ inches respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from  A to B?

• $235 \pi$
• $238\pi$
• $240 \pi$

### Key Concepts

Geometry

circumference of a semicircle

Circle

Answer:$238 \pi$

AMC-8, 2013 problem 25

Pre College Mathematics

## Try with Hints

Find the circumference of semicircle….

Can you now finish the problem ……….

Find the total distance by the ball….

can you finish the problem……..

The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses $2\pi \times \frac{2}{2}=2\pi$  inches each, and it gains $2\pi$ inches on B .

So, the departure from the length of the track means that the answer is

$\frac{200+120+160}{2} \times \pi$ + (-2-2+2) $\times \pi$=240$\pi$ -2$\pi$=238$\pi$

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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 =$\frac{10}{3}$

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## Does there exist a Magic Rectangle?

Magic Squares are infamous; so famous that even the number of letters on its Wikipedia Page is more than that of Mathematics itself. People hardly talk about Magic Rectangles.

Ya, Magic Rectangles! Have you heard of it? No, right? Not me either!

So, I set off to discover the math behind it.

#### Does there exist a Magic Rectangle?

First, we have to write the condition mathematically.

Take a table of dimension $m$ x $n$. Now fill in the tables with positive integers so that the sum of the rows, columns, and diagonals are equal. Does there exist such a rectangle?

Let’s start building it from scratch.

Now let’s check something else. Let’s calculate the sum of the elements of the table in two different ways.

Let’s say the column, row and diagonal sum be $S$. There are $m$ rows and $n$ columns.

#### Row – wala Viewpoint

The Rows say the sum of the elements of the table is $S.m$. See the picture below.

#### Column – wala Viewpoint

The Rows say the sum of the elements of the table is $S.n$. See the picture below.

Now, magically it comes that the $S.m = S.n$. Therefore the number of rows and columns must be equal.

Whoa! That was cute!

Visit this post to know about Magic Square more.

Edit 1: Look into the comments for a nice observation that if we allowed integers, and the common sum is 0, then we may not have got the result. Also we need to define the sum of the entries of a diagonal of a rectangle.