Intersection of two Squares | AMC 8, 2004 | Problem 25

Try this beautiful problem from Geometry based on Intersection of two Squares.

When 2 Squares intersect | AMC-8, 2004 | Problem 25

Two $4\times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

$28-2\pi$
$25-2\pi$
$30-2\pi$

Key Concepts

Geometry

square

Circle

Check the Answer

Answer: $28-2\pi$

AMC-8, 2004 problem 25

Pre College Mathematics

Try with Hints

Area of the square is $\pi (r)^2$,where $r$=radius of the circle

Can you now finish the problem ..........

Clearly, if 2 squares intersect, it would be a smaller square with half the side length, 2.

can you finish the problem........

Clearly the intersection of 2 squares would be a smaller square with half the side length, 2.

The area of this region =Total area of larger two squares - the area of the intersection, the smaller square i.e $4^2+4^2 -2^2=28 $

Now The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle

Using the Pythagorean th. diameter of the circle be $\sqrt{2^2 +2^2}=2\sqrt 2$

Radius=$\sqrt 2$

area of the square=$\pi (\sqrt2)^2$=$2\pi$

Area of the shaded region= 28-2$\pi$

Area of Rectangle Problem | AMC 8, 2004 | Problem 24

Try this beautiful problem from Geometry from AMC-8, 2004 ,Problem-24, based on area of Rectangle.

Rectangle | AMC-8, 2004 | Problem 24

In the figure ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to HE and FG?

$7.1$
$7.6$
$7.8$

Key Concepts

Geometry

Rectangle

Parallelogram

Check the Answer

Answer:$7.6$

AMC-8, 2004 problem 24

Pre College Mathematics

Try with Hints

Find Area of the Rectangle and area of the Triangles i.e $(\triangle AHE ,\triangle EBF , \triangle FCG , \triangle DHG) $

Can you now finish the problem ..........

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of$(\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) $

can you finish the problem........

Area of the Rectangle =$CD \times AD $=$10 \times 8$=80 sq.unit

Area of the $\triangle AHE$ =$\frac{1}{2} \times AH \times AE $= $\frac{1}{2} \times 4 \times 3$ =6 sq.unit

Area of the $\triangle EBF$ =$\frac{1}{2} \times EB \times BE $= $\frac{1}{2} \times 6 \times 5$ =15 sq.unit

Area of the $\triangle FCG$ =$\frac{1}{2} \times GC \times FC$= $\frac{1}{2} \times 4\times 3$ =6 sq.unit

Area of the $\triangle DHG$ =$\frac{1}{2} \times DG \times DH $= $\frac{1}{2} \times 6 \times 5$ =15 sq.unit

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of$(\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) $=$80-(6+15+6+15)=80-42=38$

As ABCD is a Rectangle ,$\triangle GCF$ is a Right-angle triangle,

Therefore GF=$\sqrt{4^2 + 3^2}$=5 sq.unit

Now Area of the parallelogram EFGH=$ GF \times d$=38

$\Rightarrow 5 \times d$=38

$\Rightarrow d=7.6$

Radius of the Circle | AMC-8, 2005 | Problem 25

Try this beautiful problem from Geometry: Radius of a circle

Radius of a circle - AMC-8, 2005- Problem 25

A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

$\frac{5}{\sqrt \pi}$
$ \frac{2}{\sqrt \pi} $
$\sqrt \pi$

Key Concepts

Geometry

Cube

square

Check the Answer

Answer: $ \frac{2}{\sqrt \pi} $

AMC-8 (2005) Problem 25

Pre College Mathematics

Try with Hints

The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Can you now finish the problem ..........

Region within the circle and square be $x$ i.e In other words, it is the area inside the circle and the square

The area of the circle -x=Area of the square - x

can you finish the problem........

Given that The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Let the region within the circle and square be $x$ i.e In other words, it is the area inside the circle and the square .

Let r be the radius of the circle

Therefore, The area of the circle -x=Area of the square - x

so, $\pi r^2 - x=4-x$

$\Rightarrow \pi r^2=4$

$\Rightarrow r^2 = \frac{4}{\pi}$

$\Rightarrow r=\frac{2}{\sqrt \pi}$

Area of Circle Problem | AMC 8, 2008 | Problem 25

Try this beautiful problem from Geometry based on the Area of a Circle.

Area of Circle | AMC-8, 2008 | Problem 25

Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?

$44$
$42$
$45$

Key Concepts

Geometry

Area

Circle

Check the Answer

Answer:$42$

AMC-8, 2008 problem 25

Pre College Mathematics

Try with Hints

Area of the square is $\pi (r)^2$,where $r$=radius of the circle

Can you now finish the problem ..........

Find the total area of the black region........

can you finish the problem........

Given that The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches .

The radius of the 1st circle is 2, So the area is $\pi(2)^2$=4$\pi$ sq.unit

The radius of the 2nd circle is 4, So the area is $\pi(4)^2$=16$\pi$ sq.unit

The radius of the 3rd circle is 6 So the area is $\pi(6)^2$=36$\pi$ sq.unit

The radius of the 4th circle is 8, So the area is $\pi(8)^2$=64$\pi$ sq.unit

The radius of the 5th circle is 10, So the area is $\pi(10)^2$=100$\pi$ sq.unit

The radius of the 6th circle is 12, So the area is $\pi(12)^2$=144$\pi$ sq.unit

Therefore The entire circle's area is 144$\pi$

The area of the black regions is $(100\pi-64\pi)+(36\pi-16\pi)+4\pi=60\pi $sq.unit

The percentage of the design that is black is $(\frac{60\pi}{144\pi} \times 100)\%=(\frac{5}{12} \times 100) \% \approx 42\%$

Circles and semi-circles| AMC 8, 2010|Problem 23

Try this beautiful problem from Geometry based on Ratio of the area of circle and semi-circles.

Area of circles and semi-circles - AMC-8, 2010 - Problem 23

Semicircles POQ and ROS pass through the center O. What is the ratio of the combined areas of the two semicircles to the area of circle O?

$\frac{1}{2}$
$\frac{2}{\pi}$
$ \frac{3}{2} $

Key Concepts

Geometry

Circle

co-ordinate geometry

Check the Answer

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

AMC-8 (2010) Problem 23

Pre College Mathematics

Try with Hints

Find the radius of the circle

Can you now finish the problem ..........

Join O and Q

can you finish the problem........

ratio of the areas of circles and semi-circles

The co-ordinate of Q is (1,1), So OB=1 and BQ=1

By the Pythagorean Theorem, the radius of the larger circle i.e OQ=$\sqrt{1^2+1^2}$=$\sqrt 2$.

Therefore the area of the larger circle be $\pi (\sqrt 2)^2=2\pi$

Now for the semicircles, radius OB=OC=1(as co-ordinate of P=(1,1) and S=(1,-1))

So, the area of the two semicircles is $2\times\frac{\pi(1)^2}{2}=\pi$

Finally, the ratio of the combined areas of the two semicircles to the area of circle O is

$\frac{\pi}{2\pi}$=$\frac{1}{2}$

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

Check the Answer

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}$

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

Check the Answer

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$\pi$

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

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 =$\frac{10}{3}$

Radius of a Circle - SMO 2013 - Problem 25

Try this beautiful problem from Geometry based on the radius and tangent of a circle.

SMO 2013 - Geometry (Problem 25)

As shown in the figure below ,circles $C_1 $and$ C_2$ of radius 360 are tangent to each other , and both tangent to the straight line l.if the circle$ C_3$ is tangent to $C_1$ ,$C_2$ and l ,and circle$ C_4 $is tangent to$ C_1$,$C_3$ and l ,find the radius of$ C_4$

Key Concepts

Geometry

Pythagoras theorm

Distance Formula

Check the Answer

Answer:40

SMO -Math Olympiad-2013

Pre College Mathematics

Try with Hints

Let R be the radius of $C_3$

$C_2E$ =360-R

$C_3E=360$

$C_2C_3$=360+R

Using pythagoras theorm ....

$ (360-R)^2+360^2=(360+R)^2$

i.e R=90

Can you now finish the problem ..........

Let the radius of$ C_4$ be r

then use the distacce formula and tangent property........

can you finish the problem........

Let r be the radius of $C_4$ (small triangle).

LO+OC=360

$\sqrt{(360+p)^2-(360-p)^2}+\sqrt{(90+r)^2-(90-r)^2}=360$

i.e r=40.

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 $latex m $ x $latex 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 $latex S $. There are $latex m $ rows and $latex n $ columns.

Row - wala Viewpoint

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

Column - wala Viewpoint

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

Now, magically it comes that the $latex 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.

When 2 Squares intersect | AMC-8, 2004 | Problem 25

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Rectangle | AMC-8, 2004 | Problem 24

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Radius of a circle - AMC-8, 2005- Problem 25

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Area of Circle | AMC-8, 2008 | Problem 25

Key Concepts

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Area of circles and semi-circles - AMC-8, 2010 - Problem 23

Key Concepts

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

Key Concepts

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A Ball rolling Problem from AMC-8, 2013

Key Concepts

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AMC-8(2017) - Geometry (Problem 22)

Key Concepts

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SMO 2013 - Geometry (Problem 25)

Key Concepts

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Does there exist a Magic Rectangle?

Row - wala Viewpoint

Column - wala Viewpoint