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## 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

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$\pi$

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## 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

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

$\Rightarrow d=7.6$

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## 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

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  $(\frac{60\pi}{144\pi} \times 100)\%=(\frac{5}{12} \times 100) \% \approx 42\%$

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## 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

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……..

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

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