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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 AMC-8, 2004,Problem-25. You may use sequential hints to solve the problem.

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$\pi$

## 2 replies on “Intersection of two Squares | AMC 8, 2004 | Problem 25”

Squares should be 4x4in place of 2×2 otherwise area in problem 25 will be 7-22/14.

KOUSHIK SOMsays:

yes…..

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