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 \(\sqrt{2^2 +2^2}=2\sqrt 2\)
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
Problem on Area of Circle | SMO, 2010 (Junior) | Problem 29
Try this beautiful problem on area of circle from SMO, Singapore Mathematics Olympiad, 2010.
Problem - Area of Circle (SMO Test)
Let ABCD be a rectangle with AB = 10 . Draw circles \(C_1 \) and \(C_2\) with diameters AB and CD respectively. Let P,Q be the intersection points of \(C_1\) and \(C_2\) . If the circle with diameter PQ is tangent to AB and CD , then what is the area of the shaded region ?
25
20
22
23
Key Concepts
Area of Circle
2D - Geometry
Area of Rectangle
Check the Answer
Answer: 25
Singapore Mathematics Olympiad
Challenges and thrills - Pre - college Mathematics
Try with Hints
If you are really got stuck with this sum then we can start from here:
The diagram will be like this . So let 'N' be the midpoint of CD .
so \(\angle {PNQ} = 90^\circ\)
so PQ = 5 \(\sqrt {2}\)
Now let us try to find the area of the shaded region
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 \((\frac{60\pi}{144\pi} \times 100)\%=(\frac{5}{12} \times 100) \% \approx 42\%\)
