Area of Triangle and Square | AMC 8, 2012 | Problem 25
Try this beautiful problem from AMC-8-2012 (Geometry) based on area of a Triangle and square and congruency.
Area of a Triangle- AMC 8, 2012 - Problem 25
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?
\(\frac{1}{5}\)
\(\frac{2}{5}\)
\(\frac{1}{2}\)
Key Concepts
Geometry
Square
Triangle
Check the Answer
Answer:\(\frac{1}{2}\)
AMC-8, 2014 problem 25
Challenges and Thrills of Pre College Mathematics
Try with Hints
Find the area of four triangles
Can you now finish the problem ..........
Four triangles are congruent
can you finish the problem........
Total area of the big square i.e ABCD is 5 sq.unit
and total area of the small square i.e EFGH is 4 sq.unit
So Toal area of the \((\triangle AEH + \triangle BEF + \triangle GCF + \triangle DGH)\)=\((5-4)=1\) sq.unit
Now clearly Four triangles\(( i.e \triangle AEH , \triangle BEF , \triangle GCF , \triangle DGH )\) are congruent.
Total area of the four congruent triangles formed by the squares is 5-4=1 sq.unit
So area of the one triangle is \(\frac{1}{4}\) sq.unit
Now "a" be the height and "b" be the base of one triangle
The area of one triangle be \((\frac{1}{2} \times base \times height )\)=\(\frac{1}{4}\)
i.e \((\frac{1}{2} \times b \times a)\)= \(\frac{1}{4}\)
Circumference of a Semicircle | AMC 8, 2014 | Problem 25
Try this beautiful problem from AMC-8-2014 (Geometry) based on Circumference of a Semicircle
Circumference of a Semicircle- AMC 8, 2014 - Problem 25
On A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5miles per hour, how many hours will it take to cover the one-mile stretch?
\(\frac{\pi}{11}\)
\(\frac{\pi}{10}\)
\(\frac{\pi}{5}\)
Key Concepts
Geometry
Semicircle
Distance
Check the Answer
Answer:\(\frac{\pi}{10}\)
AMC-8, 2014 problem 25
Challenges and Thrills of Pre College Mathematics
Try with Hints
Find the circumference of a semi-circle
Can you now finish the problem ..........
If Robert rides in a straight line, it will take him \(\frac {1}{5}\) hours
can you finish the problem........
If Robert rides in a straight line, it will take him \(\frac {1}{5}\) hours. When riding in semicircles, let the radius of the semicircle r, then the circumference of a semicircle is \({\pi r}\). The ratio of the circumference of the semicircle to its diameter is \(\frac {\pi}{2}\). so the time Robert takes is \(\frac{1}{5} \times \frac{\pi}{2}\). which is equal to \(\frac{\pi}{10}\)
Try this beautiful problem from AMC-8-2015 (Geometry) based on area of square.
Area of a square - AMC 8, 2015 - Problem 25
One-inch Squares are cut from the corners of this 5 inch square.what is the area in square inches of the largest square that can be fitted into the remaining space?
9
15
17
Key Concepts
Geometry
Area
Square
Check the Answer
Answer:15
AMC-8, 2015 problem 25
Challenges and Thrills of Pre College Mathematics
Try with Hints
Find the Length of HG......
Can you now finish the problem ..........
Draw the big square in the remaining space of the big sqare and find it's area .......
can you finish the problem........
We want to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base 3 and height 1 . so the combined area of the four triangles is $ 4 \times \frac {3}{2} $=6.
