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 \((\frac{1}{2} \times b \times a)\)= \(\frac{1}{4}\)
Try this beautiful problem from Geometry: Ratio of the area of the star figure to the area of the original circle
Area of the star and circle - AMC-8, 2012 - Problem 24
A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?
$\frac{1}{\pi}$
$\frac{4-\pi}{\pi}$
$\frac{\pi - 1}{\pi}$
Key Concepts
Geometry
Circle
Arc
Check the Answer
Answer:$\frac{4-\pi}{\pi}$
AMC-8 (2012) Problem 24
Pre College Mathematics
Try with Hints
Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.
Can you now finish the problem ..........
find the area of the star figure
can you finish the problem........
Draw a square around the star figure. Then the length of one side of the square be 4(as the diameter of the circle is 4)
Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.
The area of the above circle is \(\pi (2)^2 =4\pi\)
and the area of the outer square is \((4)^2=16\)
Thus, the area of the star figure is \(16-4\pi\)
Therefore \(\frac{(the \quad area \quad of \quad the \quad star \quad figure)}{(the \quad area \quad of \quad the \quad original \quad circle )}=\frac{16-4\pi}{4\pi}\)
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 Algebra based on multiplication and divisibility of two given numbers.
Multiplication and Divisibility- AMC 8, 2014
The 7-digit numbers 74A52B1 ana 326AB4C are each multiples of 3.which of the following could be the value of c ?
1
2
3
Key Concepts
Algebra
Division algorithm
Integer
Check the Answer
Answer:1
AMC-8, 2014 problem 21
Challenges and Thrills of Pre College Mathematics
Try with Hints
Use the rules of Divisibility ........
Can you now finish the problem ..........
If both numbers are divisible by 3 then the sum of their digits has to be divisible by 3......
can you finish the problem........
Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8... and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10... and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7... and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1. so the answer is 1
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
Try this beautiful problem from Geometry based on hexagon and Triangle.
Area of Triangle | AMC-8, 2015 |Problem 21
In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK $ is equilateral and FE=BC. What is the area of $\triangle KBC$?
9
12
32
Key Concepts
Geometry
Triangle
hexagon
Check the Answer
Answer:$12$
AMC-8, 2015 problem 21
Pre College Mathematics
Try with Hints
Clearly FE=BC
Can you now finish the problem ..........
$\triangle KBC$ is a Right Triangle
can you finish the problem........
Clearly ,since FE is a side of square with area 32
Therefore FE=$\sqrt 32$=$4\sqrt2$
Now since FE=BC,We have BC=$4\sqrt2$
Now JB is a side of a square with area 18
so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$
Lastly $\triangle KBC$ is a right triangle ,we see that
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} $=6.
The least common multiple of a and b is 12 .and the lest common multiple of b and c is 15.what is the least possible value of the least common multiple of a and c?
30
60
20
Key Concepts
Algebra
Division algorithm
Integer
Check the Answer
Answer:20
AMC-8, 2016 problem 20
Challenges and Thrills of Pre College Mathematics
Try with Hints
Find greatest common factors
Can you now finish the problem ..........
Find Least common multiple....
can you finish the problem........
we wish to find possible values of a,b and c .By finding the greatest common factor 12 and 15, algebrically ,it's some multiple of b and from looking at the numbers ,we are sure that it is 3.Moving on to a and c ,in order to minimize them,we wish to find the least such that the LCM of a and 3 is 12,$\to 4$.similarly with 3 and c,we obtain 5.the LCM of 4 and 5 is 20 .
Try this beautiful problem from Geometry based on Radius of a semicircle inscribed in an isosceles triangle.
Radius of a Semi circle - AMC-8, 2016 - Problem 25
A semicircle is inscribed in an isoscles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle as shown .what is the radius of the semicircle?
$\frac{110}{19}$
$\frac{120}{17}$
$\frac{9}{5}$
Key Concepts
Geometry
Area
pythagoras
Check the Answer
Answer:$\frac{120}{17}$
AMC-8, 2016 problem 25
Challenges and Thrills of Pre College Mathematics
Try with Hints
Draw a perpendicular from the point C on base AB
Can you now finish the problem ..........
D be the midpoint of the AB(since $\triangle ABC $ is an isoscles Triangle)
Find AC and area
can you finish the problem........
Area of the $\triangle ABC= \frac{1}{2} \times AB \times CD$
= $ \frac{1}{2} \times 16 \times 15 $
=120 sq.unit
Using the pythagoras th. $ AC^2= AD^2+CD^2$
i.e $AC^2=(8)^2+(15)^2$
i.e $AC=17$
Let$ ED = x$ be the radius of the semicircle
Therefore Area of $\triangle CAD = \frac{1}{2} \times AC \times ED$=$\frac {1}{2} area of \triangle ABC$
Try this beautiful problem from Geometry based on Area of a Triangle Using similarity
Area of Triangle - AMC-8, 2018 - Problem 20
In $\triangle ABC $ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.
What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?
