Try this beautiful problem from Geometry based on area of pinwheel.
Pinwheel - AMC 8, 2007
What is the area of the shaded pinwheel shown in the \(5 \times 5\) grid?
$8$
$6$
$9$
Key Concepts
Geometry
Triangle
Square
Check the Answer
Answer:$6$
AMC-8 (2007) Problem 23
Pre College Mathematics
Try with Hints
Find the area of white space
Can you now finish the problem ..........
The area of the pinwheel =Total area of the square - The white space
can you finish the problem........
Now The area of the square around the pinwheel is \(5 \times 5\)=25 sq.unit
Clearly in the above picture, there are four white triangles i.e (\(\triangle EOF, \triangle KOL, \triangle HOG, \triangle JOI)\) and four white small square(at the corner)
clearly HG=3 unit & OP=2.5 unit
Therefore the area of four Triangle = \(4 \times \frac{1}{2} \times HG \times OP= 4 \times \frac{1}{2} \times 3 \times\) 2.5=15 sq.unit
And also the area of four white corner squares =\((4 \times 1)\)=4 sq.unit
The area of the pinwheel =Total area of the square - The white space =(25-15-4)=6 sq.unit
Area of the triangle and square | AMC 8, 2008 | Problem 23
Try this beautiful problem from AMC-8, 2008 based on the Ratio of the area of the triangle and square.
Area of the star and circle - AMC- 8, 2008 - Problem 23
In square ABCE, AF=2FE and CD=2DE .what is the ratio of the area of \(\triangle BFD\) to the area of square ABCE?
$\frac{7}{20}$
$\frac{5}{18}$
$\frac{11}{20}$
Key Concepts
Geometry
Triangle
Square
Check the Answer
Answer:$\frac{5}{18}$
AMC-8 (2008) Problem 23
Pre College Mathematics
Try with Hints
Area of the square =\((side)^2\)
Area of triangle =\(\frac{1}{2} \times base \times height\)
Can you now finish the problem ..........
The area of the \(\triangle BFD\)=(The area of the square ABCE- The area of \(\triangle ABF\) -The area of\( \triangle BCD\) -Area of the\( \triangle EFD) \)
can you finish the problem........
Let us assume that the side length of the given square is 6 unit
Then clearly AB=BC=CE=EA=6 unit & AF=4 unit,EF=2 unit, CD=4 unit
Total area of the square is \(6^2\)=36 sq.unit
Area of the \(\triangle ABF=\frac{1}{2}\times AB \times AF= \frac{1}{2}\times 6 \times 4= 12\) sq.unit
Area of the \(\triangle BCD=\frac{1}{2}\times BC \times CD= \frac{1}{2}\times 6 \times 4= 12\) sq.unit
Area of the \(\triangle EFD=\frac{1}{2}\times EF \times ED= \frac{1}{2}\times 2 \times 2= 2\) sq.unit
The area of the \(\triangle BFD\)=(The area of the square ABCE- The area of \(\triangle ABF\) -The area of\( \triangle BCD\) -Area of the\( \triangle EFD)=(36-12-12-2)=10 \)sq.unit
the ratio of the area of \(\triangle BFD\) to the area of square ABCE=\(\frac{10}{36}=\frac{5}{18}\)
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
Triangle Inequality Problem - AMC 12B, 2014 - Problem 13
Try this beautiful problem from American Mathematics Competition - 12B ,2014, Problem Number - 13 based on Triangle inequality
Problem - Triangle Inequality
Real numbers a and b are chosen with 1 < a < b such that no triangle with positive area has side lengths 1,a and b or \(\frac {1}{b},\frac {1}{a}\) and 1. What is the smallest possible value of b?
$\frac{3+\sqrt{3}}{2}$
$\frac{5}{2}$
$\frac{3+\sqrt{5}}{2}$
$\frac{3+\sqrt{6}}{2}$
Key Concepts
Triangle Inequality
Inequality
Geometry
Check the Answer
Answer: \(\frac {3+\sqrt 5}{2}\)
Try with Hints
Here is the first hint where you can start this sum:
It is given $1>\frac{1}{a}>\frac{1}{b}$. Use Triangle Inequality here :
If we want to find the lowest possible value of $b$, we create we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get : $a=b-1$
Now try to do the rest of the sum……………………
Here is the rest of steps to check your problem :
$$ 2 b-1=b^2-b $$
Now Solving for $\mathbf{b}$ using the quadratic equation, we get
Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg
Try with Hints
Do you really need a hint ? You can start thinking about the Triangle Inequality...........
If you have already get the idea about the main concept for this sum then you can start the problem by taking a quadrilateral ABCD with diagonals that intersect at point 'o'.
The distance from 'o' to all vertices is equal.
Here is the diagram where OA + OB + OC + OD = AC + BD - which is sum of the diagonals. We can consider this as one case ...
As a last hint you have to take another point to o' to compare with first case
Now o' be another point inside the quadrilateral. If we use triangle inequality here we have ,
AO' + OC' > AC from Triangle AO'C
BO' + O'D > BD from Triangle BO'D
Hence AO' + O'C + BO' + O'D > AB + BD
Therefore its clear from this that the point that minimizes the sum of the distances is the point of intersection of diagonals.
Polygon means shape composed of multiple sides , for example square , triangle, trapezium pentagon etc. A regular polygon means all of its sides have same length. Square ,equilateral triangle is a regular polygon. Let us learn to find the area of polygon.
Try the problem
The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?
AMC 10B, 2019 Problem 8
Area of polygon -square and triangle
6 out of 10
challenges and thrills of pre college mathematics
Knowledge Graph
Use some hints
Split the square in 4 identical parts by drawing two perpendicular (horizontal and vertical )lines from the center. And then we will get smaller square with two similar right angle triangles in it and one fourth part of the shaded region.
When we split an equilateral triangle in half, we get two triangles with angles 30,60 and 90 degrees. Therefore, the altitude, which is also the side length of one of the smaller squares, is \(\sqrt{3}\).
the area of the two congruent triangles will be $2 \cdot \frac{1 \cdot \sqrt{3}}{2} = \sqrt{3}$.
.
The area of the each small squares is the square of the side length, i.e. $\left(\sqrt{3}\right)^2 = 3$. Therefore, the area of the shaded region in each of the four squares is $3 - \sqrt{3}$. Since there are $4$ of these squares, we multiply this by $4$ to get $4\left(3 - \sqrt{3}\right) = {\textbf{(B) } 12 - 4\sqrt{3}}$.
Problem related to triangle - AMC 10B, 2019 Problem 10
Problem related to triangle
The given problem is related to the calculation of area of triangle and distance between two points.
Try the problem
In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
Notice that it does not matter where the triangle is in the 2D plane so for our easy access we can select two points A and B in any place of choice.
So we can actually select any two points A and B such that they are 10 units apart so lets the points are \(A(0,0)\) and \(B(10,0)\) , as they are 10 units apart.
Now we can select the point C such that the perimeter of the triangle is 50 units. and then we can apply the formula of area to calculate the possible positions of C.
