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AMC 8 Geometry Math Olympiad USA Math Olympiad

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?

Area of the triangle and square
  • $\frac{7}{20}$
  • $\frac{5}{18}$
  • $\frac{11}{20}$

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 ……….

Area of the triangle and square

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……..

Square and triangle

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=\(\frac{10}{36}=\frac{5}{18}\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of trapezoid | AMC 8, 2011|Problem 20

Try this beautiful problem from Geometry: The area of trapezoid.

The area of the trapezoid – AMC-8, 2011 – Problem 20

Quadrilateral  ABCDis a trapezoid ,AD=15,AB=50,BC=20,and the altitude is 12.What is the area of the trapezoid?

area of trapezoid
  • $700$
  • $750$
  • $800$

Key Concepts

Geometry

Trapezoid

Area of Triangle

Check the Answer

Answer:$750$

AMC-8(2011) Problem 20

Pre College Mathematics

Try with Hints

Draw altitudes from the top points A and B to CD at X and Y points

Can you now finish the problem ……….

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times\) (height between AB and CD)

can you finish the problem……..

a trapezoid

Draw altitudes from the top points A and B to CD at X and Y points.Then the trapezoid will be divided into two right triangles and a rectangle .

Using The Pythagorean theorem on \(\triangle ADX and \triangle BYC\) ,

\((DX)^2+(AX)^2=(AD)^2\)

\(\Rightarrow (a)^2+(12)^2=(15)^2\)

\(\Rightarrow a=\sqrt{(15)^2-(12)^2}=\sqrt {81} =9\)

and

\((BY)^2+(YC)^2=(BC)^2\)

\(\Rightarrow (12)^2+(b)^2=(20)^2\)

\(\Rightarrow b=\sqrt{(20)^2-(12)^2}=\sqrt {256} =16\)

Now ABYX is a Rectangle so \(XY=AB=50\)

\(CD=DX+XY+YC=a+XY+b=9+50+16=75\)

The area of the trapezoid is \(\frac{1}{2} \times (AB+CD) \times (height between AB and CD)=\frac {1}{2} \times (AB+CD) \times 12=750\)

.

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AMC 8 Geometry Math Olympiad

Hexagon and Triangle |AMC 8- 2015 -|Problem 21

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 $\triangle KBC$?

hexagon and triangle

  • 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……..

hexagon and triangle

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

$\angle JBA + \angle ABC +\angle CBK +\angle KBJ$ =$360^\circ$

i.e$ 90^\circ + 120^\circ +\angle CBK + 60^\circ=360^\circ$

i.e $\angle CBK=90^\circ $

So $\triangle KBC $ is a right triangle with legs $3\sqrt 2$ and $4\sqrt2$

Now its area is $3\sqrt2 \times 4\sqrt 2 \times \frac {1}{2}$=$\frac{24}{2}$=12

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AMC 12 USA Math Olympiad

Triangle Inequality Problem – AMC 12B, 2014 – Problem 13

Try this beautiful problem from AMC 12 based on Triangle inequality problem.

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: \(\frac {3+\sqrt 5}{2}\)

American Mathematics Competition – 12B ,2014, Problem Number – 13

Secrets in Inequalities.

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 :

a+1>b

a>b-1

\(\frac {1}{a} + \frac {1}{b} >1 \)

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.………………….

We already know \(\frac {1}{a} + \frac {1}{b} = 1\)

After substituting we will get :

\(\frac {1}{b – 1} + \frac {1}{b} = \frac {b+b-1}{b(b-1)} = 1 \)

\(\frac {2b – 1}{b(b-1)} = 1 \)

Now do the rest of the calculation ………………………..

Here is the rest of steps to check your problem :

\( 2b – 1 = b^2 – b \)

Now Solving for b using the quadratic equation, we get

\(b^2 – 3b + 1 = 0 \)

\(b = \frac {3 + \sqrt 5}{2} \) (Answer)

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AMC 10

Area of polygon – AMC 10B, 2019 Problem 8

Area of polygon

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?

area of polygon- knowledge graph
$\textbf{(A) } 4 \qquad \textbf{(B) } 12 - 4\sqrt{3} \qquad \textbf{(C) } 3\sqrt{3}\qquad \textbf{(D) } 4\sqrt{3} \qquad \textbf{(E) } 16 - 4\sqrt{3}$

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 $4\left(3 – \sqrt{3}\right) = {\textbf{(B) } 12 – 4\sqrt{3}}$.

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AMC 10

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?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$

2019 AMC 10B Problem 10

Problem related to triangle

6 out of 10

Secrets in Inequalities.

Knowledge Graph

Problem related to triangle- knowledge graph

Use some hints

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.

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