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## Area of The Region | AMC-8, 2017 | Problem 25

Try this beautiful problem from Geometry based on the area of Region

## Find the area – AMC-8, 2017- Problem 25

In the figure shown, US and UT and are line segments each of length 2, and$\angle TUS=60^{\circ}$. Arcs TR and SR  and are each one-sixth of a circle with radius 2. What is the area of the figure shown?

• (4 – $\frac{4\pi}{3})$
• ($4\sqrt3$ – $\frac{4\pi}{3})$
• ($4\sqrt3$ – $4\pi$

### Key Concepts

Geometry

Triangle

Circle

Answer: ($4\sqrt3$ – $\frac{4\pi}{3})$

AMC-8 (2017) Problem 25

Pre College Mathematics

## Try with Hints

We have to find out the shaded region. The above diagram is not a standard geometrical figure (such as triangle, square or circle, etc.). So we can not find out the area of the shaded region by any standard formula.

Now if we extend US and UT and joined XY( shown in the above figure ) then it becomes a Triangle shape i.e $\triangle UXY$ and region XSR and region TRY are circular shapes. Now you have to find out the area of these geometrical figures…

Can you now finish the problem ……….

Given that US and UT and are line segments each of length 2 and Arcs TR and SR  and are each one-sixth of a circle with radius 2. Therefore UX=2+2=4,UY=2+2=4 and SX=2+2=4.Therefore $\triangle UXY$ is an equilateral triangle with side length 4 and area of equilateral triangle =$\frac{\sqrt 3}{4} (side)^2$ and the region XSR and region TRY are each one-sixth of a circle with radius 2.nor area of the circle =$\pi (radius)^2$

can you finish the problem……..

The area of the $\triangle UXY=\frac{\sqrt 3}{4} (4)^2=4\sqrt3$ sq.unit

Now area of (region SXR + region TRY)=$2 \times \frac{\pi (2)^2}{6}=\frac{4\pi}{3}$

Therefore the area of the region USRT=Area of $\triangle UXY$- (region SXR + region TRY)=($4\sqrt3$ – $\frac{4\pi}{3})$

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## Area of the figure | AMC-8, 2014 | Problem 20

Try this beautiful problem from Geometry based on the Area of the figure.

## Area of the figure – AMC-8, 2014- Problem 20

Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle of radius $1$ is centered at $A$, a circle of radius $2$ is centered at $B$, and a circle of radius $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?

• $3.5$
• $4.0$
• $4.5$

### Key Concepts

Geometry

Rectangle

Circle

Answer: $4.0$

AMC-8 (2014) Problem 20

Pre College Mathematics

## Try with Hints

To Find out the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure), we have to find out the area of the Rectangle -the area of three quarter circle inside the circle(i.e green shaded region)

Can you now finish the problem ……….

To find out the area of the rectangle, AD=5 and CD=3 are given. to find out the area of the three-quarter circles, the radii are 1,2 & 3 respectively.

Now area of Rectangle=$AD \times CD$ and area of  quarter circles =$\frac{\pi r^2}{4}$,where $r$=Radius of the circle

can you finish the problem……..

Area of the rectangle=$5 \times 3$=15 sq.unit

Area of the quarter circle with the center C=$\frac{\pi (3)^2}{4}$=$\frac{9 \pi}{4}$ sq.unit

Area of the quarter circle with the center B= $\frac{\pi (2)^2}{4}$ =$\frac{4 \pi}{4}$=$\pi$ sq.unit

Area of the quarter circle with the center C= $\frac{\pi (1)^2}{4}$ =$\frac{\pi}{4}$ sq.unit

Therefore the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure) =(15- $\frac{9 \pi}{4}$- $\pi$ – $\frac{\pi}{4}$ )=15-$\frac{7\pi}{2}$=15-11=4 sq.unit (Taking $\pi =\frac{22}{7}$)

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## Linear Equations | AMC 8, 2007 | Problem 20

Try this beautiful problem from Algebra based on Linear equations from AMC-8, 2007.

## Linear equations – AMC 8, 2007

Before the district play, the Unicorns had won $45$% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

• $40$
• $48$
• $58$

### Key Concepts

Algebra

linear equation

multiplication

AMC-8, 2007 problem 20

Challenges and Thrills of Pre College Mathematics

## Try with Hints

At first, we have to Calculate the number of won games and lost games. Unicorns had won $45$% of their basketball game.so we may assume that out of 20 unicorns woned 9.

Can you now finish the problem ……….

Next unicorns won six more games and lost two.so find out the total numbers of won game and total numbers of games i.e won=$9x+6$ and the total number of games become $20x+8$

can you finish the problem……..

Given that Unicorns had won $45$% of their basketball games i.e $\frac{45}{100}=\frac{9}{20}$

During district play, they won six more games and lost two,

Therefore they won$9x+6$ and the total number of games becomes $20x+8$

According to the question, Unicorns finish the season having won half their games. …

Therefore,$\frac{9x+6}{20x+8}=\frac{1}{2}$

$\Rightarrow 18x+12=20x+8$

$\Rightarrow 2x=4$

$\Rightarrow x=2$

Total number of games becomes $20x+8$ =$(20 \times 2) +8=48$

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## Problem on Semicircle | AMC 8, 2013 | Problem 20

Try this beautiful problem from Geometry based on Semicircle.

## Area of the Semicircle – AMC 8, 2013 – Problem 20

A $1\times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?

• $\frac{\pi}{2}$
• $\pi$
• $\frac{\pi}{3}$

### Key Concepts

Geometry

Square

semi circle

Answer:$\pi$

AMC-8 (2013) Problem 20

Pre College Mathematics

## Try with Hints

At first we have to find out the radius of the semicircle for the area of the semicircle.Now in the diagram,AC is the radius of the semicircle and also AC is the hypotenuse of the right Triangle ABC.

Can you now finish the problem ……….

Now AB=1 AND AC=1 (As ABDE $1\times 2$ rectangle .So using pythagorean theorm we can eassily get the value of AC .and area of semicircle =$\frac{\pi r^2}{2}$

can you finish the problem……..

Given that ABDE is a square whose AB=1 and BD=2

Therefore BC=1

Clearly AC be the radius of the given semi circle

From the $\triangle ABC$,$(AB)^2+(BC)^2=(AC)^2\Rightarrow AC=\sqrt{(1^2+1^2)}=\sqrt2$

Therefore the area of the semicircle=$\frac{1}{2}\times \pi(\sqrt 2)^2$=$\pi$

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## Radius of semicircle | AMC-8, 2013 | Problem 23

Try this beautiful problem from Geometry: Radius of the semicircle

## Radius of the semicircle- AMC-8, 2013- Problem 23

Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$?

• $9$
• $7.5$
• $6$

### Key Concepts

Geometry

Triangle

Semi-circle

Answer: $7.5$

AMC-8 (2013) Problem 23

Pre College Mathematics

## Try with Hints

We have to find out the radius of the semi-circle on ${BC}$? Now ABC is a Right angle Triangle.so if you find out AB and AC then BC will be easily calculated…..

Can you now finish the problem ……….

To find the value of AB and AC, notice that area of the semi-circle on AB is given and length of the arc of AC is given….

can you finish the problem……..

Let the length of AB=$2x$,So the radius of semi-circle on AB=$x$.Therefore the area =$\frac{1}{2}\times \pi (\frac{x}{2})^2=8\pi$$\Rightarrow x=4$

Theregfore length of AB=$2x$=8

Given that the arc of the semi-circle on $\overline{AC}$ has length $8.5\pi$,let us take the radius of the semicircle on AB =$r$.now length of the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ i.e half perimeter=$8.5$

so $\frac{1}{2}\times {2\pi r}=8.5\pi$$\Rightarrow r=8.5$$\Rightarrow 2r=17$.so AC=17

The triangle ABC is a Right Triangle,using pythagorean theorm……

$(AB)^2 + (BC)^2=(AC)^2$$\Rightarrow BC=\sqrt{(AC)^2 -(AB)^2}$$\Rightarrow BC =\sqrt{(17)^2 -(8)^2}$$\Rightarrow BC= \sqrt{289 -64 }$$\Rightarrow BC =\sqrt {225}=15$

Therefore the radius of semicircle on BC =$\frac{15}{2}=7.5$

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## Area of a Regular Hexagon | AMC-8, 2012 | Problem 23

Try this beautiful problem from Geometry: Area of the Regular Hexagon – AMC-8, 2012 – Problem 23.

## Area of the Regular Hexagon – AMC-8, 2012- Problem 23

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle’s area is 4, what is the area of the hexagon?

• $8$
• $6$
• $10$

### Key Concepts

Geometry

Triangle

Hexagon

Answer: $6$

AMC-8 (2012) Problem 23

Pre College Mathematics

## Try with Hints

To find out the area of the Regular hexagon,we have to find out the side length of it.Now the perimeter of the triangle and Regular Hexagon are same….from this condition you can easily find out the side length of the regular Hexagon

Can you now finish the problem ……….

Let the side length of an equilateral triangle is$x$.so the perimeter will be $3x$ .Now according to the problem the perimeter of the equiliteral triangle and regular hexagon are same,i.e the perimeter of regular hexagon=$3x$

So the side length of be $\frac{3x}{6}=\frac{x}{2}$

can you finish the problem……..

Now area of the triangle $\frac{\sqrt 3}{4}x^2=4$

Now the area of the Regular Hexagon=$\frac{3\sqrt3}{2} (\frac{x}{2})^2=\frac{3}{2} \times \frac{\sqrt{3}}{4}x^2=\frac{3}{2} \times 4$=6

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## Area of Triangle Problem | AMC-8, 2019 | Problem 21

Try this beautiful problem from Geometry based on the area of the triangle.

## Area of Triangle – AMC-8, 2019- Problem 21

What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$

• $8$
• $16$
• $15$

### Key Concepts

Geometry

Triangle

Linear equation

Answer: $16$

AMC-8 (2019) Problem 21

Pre College Mathematics

## Try with Hints

Find the three vertex of the triangle

Can you now finish the problem ……….

The area of the Triangle =$\frac{1}{2} \times \{x_1(y_2 – y_3)+x_2(y+3 -y_1)+x_3(y_1 -y_2)\}$

can you finish the problem……..

Solving two The lines y=5 and y=1+x are intersect at (4,5)=$(x_1,y_1)$(say)

Solving two The lines y=5 and y=1-x are intersect at (-4,5)=$(x_2,y_2)$(say)

Solving two The lines y=1-x and y=1+x are intersect at (0,1)=$(x_1,y_1)$(say)

Then the area of Triangle =$\frac{1}{2} \times\{ x_1(y_2 – y_3)+x_2(y+3 -y_1)+x_3(y_1 -y_2)\}$

= $\frac{1}{2} \times \{4(5-1)+(-4)(1-5)+0(5-5)=\frac{1}{2} (16+16)=16\}$

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## Angles of Star | AMC 8, 2000 | Problem 24

Try this beautiful problem from Geometry from AMC-8, 2000, Problem-24, based on angles of Star

## Angles of Star | AMC-8, 2000 | Problem 24

If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$, then $\angle B+\angle D =$

• $90$
• $70$
• $80$

### Key Concepts

Geometry

Star

Triangle

Answer:$80$

AMC-8, 2000 problem 24

Pre College Mathematics

## Try with Hints

Find the $\angle AFG$

Can you now finish the problem ……….

sum of the angles of a Triangle is $180^\circ$

can you finish the problem……..

we know that the sum of the angles of a Triangle is $180^\circ$

In the $\triangle AGF$ we have,$(\angle A +\angle AGF +\angle AFG) =180^\circ$

$\Rightarrow 20^\circ +2\angle AFG=180^\circ$(as $\angle A =20^\circ$ & $\angle AFG=\angle AGF$)

$\Rightarrow \angle AFG=80^\circ$ i.e $\angle EFD=\angle 80^\circ$

So the $\angle BFD=\frac{360^\circ -80^\circ-80^\circ}{2}=100^\circ$

Now in the $\triangle BFD$,$(\angle BFD +\angle B +\angle D$)=$180^\circ$

$\Rightarrow \angle B +\angle D=180^\circ -100^\circ$

$\Rightarrow \angle B +\angle D=80^\circ$

Categories

## Area of a Triangle | AMC-8, 2000 | Problem 25

Try this beautiful problem from Geometry: Area of a Triangle

## Area of the Triangle- AMC-8, 2000- Problem 25

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $BC$ and $CD$ are joined to form a triangle, the area of that triangle is

• $25$
• $27$
• $29$

### Key Concepts

Geometry

Triangle

square

Answer: $27$

AMC-8 (2000) Problem 25

Pre College Mathematics

## Try with Hints

Area of the triangle =$\frac{1}{2} \times base \times height$

Can you now finish the problem ……….

Therefore area of the shaded region i.e area of the $\triangle AEF$=area of the square- area of $(\triangle ADE +\triangle EFC +\triangle ABF)$

can you finish the problem……..

Given that area of the rectangle ABCD=72

Let length AB=$x$ and length of CD=$y$

Therefore DE=EC=$\frac{y}{2}$ and BF=FC=$\frac{x}{2}$

Area of ABCD=$xy$=72

Area of the $\triangle ADE=\frac{1}{2}\times DE \times AD= \frac{1}{2}\times \frac{x}{2} \times y =\frac{xy}{4}=\frac{72}{4}=18$

Area of the $\triangle EFC=\frac{1}{2}\times EC \times FC= \frac{1}{2}\times \frac{x}{2} \times \frac{y}{2} =\frac{xy}{8}=\frac{72}{8}=9$

Area of the $\triangle ABF=\frac{1}{2}\times AB \times BF= \frac{1}{2}\times y\times \frac{x}{2}=\frac{xy}{4}=\frac{72}{4}=18$

Therefore area of the shaded region i.e area of the $\triangle AEF$=area of the square- area of $(\triangle ADE +\triangle EFC +\triangle ABF)=72-(18+8+18)=27$

Categories

## Unit digit | Algebra | AMC 8, 2014 | Problem 22

Try this beautiful problem from Algebra about unit digit from AMC-8, 2014.

## Unit digit | AMC-8, 2014|Problem 22

A $2$-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the unit digit of the number?

• 7
• 9
• 5

### Key Concepts

Algebra

Multiplication

integer

Answer:$9$

AMC-8, 2014 problem 22

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Let the ones digit place be y and ten’s place be x

Therefore the number be $10x+y$

Can you now finish the problem ……….

Given that the product of the digits plus the sum of the digits is equal to the number

can you finish the problem……..

Let the ones digit place be y and ten’s place be x

Therefore the number be $10x+y$

Now the product of the digits=$xy$

Given that the product of the digits plus the sum of the digits is equal to the number

Therefore $10x+y=(x\times y)+(x+y)$

$\Rightarrow 9x=xy$

$\Rightarrow y=9$

Therefore the unit digit =$y$=9