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## Area of the Trapezium | PRMO-2017 | Question 30

Try this beautiful Problem from Geometry based on Area of the Trapezium from PRMO 2017.

## Area of the Trapezium – PRMO 2017, Problem 30

Consider the areas of the four triangles obtained by drawing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576 , determine the square root of the maximum possible area of the trapezium to the nearest integer.

• $9$
• $40$
• $13$
• $20$

### Key Concepts

Geometry

Triangle

Trapezium

Answer:$13$

PRMO-2017, Problem 30

Pre College Mathematics

## Try with Hints

Let $x, y, z, w$ be areas of the four triangles as shown in figure.
then area of $\triangle ADB$= Area of $\triangle ACB$
$\Rightarrow x+y=x+w \Rightarrow y=w$

Also $\frac{AE}{EC}$=$\frac{area of \triangle ADE}{area of \triangle DEF}$=$\frac{area of \triangle AEB}{area of \triangle BEC}$
$\Rightarrow \frac{y}{z}=\frac{x}{w}=\frac{x}{y} \Rightarrow y^{2}=z x$
$\Rightarrow z, y, x$ are in G.P.

Can you now finish the problem ……….

Let $y=z r$ and $x=z r^{2},$ where $r \geq 1$ To make area of trapezium ABCD maximum, we take $z y=z^{2} r=576$
and $y w=z^{2} r^{2}=1296$
As $( z \leq y \leq x)$
Therefore $\frac{z^{2} r^{2}}{z^{2} r}=\frac{1296}{576} \Rightarrow r=\frac{9}{4} \Rightarrow z=16$

Can you finish the problem……..

Therefore area of trapezium $\mathrm{ABCD}$
$=x+y+z+w=z r^{2}+2 z r+z$
$=z(1+r)^{2}=16\left(1+\frac{9}{4}\right)^{2}=13^{2}$
Therefore Answer is $13 .$

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## Problem on Circle and Triangle | AMC 10A, 2016 | Problem 21

Try this beautiful problem from Geometry: Problem on Circle and Triangle

## Problem on Circle and Triangle – AMC-10A, 2016- Question 21

Circles with centers $P, Q$ and $R,$ having radii 1,2 and 3 , respectively, lie on the same side of line $l$ and are tangent to $l$ at $P^{\prime}, Q^{\prime}$ and $R^{\prime}$ respectively, with $Q^{\prime}$ between $P^{\prime}$ and $R^{\prime}$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $P Q R ?$

,

• $0$
• $\sqrt{6} / 3$
• $1$
• $\sqrt{6}-\sqrt{2}$
• $\sqrt{6} / 2$

### Key Concepts

Geometry

Circle

Triangle

Answer: $\sqrt{6}-\sqrt{2}$

AMC-10A (2016) Problem 21

Pre College Mathematics

## Try with Hints

We have to find out area of the Triangle PQR. But PQR is not a Standard Triangle that we can find out eassily. Join $PP^{\prime}$, $QQ^{\prime}$, $RR^{\prime}$. Now we can find out PQR such that $\left[P^{\prime} P Q R R^{\prime}\right]$ in two different ways: $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]$ and $[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$, so $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

Can you now finish the problem ……….

$P^{\prime} Q^{\prime}=\sqrt{P Q^{2}-\left(Q Q^{\prime}-P P^{\prime}\right)^{2}}=\sqrt{9-1}=\sqrt{8}=2 \sqrt{2}$

$Q^{\prime} R^{\prime}=\sqrt{Q R^{2}-\left(R R^{\prime}-Q Q^{\prime}\right)^{2}}=\sqrt{5^{2}-1^{2}}=\sqrt{24}=2 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]=\frac{P^{\prime} P+Q^{\prime} Q}{2} * 2 \sqrt{2}=\frac{1+2}{2} * 2 \sqrt{2}=3 \sqrt{2}$

$\left[Q^{\prime} Q R R^{\prime}\right]=5 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]$ = $P^{\prime} R^{\prime}=P^{\prime} Q^{\prime}+Q^{\prime} R^{\prime}=2 \sqrt{2}+2 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]=4 \sqrt{2}+4 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

$3 \sqrt{2}+5 \sqrt{6}=4 \sqrt{2}+4 \sqrt{6}+[P Q R]$

$[P Q R]=\sqrt{6}-\sqrt{2}$

Categories

## Median of numbers | AMC-10A, 2020 | Problem 11

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020.

## Median of numbers – AMC-10A, 2020- Problem 11

What is the median of the following list of $4040$ numbers$?$

$1,2,3,…….2020,1^2,2^2,3^2………..{2020}^2$

• $1989.5$
• $1976.5$
• $1972.5$

### Key Concepts

Median

Algebra

square numbers

Answer: $1976.5$

AMC-10A (2020) Problem 11

Pre College Mathematics

## Try with Hints

To find the median we need to know how many terms are there and the position of the numbers .here two types of numbers, first nonsquare i.e (1,2,3……2020) and squares numbers i.e $(1^2,2^2,3^2……2020^2)$.so We want to know the $2020$th term and the $2021$st term to get the median.

Can you now finish the problem ……….

Now less than 2020 the square number is ${44}^2$=1936 and if we take ${45}^2$=2025 which is greater than 2020.therefore we take the term that $1,2,3…2020$ trms + 44 terms=$2064$ terms.

can you finish the problem……..

since $44^{2}$ is $44+45=89$ less than $45^{2}=2025$ and 84 less than 2020 we will only need to consider the perfect square terms going down from the 2064 th term, 2020, after going down $84$ terms. Since the $2020$th and $2021$st terms are only $44$ and $43$ terms away from the $2064$th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$. Averaging the two,=$1976.5$

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## Circle | Geometry Problem | PRMO-2017 | Question 27

Try this beautiful Problem from Geometry based on Circle from PRMO 2017.

## Circle – PRMO 2017, Problem 27

Let $\Omega_{1}$ be a circle with centre 0 and let $A B$ be a diameter of $\Omega_{1} .$ Le $P$ be a point on the segment $O B$ different from 0. Suppose another circle $\Omega_{2}$ with centre P lies in the interior of $\Omega_{1}$. Tangents are drawn from A and B to the circle $\Omega_{2}$ intersecting $\Omega_{1}$ again at $A_{1}$ and $B_{1}$ respectively such that $A_{1}$, and $B_{1}$ are on the opposite sides of $A B$. Given that $A_{1} B=5, A B_{1}=15$ and $O P=10,$ find the radius of $\Omega_{1}$

• $9$
• $40$
• $34$
• $20$

### Key Concepts

Geometry

Circle

Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

## Try with Hints

Let radius of $\Omega_{1}$ be $R$ and that of $\Omega_{2}$ be $r$
From figure, $\Delta \mathrm{ADP} \sim \Delta \mathrm{AA}_{1} \mathrm{B}$
[
\begin{array}{l}
\Rightarrow \frac{D P}{A, B}=\frac{A P}{A B} \
\Rightarrow \frac{r}{5}=\frac{R+10}{2 R}
\end{array}
]

Can you now finish the problem ……….

Again, $\Delta B P E \sim \Delta B A B_{1}$
Therefore $\frac{P E}{A B_{1}}=\frac{B P}{B A}$
$\Rightarrow \frac{r}{15}=\frac{R-10}{2 R}$

Can you finish the problem……..

Dividing (1) by (2)

$3=\frac{R+10}{R-10} \Rightarrow R=20$

Categories

## Side of Square | AMC 10A, 2013 | Problem 3

Try this beautiful problem from Geometry: Side of Square.

## Sides of Square – AMC-10A, 2013- Problem 3

Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?

,

i

• $4$
• $5$
• $6$
• $7$
• $8$

### Key Concepts

Geometry

Square

Triangle

Answer: $8$

AMC-10A (2013) Problem 3

Pre College Mathematics

## Try with Hints

Given that Square $ABCD$ has side length $10$ and area of $\triangle ABE$ is $40$.we have to find out length of $BE$ where $E$ is the point on $BC$. we know area of the $\triangle ABE=\frac{1}{2} AB.BE=40$

Can you find out the side length of $BE$?

Can you now finish the problem ……….

$\triangle ABE=\frac{1}{2} AB.BE=40$

$\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40$

$\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40$

$\Rightarrow BE=8$

Categories

## Triangle Area Problem | AMC-10A, 2009 | Problem 10

Try this beautiful problem from Geometry based on Area of Triangle.

## Area of Triangle – AMC-10A, 2009- Problem 10

Triangle $A B C$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B, A D=3$, and $D C=4 .$ What is the area of $\triangle A B C$ ?

• $4 \sqrt{3}$
• $7 \sqrt{3}$
• $14 \sqrt{3}$
• $21$
• $42$

### Key Concepts

Triangle

Similarity

Geometry

Answer: $7 \sqrt{3}$

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

We have to find out the area of the Triangle ABC where $\angle B=90^{\circ}$ and $BD \perp AC$

Area of a Triangle = $\frac{1}{2}\times$ Base $\times$ Height.But we don know the value of $AB$ & $BC$. But we know $AC=7$. So if we can find out the value of $BD$ then we can find out the are of $\triangle ABC$ by $\frac{1}{2}\times AC \times BD$

Can you now finish the problem ……….

Let $\angle C=\theta$, then $\angle A=(90-\theta)$ (as $\angle B=90^{\circ}$, Sum of the angles in a triangle is $180^{\circ}$)

In $\triangle ABD$, $\angle ABD=\theta$ $\Rightarrow \angle A=(90-\theta$)

Again In $\triangle DBC$, $\angle DBC$=($90-\theta$) $\Rightarrow \angle C=\theta$

From the above condition we say that , $\triangle ABD \sim \triangle BDC$

Therefore , $\frac{BD}{CD}=\frac{AD}{BD}$ $\Rightarrow {BD}^2=AD.CD=4\times 3$

$\Rightarrow BD=\sqrt {12}$

can you finish the problem……..

Therefore area of the $\triangle ABC=\frac{1}{2}\times AC \times BD=\frac{1}{2}\times 7 \times \sqrt{12}=7 \sqrt{3}$

Categories

## Side Length of Rectangle | AMC-10A, 2009 | Problem 17

Try this beautiful problem from Geometry based on Side Length of Rectangle.

## Side Length of Rectangle – AMC-10A, 2009- Problem 17

Rectangle $A B C D$ has $A B=4$ and $B C=3 .$ Segment $E F$ is constructed through $B$ so that $E F$ isperpendicular to $D B$, and $A$ and $C$ lie on $D E$ and $D F$, respectively. What is $E F$ ?

• $9$
• $10$
• $\frac{125}{12}$
• $\frac{103}{9}$
• $12$

### Key Concepts

Triangle

Rectangle

Geometry

Answer: $\frac{125}{12}$

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

We have to find out the length of $EF$

Now $BD$ is the altitude from $B$ to $EF$, we can use the equation $BD^2 = EB\cdot BF$. ( as $\triangle BDE \sim \triangle BDF$).so we have to find out $BE$ and $BF$

Can you now finish the problem ……….

Now Clearly, $\triangle BDE \sim \triangle DCB$. Because of this, $\frac{A B}{C B}=\frac{E B}{D B}$. From the given information and the Pythagorean theorem, $A B=4, C B=3$, and $D B=5 .$ Solving gives $E B=20 / 3$
We can use the above formula to solve for $B F . B D^{2}=20 / 3 \cdot B F$. Solve to obtain $B F=15 / 4$

can you finish the problem……..

Therefore $E F=E B+B F=\frac{20}{3}+\frac{15}{4}=\frac{80+45}{12}$

Categories

## Problem on Cube | AMC 10A, 2008 | Problem 21

Try this beautiful problem from Geometry: Problem on Cube.

## Problem on Cube – AMC-10A, 2008- Problem 21

A cube with side length 1 is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $A B C D ?$

,

i

• $\frac{\sqrt{6}}{2}$
• $\frac{5}{4}$
• $\sqrt{2}$
• $\frac{5}{8}$
• $\frac{3}{4}$

### Key Concepts

Geometry

Square

Pythagoras

Answer: $\frac{\sqrt{6}}{2}$

AMC-10A (2008) Problem 21

Pre College Mathematics

## Try with Hints

The above diagram is a cube and given that side length $1$ and $B$ and $D$ are the mid points .we have to find out area of the $ABCD$.Now since $A B=A D=C B=C D=\sqrt{\frac{1}{2}^{2}}+1^{2},$ it follows that $A B C D$ is a rhombus. can you find out area of the rhombus?

Can you now finish the problem ……….

The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^{2}+1^{2}}=\sqrt{2} \cdot A C$ is a space diagonal, with length $\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$

can you finish the problem……..

Therefore area $A=\frac{1}{2} \times \sqrt{2} \times \sqrt{3}=\frac{\sqrt{6}}{2}$

Categories

## Television Problem | AMC 10A, 2008 | Problem 14

Try this beautiful Television Problem from AMC – 10A, 2008.

## Television Problem – AMC-10A, 2008- Problem 14

Older television screens have an aspect ratio of 4: 3 . That is, the ratio of the width to the height is 4: 3 . The aspect ratio of many movies is not $4: 3,$ so they are sometimes shown on a television screen by “letterboxing” – darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of 2: 1 and is shown on an older television screen with a 27 -inch diagonal. What is the height, in inches, of each darkened strip?

,

i

• $2$
• $2.25$
• $2.5$
• $2.7$
• $3$

### Key Concepts

Geometry

Square

Pythagoras

Answer: $2.7$

AMC-10A (2008) Problem 14

Pre College Mathematics

## Try with Hints

The above diagram is a diagram of Television set whose aspect ratio of $4: 3$.Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$-inch diagonal. Then we have to find the height, in inches, of each darkened strip.

we assume that the width and height of the screen be $4x$ and $3x$ respectively, and let the width and height of the movie be $2y$ and $y$ respectively. If we can find out the value of $x$ and $y$ then  the height of each strip can be calculate eassily

Can you now finish the problem ……….

By the Pythagorean Theorem, the diagonal is $\sqrt{(3 x)^{2}+(4 x)^{2}}=5 x=27 .$ So $x=\frac{27}{5}$

Now the movie and the screen have the same width, $2 y=4 x \Rightarrow y=2 x$

can you finish the problem……..

Thus, the height of each strip is $\frac{3 x-y}{2}=\frac{3 x-2 x}{2}=\frac{x}{2}=\frac{27}{10}=2.7$

Categories

## Area of Triangle | AMC 10A, 2006 | Problem 21

Try this beautiful problem from Geometry: Area of a triangle

## Triangle – AMC-10A, 2006- Problem 21

A circle of radius 1 is tangent to a circle of radius 2 . The sides of $\triangle A B C$ are tangent to the circles as shown, and the sides $\overline{A B}$ and $\overline{A C}$ are congruent. What is the area of $\triangle A B C ?$

,

i

• $15 \sqrt{2}$
• $\frac{35}{2}$
• $\frac{64}{3}$
• $16 \sqrt{2}$
• $24$

### Key Concepts

Geometry

Circle

Triangle

Answer: $16 \sqrt{2}$

AMC-10A (2006) Problem 21

Pre College Mathematics

## Try with Hints

Given that there are two circle of radius 1 is tangent to a circle of radius 2.we have to find out the area of the $\triangle ABC$.Now draw a perpendicular line $AF$ on $BC$.Clearly it will pass through two centers $O_1$ and $O_2$. and $\overline{A B}$ and $\overline{A C}$ are congruent i.e $\triangle ABC$ is an Isosceles triangle. Therefore $BF=FC$

So if we can find out $AF$ and $BC$ then we can find out the area of the $\triangle ABC$.can you find out $AF$ and $BC$?

Can you now finish the problem ……….

Now clearly $\triangle A D O_{1} \sim \triangle A E O_{2} \sim \triangle A F C$ ( as $O_1D$ and $O_2E$ are perpendicular on $AC$ , R-H-S law )

From Similarity we can say that , $\frac{A O_{1}}{A O_{2}}=\frac{D O_{1}}{E O_{2}} \Rightarrow \frac{A O_{1}}{A O_{1}+3}=\frac{1}{2} \Longrightarrow A O_{1}=3$

By the Pythagorean Theorem we have that $A D=\sqrt{3^{2}-1^{2}}=\sqrt{8}$

Again from $\triangle A D O_{1} \sim \triangle A F C$
$\frac{A D}{A F}=\frac{D O_{1}}{C F} \Longrightarrow \frac{2 \sqrt{2}}{8}=\frac{1}{C F} \Rightarrow C F=2 \sqrt{2}$

can you finish the problem……..

The area of the triangle is $\frac{1}{2} \cdot A F \cdot B C=\frac{1}{2} \cdot A F \cdot(2 \cdot C F)=A F \cdot C F=8(2 \sqrt{2})$=$16\sqrt2$