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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.

• $9$
• $40$
• $13$
• $20$

Geometry

Triangle

Trapezium

## Check the Answer

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 $P Q R ?$

,

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

Geometry

Circle

Triangle

## Check the Answer

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

## 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 $\Omega_{1}$

• $9$
• $40$
• $34$
• $20$

Geometry

Circle

## Check the Answer

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

## 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

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

Geometry

Circle

Triangle

## Check the Answer

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 $\frac{1}{2} \cdot A F \cdot B C=\frac{1}{2} \cdot A F \cdot(2 \cdot C F)=A F \cdot C F=8(2 \sqrt{2})$=$16\sqrt2$

Categories

## Circle Problem | AMC 10A, 2006 | Problem 23

Try this beautiful problem from Geometry: Circle

## Circle Problem – AMC-10A, 2006- Problem 23

Circles with centers $A$ and $B$ have radii 3 and 8 , respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at $E,$ and $A E=5 .$ What is $C D ?$

,

i

• $13$
• $\frac{44}{3}$
• $\sqrt{221}$
• $\sqrt{255}$
• $\frac{55}{3}$

Geometry

Circle

Tangents

## Check the Answer

Answer: $\frac{44}{3}$

AMC-10 (2006) Problem 23

Pre College Mathematics

## Try with Hints

Given that Circles with centers $A$ and $B$ have radii 3 and 8 and $A E=5 .$.we have to find out $CD$.So join $BC$ and $AD$.then clearly $\triangle BCE$ and $\triangle ADE$ are Right-Triangle(as $CD$ is the common tangent ).Now $\triangle BCE$ and $\triangle ADE$ are similar.Can you proof $\triangle BCE$ and $\triangle ADE$?

Can you now finish the problem ……….

$\angle A E D$ and $\angle B E C$ are vertical angles so they are congruent, as are angles $\angle A D E$ and $\angle B C E$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle A C E \sim \triangle B D E$.

By the Pythagorean Theorem, line segment $DE=4$

Therefore from the similarity we can say that $\frac{D E}{A D}=\frac{C E}{B C} \Rightarrow \frac{4}{3}=\frac{C E}{8}$ .

Therefore $C E=\frac{32}{3}$

can you finish the problem……..

Therefore $CD=CE+DE=4+\frac{32}{3}=\frac{44}{3}$

Categories

## Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

## Finding side of Triangle | PRMO | Problem 15

Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

• $28$
• $26$
• $30$

Geometry

Triangle

Pythagoras

## Check the Answer

Answer:$26$

PRMO-2014, Problem 15

Pre College Mathematics

## Try with Hints

Given that $\angle XOY=90^{\circ}$ .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now $\triangle XON$ & $\triangle MOY$ are Right angle Triangle. Use Pythagoras theorem …….

Can you now finish the problem ……….

Let $XM=MO=p$ and $ON=NY=q$.Now using Pythagoras theorm on $\triangle XON$ & $\triangle MOY$ we have…

$OX^2 +ON^2=XN^2$ $\Rightarrow 4p^2 +q^2=19^2$ $\Rightarrow 4p^2 +q^2=361$………..(1) and $OM^2 +OY^2=MY^2$ $\Rightarrow p^2 +4q^2=22^2$ $\Rightarrow p^2 +4q^2=484$……(2)

Now Adding (1)+(2)=$(4p^2 +q^2=361)$+$(p^2 +4q^2=484$ $\Rightarrow 5(p^2+q^2)=845$ $\Rightarrow (p^2+q^2)=169$ $\Rightarrow 4(p^2+q^2)=676$ $\Rightarrow (OX)^2+(OY)^2=(26)^2$ $\Rightarrow (XY)^2=(26)^2$ $\Rightarrow XY=26$.

Categories

## The area of trapezoid | AMC 8, 2003 | Problem 21

Try this beautiful problem from Geometry: The area of trapezoid

## The area of trapezoid – AMC-8, 2003- Problem 21

The area of trapezoid ABCD is 164 $cm^2$. The altitude is  8 cm, AB is 10 cm, and CD is 17 cm. What is BC in centimeters?

,

i

• $8$
• $10$
• $15$

Geometry

trapezoid

Triangle

## Check the Answer

Answer: $10$

AMC-8 (2003) Problem 21

Pre College Mathematics

## Try with Hints

Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Can you now finish the problem ……….

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

can you finish the problem……..

Given that the area of the trapezoid is 164 sq.unit

Draw two altitudes from the points B and C On the straight line AD at D and E respectively.

Now the Triangle ABD and Triangle CED, are right angle triangle and BD=CE= 8 cm

Using Pythagorean rules on the triangle ABD,we have…

$(AD)^2 + (BD)^2 =(AB)^2$

$\Rightarrow (AD)^2 + (8)^2 =(10)^2$

$\Rightarrow (AD)^2 =(10)^2 – (8)^2$

$\Rightarrow (AD)^2 = 36$

$\Rightarrow (AD) =6$

Using Pythagorean rules on the triangle CED,we have…

$(CE)^2 + (DE)^2 =(DC)^2$

$\Rightarrow (CE)^2 + (8)^2 =(17)^2$

$\Rightarrow (CE)^2 =(17)^2 – (8)^2$

$\Rightarrow (CE)^2 = 225$

$\Rightarrow (CE) =15$

Let BC= DE=x

Therefore area of the trapezoid=$\frac{1}{2} \times (AD+BC) \times 8$=164

$\Rightarrow \frac{1}{2} \times (6+x+15) \times 8$ =164

$\Rightarrow x=10$

Therefore BC=10 cm

Categories

## Largest area Problem | AMC 8, 2003 | Problem 22

Try this beautiful problem from Geometry based Largest area.

## Largest area – AMC-8, 2003 – Problem 22

The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

• $A$
• $B$
• $C$

Geometry

Circle

Square

## Check the Answer

Answer:$C$

AMC-8 (2003) Problem 22

Pre College Mathematics

## Try with Hints

To find out the largest area at first we have to find out the radius of the circles . all the circles are inscribed ito the squares .now there is a relation between the radius and the side length of the squares….

Can you now finish the problem ……….

area of circle =$\pi r^2$

can you finish the problem……..

In A:

Total area of the square =$2^2=4$

Now the radius of the inscribed be 1(as the diameter of circle = side length of the side =2)

Area of the inscribed circle is $\pi (1)^2=\pi$

Therefore the shaded area =$4- \pi$

In B:

Total area of the square =$2^2=4$

There are 4 circle and radius of one circle be $\frac{1}{2}$

Total area pf 4 circles be $4 \times \pi \times (\frac{1}{2})^2=\pi$

Therefore the shaded area =$4- \pi$

In C:

Total area of the square =$2^2=4$

Now the length of the diameter = length of the diagonal of the square=2

Therefore radius of the circle=$\pi$ and lengthe of the side of the square=$\sqrt 2$

Thertefore area of the shaded region=Area of the square-Area of the circle=$\pi (1)^2-(\sqrt 2)^2$=$\pi – 2$

Therefore the answer is  C

Categories

## Ratio of Circles | AMC-10A, 2009 | Problem 21

Try this beautiful problem from Geometry based on ratio of Circles from AMC 10A, 2009, Problem 21.

## Ratio of Circles – AMC-10A, 2009- Problem 21

Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

• $3-2 \sqrt{2}$
• $2-\sqrt{2}$
• $4(3-2 \sqrt{2})$
• $\frac{1}{2}(3-\sqrt{2})$
• $2 \sqrt{2}-2$

Geometry

Circle

Pythagoras

## Check the Answer

Answer: $4(3-2 \sqrt{2})$

AMC-10A (2009) Problem 21

Pre College Mathematics

## Try with Hints

We have to find out the ratio of the sum of the areas of the four smaller circles to the area of the larger circle. To find out the area any circle,we need radius.so at first we have to find out radius of two types circle.Can you find out the radius of two type circle i.e small circle and big circle………………

Can you now finish the problem ……….

Let the radius of the Four small circles be $r$.Therfore from the above diagram we can say $CD=DE=EF=CF=2r$. Now the quadrilateral $CDEF$ in the center must be a square. Therefore from Pythagoras theorm we can say $DF=\sqrt{(2r)^2 + (2r)^2}=2r\sqrt 2$. So $AB=AD+DF+BF=r+2r\sqrt 2+r=2r+2r\sqrt 2$

Therefore radius of the small circle is $r$ and big circle is$R=r+r \sqrt{2}=r(1+\sqrt{2})$

Can you now finish the problem ……….

Therefore the area of the large circle is $L=\pi R^{2}=\pi r^{2}(1+\sqrt{2})^{2}=\pi r^{2}(3+2 \sqrt{2})$and the The area of four small circles is $S=4 \pi r^{2}$

The ratio of the area will be $\frac{S}{L}=\frac{4 \pi r^{2}}{\pi r^{2}(3+2 \sqrt{2})}$

=$\frac{4}{3+2 \sqrt{2}}$

=$\frac{4}{3+2 \sqrt{2}} \cdot \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$

=$\frac{4(3-2 \sqrt{2})}{3^{2}-(2 \sqrt{2})^{2}}$

=$\frac{4(3-2 \sqrt{2})}{1}$

=$4(3-2 \sqrt{2})$

Categories

## Area of the Region Problem | AMC-10A, 2007 | Problem 24

Try this beautiful problem from Geometry: Area of the region

## Problem on Area of the Region – AMC-10A, 2007- Problem 24

Circle centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the midpoint of $\overline{AB}$, and $OA = 2\sqrt {2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$?

• $\pi$
• $7\sqrt 3 -\pi$
• $8\sqrt 2 -4-\pi$

Geometry

Triangle

similarity

## Check the Answer

Answer: $8\sqrt 2 -4-\pi$

AMC-10A (2007) Problem 24

Pre College Mathematics

## Try with Hints

We have to find out the area of the region $ECODF$ i.e of gray shaded region.this is not any standard geometrical figure (such as circle,triangle…etc).so we can not find out the value easily.Now if we join $AC$,$AE$,$BD$,$BF$.Then $ABFE$ is a rectangle.then we can find out the required area by [ area of rectangle $ABEF$- (area of arc $AEC$+area of $\triangle ACO$+area of $\triangle BDO$+ area of arc $BFD$)]

Can you find out the required area…..?

Given that Circle centered at $A$ and $B$ each have radius $2$ and Point $O$ is the midpoint of $\overline{AB}$, and $OA = 2\sqrt {2}$

Area of $ABEF$=$2 \times 2 \times 2\sqrt 2$=$8\sqrt 2$

Now $\triangle{ACO}$ is a right triangle. We know $AO=2\sqrt{2}$and $AC=2$, so $\triangle{ACO}$ is isosceles, a $45$-$45$ right triangle.$\overline{CO}$ with length $2$. The area of $\triangle{ACO}=\frac{1}{2} \times base \times height=2$. By symmetry, $\triangle{ACO}\cong\triangle{BDO}$, and so the area of $\triangle{BDO}$ is also $2$.now the $\angle CAO$ = $\angle DBO$=$45^{\circ}$. therefore $\frac{360}{45}=8$

So the area of arc $AEC$ and arc $BFD$=$\frac{1}{8} \times$ area of the circle=$\frac{\pi 2^2}{8}$=$\frac{\pi}{2}$

can you finish the problem……..

Therefore the required area by [ area of rectangle $ABEF$- (area of arc $AEC$+area of $\triangle ACO$+area of $\triangle BDO$+ area of arc $BFD$)]=$8\sqrt 2-(\frac{\pi}{2}+2+2+\frac{\pi}{2}$)=$8\sqrt 2 -4-\pi$