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## Application of Pythagoras Theorem | SMO, 2010 | Problem 22

Try this problem from the Singapore Mathematics Olympiad, SMO, 2010 based on the application of the Pythagoras Theorem.

## Application of Pythagoras Theorem- (SMO Test)

The figure below shows a circle with diameter AB. C ad D are points on the circle on the same side of AB such that BD bisects $\angle {CBA}$. The chords AC and BD intersect at E. It is given that AE = 169 cm and EC = 119 cm. If ED = x cm, find the value of x.

• 65
• 55
• 56
• 60

### Key Concepts

Circle

Pythagoras Theorem

2D – Geometry

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

If you get stuck in this problem this is the first hint we can start with :

As BE intersect $\angle {CBA}$ we have $\frac {BC}{BA} = \frac {EC}{EA} = \frac {119}{169}$

Thus we can let BC = 119 y and BA = 169 y .

Since $\angle {BCA} = 90 ^\circ$.

Then try to do the rest of the problem ………………………………………………

If we want to continue from the last hint we have :

Apply Pythagoras Theorem ,

$AB ^2 = AC^2 + BC ^2$

$(169y)^2 = (169 + 119)^2 + (119y)^2$

$y^2 (169-119)(169+119) = (169+119)^2$

$y^2 = \frac {169+119}{169-119} = \frac {144}{25}$

$y = \frac {12}{5}$

In the last hint:

Hence , from triangle BCE , we have BE = $\sqrt{119^2 + (119y)^2} = 119 \times \frac {13}{5}$

Finally , note that $\triangle {ADE}$ and $\triangle {BCE}$ are similar , so we have

ED = $\frac {AE \times CE}{BE} = \frac {169 \times 119}{119 \times \frac {13}{5}} = 65$ cm .

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## Problem on Area of Triangle | SMO, 2010 | Problem 32

Try this beautiful problem from Singapore Mathematics Olympiad based on area of triangle.

## Problem – Area of Triangle (SMO Entrance)

Given that ABCD is a square . Points E and F lie on the side BC and CD respectively, such that BE = $\frac {1}{3} AB$ = CF. G is the intersection of BF and DE . If

$\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$ is in its lowest term find the value of m+n.

• 12
• 19
• 21
• 23

### Key Concepts

2D – Geometry

Area of Triangle

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

Here is the 1st hint for them who got stuck in this problem:

At 1st we will join the points BD and CG .

Then to proceed with this sum we will assume the length of AB to be 1. The area of $\triangle {BGE}$ and $\triangle {FGC}$ are a and b respectively.

Try to find the area of $\triangle {EGC}$ and $\triangle {DGF}$……………………..

Now for the 2nd hint let’s start from the previous hint:

So the area of $\triangle {EGC}$ and $\triangle {DGF}$ are 2a and 2b .From the given value in the question we can say the area of $\triangle {BFC} = \frac {1}{3}$(as BE = CF = 1/3 AB\).

We can again write 3a + b = $\frac {1}{6}$

Similarly 3b + 2a = area of the $\triangle DEC = \frac {1}{3}$.

Now solve this two equation and find a and b…………..

In the last hint :

2a + 3b = 1/3……………………(1)

3a + b = 1/6……………(2)

so in $(1) \times 3$ and in $(2) \times 2$

6a + 9b = 1

6a + 2b = 1/3

so , 7 b = 2/3 (subtracting above 2 equation)

b = $\frac {2}{21}$ and a = $\frac {1}{42}$

So,$\frac {Area of ABGD}{Area of ABCD} = 1-3(a+b) = 1- \frac {15}{42} = \frac {9}{14}$

Comparing the given values from the question , $\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$

m = 9 and n = 14

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## Problem on Area of Circle | SMO, 2010 (Junior) | Problem 29

Try this beautiful problem on area of circle from SMO, Singapore Mathematics Olympiad, 2010.

## Problem – Area of Circle (SMO Test)

Let ABCD be a rectangle with AB = 10 . Draw circles $C_1$ and $C_2$ with diameters AB and CD respectively. Let P,Q be the intersection points of $C_1$ and $C_2$ . If the circle with diameter PQ is tangent to AB and CD , then what is the area of the shaded region ?

• 25
• 20
• 22
• 23

### Key Concepts

Area of Circle

2D – Geometry

Area of Rectangle

Challenges and thrills – Pre – college Mathematics

## Try with Hints

If you are really got stuck with this sum then we can start from here:

The diagram will be like this . So let ‘N’ be the midpoint of CD .

so $\angle {PNQ} = 90^\circ$

so PQ = 5 $\sqrt {2}$

Now let us try to find the area of the shaded region

A = $2 [ \frac {1}{2}\pi (\frac {PQ)}{2})^2 + \frac {1}{2}(PN)^2 – \frac {1}{4} \pi (PN)^2]$

= $2 [\frac {1}{2}\pi (\frac {5\sqrt{2}}{2})^2+\frac {1}{2}.5^2 – \frac {1}{4}\pi . 5^2]$

= 25

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## Area of Square – Singapore Mathematical Olympiad – 2013 – Problem No.17

Try this beautiful problem from Singapore Mathematical Olympiad. 2013 based on the area of Square.

## Problem – Area of Square

Let ABCD be a square and X and Y be points such that the lengths of XY, AX, and AY are 6,8 and 10 respectively. The area of ABCD can be expressed as $\frac{m}{n}$ units where m and n are positive integers without common factors. Find the value of m+n.

• 1215
• 1041
• 2001
• 1001

### Key Concepts

2D Geometry

Area of Square

Singapore Mathematical Olympiad – 2013 – Junior Section – Problem Number 17

Challenges and Thrills –

## Try with Hints

This can the very first hint to start this sum:

Assume the length of the side is a.

Now from the given data we can apply Pythagoras’ Theorem :

Since, $6^2+8^2 = 10^2$

so $\angle AXY = 90^\circ$.

From this, we can understand that $\triangle ABX$ is similar to $\triangle XCY$

Try to do the rest of the sum……………………

From the previous hint we find that :

$\triangle ABX \sim \triangle XCY$

From this we can find $\frac {AX}{XY} = \frac {AB}{XC}$

$\frac {8}{6} = \frac {a}{a – BX}$

Can you now solve this equation ?????????????

This is the very last part of this sum :

Solving the equation from last hint we get :

a = 4BX and from this we can compute :

$8^2 = {AB}^2 +{BX}^2 = {16BX}^2 + {BX}^2$

so , $BX = \frac {8}{\sqrt {17}} and \(a^2 = 16 \times \frac {64}{17} = \frac {1024}{17}$

Thus m + n = 1041 (Answer).

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## Objective Problem Geometry (ISI Entrance)

Find the radius of smaller circle.

• 0
• 1
• $\frac{3}{4}$
• 2

### Key Concepts

2D Geometry

Similar Triangles

Linear Equations

Answer: $\frac{3}{4}$

ISI Entrance B.Stat Objective Problem, India

Test of Mathematics at 10+2 Level by East West Press

## Try with Hints

First hint

Given triangle to smaller circle and given triangle to bigger circle are similar.

Second Hint

By similarity,

$\frac{2-x}{5}=\frac{x}{3}$

Final Step

Solving we get x=$\frac{3}{4}$

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## Problem related to triangle – AMC 10B, 2019 Problem 10

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.

## 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.