# 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 .

# 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

# 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

# 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).

# Geometry of Tangents | ISI Entrance B.Stat 2009

## 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=$\frac{3}{4}$

# 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.