# Triangle and Quadrilateral | AMC-10A, 2005 | Problem 25

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral from AMC 10A, 2005.

## Ratios of the areas of Triangle and Quadrilateral - AMC-10A, 2005- Problem 25

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?

• $$\frac{19}{56}$$
• $$\frac{19}{66}$$
• $$\frac{17}{56}$$
• $$\frac{11}{56}$$
• $$\frac{19}{37}$$

### Key Concepts

Geometry

Triangle

Answer: $$\frac{19}{56}$$

AMC-10A (2005) Problem 25

Pre College Mathematics

## Try with Hints

Given that $AB = 25$, $BC = 39$, and $AC=42$.we have to find out Ratios of the areas of Triangle$$\triangle ADE$$ and the quadrilateral $$CBED$$.So if we can find out the area the $$\triangle ADE$$ and area of the $$\triangle ABC$$ ,and subtract $$\triangle ADE$$ from $$\triangle ABC$$ then we will get area of the region $$CBDE$$.Can you find out the area of $$CBDE$$?

Can you find out the required area.....?

Now $$\frac{\triangle ADE}{\triangle ABC}=\frac{AD}{AB}.\frac{AE}{AC}=\frac{19}{25}.\frac{14}{42}=\frac{19}{75}$$

Therefore area of $$BCED$$=area of $$\triangle ABC$$-area of $$\triangle ADE$$.Now can you find out Ratios of the areas of Triangle and the quadrilateral?

can you finish the problem........

$$\frac{[A D E]}{[B C E D]}=\frac{[A D E]}{[A B C]-[A D E]}$$
=$$\frac{1}{[A B C] /[A D E]-1}$$
=$$\frac{1}{75 / 19-1}$$

=$$\frac{19}{56}$$

# Covex Cyclic Quadrilateral | PRMO 2019 | Question 23

Try this beautiful problem from the Pre-RMO, 2019 based on Covex Cyclic Quadrilateral.

## Covex Cyclic Quadrilateral - PRMO 2019

Let ABCD be a convex cyclic quadrilateral. Suppose P is a point in the plane of the quadrilateral such that the sum of its distance from the vertices of ABCD is the least. If {PA,PB,PC,PD}={3,4,6,8}, find the maximum possible area of ABCD.

• is 107
• is 55
• is 840
• cannot be determined from the given information

### Key Concepts

Largest Area

Distance

PRMO, 2019, Question 23

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

Let $$\angle APB= \theta$$

area of $$\Delta PAB$$=$$\frac{1}{2}(3)(4)sin\theta$$

Second Hint

area of $$\Delta PAD$$=$$\frac{1}{2}(3)(6)sin(\pi-\theta)$$

area of $$\Delta PDC$$=$$\frac{1}{2}(8)(6)sin{\theta}$$

area of $$\Delta PCD$$=$$\frac{1}{2}(8)(4)sin(\pi-\theta)$$

Final Step

or, area of quadrilateral ABCD is $$\frac{1}{2}(12+18+48+32)sin{\theta}$$

or, maximum area of ABCD=6+9+24+16=55.

# Triangle and Quadrilateral | AMC-10A, 2005 | Problem 25

Try this beautiful problem from Geometry: Area of Triangle and Quadrilateral

## Ratios of the areas of Triangle and Quadrilateral - AMC-10A, 2005- Problem 25

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?

• $$\frac{19}{56}$$
• $$\frac{19}{66}$$
• $$\frac{17}{56}$$
• $$\frac{11}{56}$$
• $$\frac{19}{37}$$

### Key Concepts

Geometry

Triangle

Answer: $$\frac{19}{56}$$

AMC-10A (2005) Problem 25

Pre College Mathematics

## Try with Hints

Given that $AB = 25$, $BC = 39$, and $AC=42$.we have to find out Ratios of the areas of Triangle$$\triangle ADE$$ and the quadrilateral $$CBED$$.So if we can find out the area the $$\triangle ADE$$ and area of the $$\triangle ABC$$ ,and subtract $$\triangle ADE$$ from $$\triangle ABC$$ then we will get area of the region $$CBDE$$.Can you find out the area of $$CBDE$$?

Can you find out the required area.....?

Now $$\frac{\triangle ADE}{\triangle ABC}=\frac{AD}{AB}.\frac{AE}{AC}=\frac{19}{25}.\frac{14}{42}=\frac{19}{75}$$

Therefore area of $$BCED$$=area of $$\triangle ABC$$-area of $$\triangle ADE$$.Now can you find out Ratios of the areas of Triangle and the quadrilateral?

can you finish the problem........

Now $$\frac{\triangle ADE}{quad.BCED}$$=$$\frac{\triangle ADE}{{\triangle ABC}-{\triangle ADE}}$$=$$\frac{1}{\frac{\triangle ABC}{\triangle ADE}-1}=\frac{1}{\frac{75}{19}-1}=\frac{19}{56}$$

# Sides of Quadrilateral | PRMO 2017 | Question 20

Try this beautiful problem from the Pre-RMO, 2017 based on Sides of Quadrilateral.

## Sides of Quadrilateral - PRMO 2017

What is the number of triples (a,b,c) of positive integers such that (i) a<b<c<10 and (ii) a,b,c,10 form the sides of a quadrilateral?

• is 107
• is 73
• is 840
• cannot be determined from the given information

### Key Concepts

Largest number of triples

Distance

PRMO, 2017, Question 20

Geometry Vol I to IV by Hall and Stevens

First hint

a+b+c>10

(a+b+c) can be

a b c

1 2 8,9

1 3 7,8,9

1 4 6 ,7,8,9

1 5 6,7,8,9

1 6 7,8,9

1 7 8,9

1 8 9

Second Hint

2 3 6,7,8,9

2 4 5,6,7,8,9

2 5 6,7,8,9

2 6 7,8,9

2 7 8,9

2 8 9

Final Step

3 4 5,6,7,8,9

3 5 6,7,8,9

3 6 7,8,9

3 7 8,9

3 8 9

4 5 6,7,8,9

4 6 7,8,9

4 7 8,9

4 8 9

5 6 7,8,9

5 7 8,9

5 8 9

6 7 8,9

6 8 9

7 8 9

Total 73 cases.