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

quadrilateral

Check the Answer


Answer: \(\frac{19}{56}\)

AMC-10A (2005) Problem 25

Pre College Mathematics

Try with Hints


Triangle and Rectangle Figure

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.....?

Triangle and Rectangle Figure

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........


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

=\(\frac{19}{56}\)

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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

Quadrilateral

Distance

Check the Answer


Answer: is 55.

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.

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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

quadrilateral

Check the Answer


Answer: \(\frac{19}{56}\)

AMC-10A (2005) Problem 25

Pre College Mathematics

Try with Hints


Triangle and Rectangle Figure

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.....?

Triangle and Rectangle Figure

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

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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

Quadrilateral

Distance

Check the Answer


Answer: is 73.

PRMO, 2017, Question 20

Geometry Vol I to IV by Hall and Stevens

Try with Hints


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.

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