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
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........
\(\frac{[A D E]}{[B C E D]}=\frac{[A D E]}{[A B C]-[A D E]}\) =\(\frac{1}{[A B C] /[A D E]-1}\) =\(\frac{1}{75 / 19-1}\)
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
Let \(\angle APB= \theta\)
area of \(\Delta PAB\)=\(\frac{1}{2}(3)(4)sin\theta\)
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 \(\Delta PCD\)=\(\frac{1}{2}(8)(4)sin(\pi-\theta)\)
or, area of quadrilateral ABCD is \(\frac{1}{2}(12+18+48+32)sin{\theta}\)
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
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 \(\frac{\triangle ADE}{quad.BCED}\)=\(\frac{\triangle ADE}{{\triangle ABC}-{\triangle ADE}}\)=\(\frac{1}{\frac{\triangle ABC}{\triangle ADE}-1}=\frac{1}{\frac{75}{19}-1}=\frac{19}{56}\)
