Centroid Problem: Ratio of the areas of two Triangles
Try this beautiful problem from Geometry based on Centroid.
Centroid Problem: Ratio of the areas of two Triangles
\(\triangle ABC\) has centroid \(G\).\(\triangle ABG\),\(triangle BCG\), and \(\triangle CAG\) have centroids \(G_1\), \(G_2\), \(G_3\) respectively. The value of \(\frac{[G1G2G3]}{[ABC]}\) can BE represented by \(\frac{p}{q}\) for positive integers \(p\) and \(q\).
Find \(p+q\) where\([ABCD]\) denotes the area of ABCD.
$14$
$ 10$
$7$
Key Concepts
Geometry
Triangle
centroid
Check the Answer
Answer: \(10\)
Question Papers
Pre College Mathematics
Try with Hints
\(\triangle ABC\) has centroid \(G\).\(\triangle ABG\),\(triangle BCG\), and \(\triangle CAG\) have centroids \(G1\),\(G2\),\(G3\) respectively.we have to find out value of \(\frac{[G1G2G3]}{[ABC]}\) i.e area of \(\frac{[G1G2G3]}{[ABC]}\)
Let D, E, F be the midpoints of BC, CA, AB respectively. Area of $\frac{[DEF]}{[ABC]}$=$\frac{1}{4}$
we know that any median is divided at the centroid $2:1$. Now can you find out \(GG_1,GG_2,GG_3\) ?
Can you now finish the problem ..........
we know that any median is divided at the centroid $2:1$ Now $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G_1F$ Similarly,$GG_2 = 2G_2D$ and$GG_3 = 2G_3E$ Thus, From homothetic transformation $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$ Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$
can you finish the problem........
Therefore we say that $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]} $= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$
Try this beautiful problem from Geometry based on the Area of the Trapezium
Area of the Trapezium - AMC-10A, 2018- Problem 24
Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?
$79$
$ 75$
$82$
Key Concepts
Geometry
Triangle
Trapezium
Check the Answer
Answer: $75$
AMC-10A (2018) Problem 24
Pre College Mathematics
Try with Hints
We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm...
So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)
can you find out the value of \(BG,DF \) and height between them....?
Can you now finish the problem ..........
Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.
Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)....................(1)
From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)
Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)
now can you find out the area of Trapezium ?
can you finish the problem........
Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ........(1))\)
Area of the Trapezium | AMC-10A, 2018 | Problem 24
Try this beautiful problem from Geometry based on the Area of the Trapezium.
Area of the Trapezium - AMC-10A, 2018- Problem 24
Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?
$79$
$ 75$
$82$
Key Concepts
Geometry
Triangle
Trapezium
Check the Answer
Answer: $75$
AMC-10A (2018) Problem 24
Pre College Mathematics
Try with Hints
We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm...
So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)
can you find out the value of \(BG,DF \) and height between them....?
Can you now finish the problem ..........
Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.
Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)....................(1)
From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)
Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)
now can you find out the area of Trapezium and area of Triangle?
can you finish the problem........
Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ........(1))\)
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Squares and Triangles.
Squares, triangles and Trapezium - AIME I, 2008
Square AIME has sides of length 10 units, isosceles triangle GEM has base EM, and the area common to triangle GEM and square AIME is 80 square units.Find the length of the altitude to EM in triangle GEM.
is 107
is 25
is 840
cannot be determined from the given information
Key Concepts
Squares
Trapezium
Triangles
Check the Answer
Answer: is 25.
AIME I, 2008, Question 2
Geometry Revisited by Coxeter
Try with Hints
let X and Y be points where the triangle intersects the square and [AXE]=[YIM]=\(\frac{100-80}{2}\)=10 then AX=YI=2 units then XY=10-4=6 units
triangle GXY is similar to triangle GEM where h=height of triangle GXY then by similarity \(\frac{h+10}{10}=\frac{h}{6}\)
Try this beautiful problem from Geometry based on Trapezium.
Geometry Problem based on Trapezium | PRMO-2018 | Problem 5
Let ABCD be a trapezium in which AB||CD and AD is perpendicular on AB .suppose has an incircle which touches AB at Q and CD at P.Given that PC=36 and QB=49. Find PQ?
$64$
$81$
$84$
Key Concepts
Geometry
Trapezoid
Circle
Check the Answer
Answer:$84$
PRMO-2018, Problem 5
Pre College Mathematics
Try with Hints
Let the radius of the inner circle be r
Therefore AQ=PD=r and AD=2r
Can you now finish the problem ..........
Draw a perpendicular from C on AB at the point F
Can you finish the problem........
Let the inner circle touches BC at E
Then CE=36 (as BE & BQ are tangents)
BE=49 (as CE & PC are tangents)
Let the radius of the inner circle be r
Therefore AQ=PD=r and AD=2r
Let draw a perpendicular from C on AB at the point F
