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## 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.

• $14$
• $10$
• $7$

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 $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]}$= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$

So $p+q$=$9+1$=$10$

Categories

## Area of 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 $FDBG$?

• $79$
• $75$
• $82$

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=$\frac{1}{2} (BG+DF) \times FG$=$\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}$=$\frac{ah}{2} \times \frac{15}{24}$=$120 \times \frac{15}{24}$=$75$ $(from ……..(1))$

Categories

## 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 $FDBG$?

• $79$
• $75$
• $82$

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=$\frac{1}{2} (BG+DF) \times FG$=$\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}$=$\frac{ah}{2} \times \frac{15}{24}$=$120 \times \frac{15}{24}$=$75$ $(from ……..(1))$

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## Squares and Triangles | AIME I, 2008 | Question 2

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

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 $\frac{h+10}{10}=\frac{h}{6}$

then h=15 and h+10=25.