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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.

• $14$
• $10$
• $7$

### Key Concepts

Geometry

Triangle

centroid

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 $FDBG$?

• $79$
• $75$
• $82$

### Key Concepts

Geometry

Triangle

Trapezium

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 Octagon | AMC-10A, 2005 | Problem 20

Try this beautiful problem from Geometry based on the Area of the Octagon from AMC 10A, 2005, Problem 2005

## Area of the octagon – AMC-10A, 2005- Problem 20

An equiangular octagon has four sides of length 1 and four sides of length $\frac{\sqrt{2}}{2}$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?

• $\frac{4+5\sqrt 2}{2}$
• $\frac{7}{2}$
• $7$

### Key Concepts

Geometry

Triangle

Octagon

Answer: $\frac{7}{2}$

AMC-10A (2005) Problem 20

Pre College Mathematics

## Try with Hints

We have to find out the equiangular octagon whose four sides of length 1 and four sides of length $\frac{\sqrt{2}}{2}$,

we join $AD$,$HE$,$BG$ and $CF$.We assume that side lengths of $AB=CD=EF=GH=1$ and side lengths of $AH=BC=DE=GF=\frac{\sqrt{2}}{2}$( As no two consecutive sides have the same length). Now

Can you now finish the problem ……….

There are 5 squares with side lengths $\frac{\sqrt{2}}{2}$ and 4 Triangles of side lengths $1$

Now area of $5$ squares=$5 \times (\frac{\sqrt{2}}{2})^2$=$\frac{5}{2}$ and area of each Triangle is half of the area of a square.so the area of $4$ Triangles=$4 \times \frac{1}{2} \times \frac{1}{2}$=$1$

can you finish the problem……..

Therefore the Total area of the required octagon=Total area of Five squares + Total areas of Four Triangles=$\frac{5}{2} +1$=$\frac{7}{2}$

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 $FDBG$?

• $79$
• $75$
• $82$

### Key Concepts

Geometry

Triangle

Trapezium

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