How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

# Problem based on Triangle | PRMO-2018 | Problem 21

Try this beautiful problem from Geometry based on Triangle.

## Problem based on Triangle | PRMO | Problem 21

Let ABC be an acute-angled triangle and let H be its orthocentre. Let G1, G2 and G3 be the
centroids of the triangles HBC , HCA and HAB respectively. If the area of triangle G1G2G3 is 7
units, what is the area of triangle ABC ?

• $24$
• $63$
• $34$

Geometry

Triangle

Centroid

## Check the Answer

Answer:$63$

PRMO-2018, Problem 21

Pre College Mathematics

## Try with Hints

Arrange the given number

we have to find out the area of triangle ABC.but there is no given data such as side length of AB or BC ..etc so that we can find out the value of the Triangle, but given that H be its orthocentre and G1, G2 and G3 be the centroids of the triangles HBC, HCA and HAB respectively .so use the orthocentre and centroid property.....

Can you now finish the problem ..........

If we see very carefully $\triangle HG_1G_2$ and $\triangle HDE$ are similar...... and $\triangle G_1G_2G_3$ and $\triangle ABC$ are similar....

Can you finish the problem........

AB = 2DE …..(1)
In $\triangle H G_1 G_2$ & $\triangle H D E$

$\frac{HG_1}{HD}=\frac{G_1G_2}{DE}=\frac{2}{3}$

$G_1G_2=\frac{2}{3} DE =\frac{2}{3}(\frac{AB}{2})=\frac{AB}{3}$

Therefore $\triangle G_1G_2G_3 \sim \triangle ABC$
So we have $\frac {Area of \triangle ABC}{Area of \triangle G_1G_2G_3}$=$\frac{(AB)^2}{(G_1G_2)^2}$=$(\frac{AB}{G_1G_2})^2$=$(\frac{3}{1})^2$

$\Rightarrow Area of \triangle ABC=9 \times (Area of \triangle G_1G_2G_3)$

$\Rightarrow Area of \triangle ABC =9\times 7=63$

# Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

### Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
HALL OF FAMESUPER STARSBOSE OLYMPIADBLOG
CAREERTEAM
support@cheenta.com