Area, Area and Area.

Geometry Level 5

In a Δ A B C \Delta ABC D, E and F are mid-points of BC, AC and AB respectively. If P, Q, R are the points on AD, BE and CF such that A P A G = α \frac{AP}{AG} = \alpha , B Q B G = β \frac{BQ}{BG} = \beta and C R C G = γ \frac{CR}{CG} = \gamma . Find area of Δ P Q R \Delta PQR .

Given that G G is the centroid of Δ A B C \Delta ABC , α = 1 2 \alpha = \frac{1}{2} , β = 2 3 \beta = \frac{2}{3} , γ = 3 4 \gamma = \frac{3}{4} and area of Δ A B C = 4 unit 2 \Delta ABC = 4 \text{ unit}^2 .

Try more from my set Geometry Problems .


The answer is 0.5.

View wiki
Purushottam Abhisheikh by Purushottam Abhisheikh , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Since G is the centroid of Δ A B C \Delta ABC so area of Δ A G B \Delta AGB = area of Δ B G C \Delta BGC = area of Δ C G A \Delta CGA = area of Δ A B C 3 \frac{{\text{area of}}\Delta ABC}{3}

In Δ A G B \Delta AGB

A P A G = α \frac{AP}{AG} = \alpha ⟹ \implies P G A G = ( 1 − α ) \frac{PG}{AG} = (1-\alpha) or P G = ( 1 − α ) A G PG = (1-\alpha)AG . Similarly Q G = ( 1 − β ) B G QG = (1-\beta)BG

Now area of Δ P Q G = 1 2 P G × Q G × s i n ( ∠ P G Q ) \Delta PQG = \frac{1}{2} PG\times QG\times sin(\angle PGQ)

= 1 2 ( 1 − α ) A G × ( 1 − β ) B G × s i n ( ∠ P G Q ) =\frac{1}{2} (1-\alpha)AG\times (1-\beta)BG\times sin(\angle PGQ)

So, area of Δ P Q G = ( 1 − α ) ( 1 − β ) × ( a r e a o f Δ A G B ) \Delta PQG = (1-\alpha)(1-\beta)\times (area~ of ~\Delta AGB)

or area of Δ P Q G = ( 1 − α ) ( 1 − β ) × ( a r e a o f Δ A B C ) 3 \boxed{\Delta PQG = (1-\alpha)(1-\beta)\times \frac{(area ~ of~\Delta ABC)}{3}} .

Similarly area of Δ Q G R = ( 1 − β ) ( 1 − γ ) × ( a r e a o f Δ A B C ) 3 \Delta QGR = (1-\beta)(1-\gamma)\times \frac{(area ~ of~\Delta ABC)}{3} and area of Δ P G R = ( 1 − γ ) ( 1 − α ) × ( a r e a o f Δ A B C ) 3 \Delta PGR = (1-\gamma)(1-\alpha)\times \frac{(area ~ of~\Delta ABC)}{3} .

Now area of Δ P Q R \Delta PQR = area of Δ P Q G \Delta PQG + area of Δ Q G R \Delta QGR + area of Δ P G R \Delta PGR

⟹ \implies area of Δ P Q R = ∑ ( 1 − α ) ( 1 − β ) 3 × a r e a o f Δ A B C \Delta PQR = \frac{\sum(1-\alpha)(1-\beta)}{3}\times ~area ~of~ \Delta ABC

Putting all the given values, a r e a o f Δ P Q R = 0.5 s q . u n i t s \boxed{area ~of~\Delta PQR = 0.5 sq. units}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...