Question

Let P be an interior point of ∆ABC . Match the correct entries for the ratios of the Area of ∆PBC : Area of ∆ PCA: Area of ∆ PAB depending on the position of the point P w.r.t. ∆ ABC.

	Column I		Column II
(A)	If P is centroid (G)	(p)	tanA : tanB : tanC
(B)	If P is in centre (I)	(q)	sin2A : sin 2B : sin2C
(C)	If P is orthocenter (H)	(r)	sinA : sinB : sinC
(D)	If P is circum centre (S)	(s)	1 : 1 : 1
		(t)	cosA : cosB : cosC

Answer 1

Correct option A → s; B → r; C → p; D → q

Explanation:

(A) Centroid divides the triangle into three equal part so 1 : 1 : 1

(B) P is incentre 

PD = r 

BC = a 

Area (ΔPBC) = ar 

hence ratio = a : b : c = sinA : sinB : sinC 

(C) P is orthocentre

Area of (ΔPBC) = 1/2 x PD x BC = a/2 x (AD + AP)

= a/2(2RsinBsinC - 2RcosA)

= aR(sinB sinC – cosA) = aR[sinB sinC + cos(B + C)

= 2R cosB cos C = 2R sinA cosB cosC

Hence ratio tanA : tanB: tanC

(D) If P is circumcentre

PB = R 

BD = a/2

PD = √(R² - a²/4)

Area ΔPBC = 1/2 x PD x BC

= a/2√(R² - a²/4) = a/2√(a²/4sin2A - a²/4)

= a⁴/4((cosA/sinB)) = (2R)²sin2A

Ratio = sin2A : sin2B : sin2C