In ∆LMN, medians MX and NY are perpendicular to each other and intersect at Z. If MX = 20 cm and NY = 30 cm, what is the area of ∆LMN (in cm²)?

Given:

∆LMN with medians MX and NY

MX ⊥ NY

MX intersects NY at Z

MX = 20 cm

NY = 30 cm

Formula Used:

The centroid of a triangle divides each median in the ratio 2:1.

Area of a triangle = 1/2 × base × height

Area of ∆LMN = 3 × Area of ∆MNZ (since Z is the centroid)

Calculations:

Since Z is the centroid, it divides the medians in the ratio 2:1.

MZ : ZX = 2 : 1

⇒ MZ = (2/3) × MX = (2/3) × 20 = 40/3 cm

⇒ ZX = (1/3) × MX = (1/3) × 20 = 20/3 cm

NZ : ZY = 2 : 1

⇒ NZ = (2/3) × NY = (2/3) × 30 = 20 cm

⇒ ZY = (1/3) × NY = (1/3) × 30 = 10 cm

Since MX ⊥ NY, ∆MNZ is a right-angled triangle with legs NZ and MZ.

Area of ∆MNZ = 1/2 × base × height = 1/2 × NZ × MZ

⇒ Area of ∆MNZ = 1/2 × 20 × (40/3)

⇒ Area of ∆MNZ = 10 × (40/3) = 400/3 cm²

Area of ∆LMN = 3 × Area of ∆MNZ

⇒ Area of ∆LMN = 3 × (400/3)

⇒ Area of ∆LMN = 400 cm²

∴ The area of ∆LMN is 400 cm².