Centers of Triangle MCQ Quiz - Objective Question with Answer for Centers of Triangle - Download Free PDF
Last updated on Jun 2, 2025
Latest Centers of Triangle MCQ Objective Questions
Centers of Triangle Question 1:
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)?
Answer (Detailed Solution Below)
Centers of Triangle Question 1 Detailed Solution
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.
Top Centers of Triangle MCQ Objective Questions
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)?
Answer (Detailed Solution Below)
Centers of Triangle Question 2 Detailed Solution
Download Solution PDFGiven:
∆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.
Centers of Triangle Question 3:
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)?
Answer (Detailed Solution Below)
Centers of Triangle Question 3 Detailed Solution
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.