Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1), taken in order, form a rhombus. Also, find its area.
Answers
The area of the rhombus is 24 square units.
Step-by-step explanation:
Let us assume P(x1,y1), Q(x2,y2), R(x3,y3) and S(x4,y4) to be P(3,0) Q(4,5) R(-1,4) and S(-2,-1) which are the vertices of the rhombus.
The length of diagonal PR = √(x₂ - x₁)² + (y₂ - y₁)²
x₁ = 3 y₁ = 0 x₂ =-1 y₂ = 4
= √(-1 - 3)² + (4 - 0)²
= √(-4)² + (4)²
= √16 + 16
PR = √32
The length of diagonal QS = √(x₂ - x₁)² + (y₂ - y₁)²
x₁ = 4 y₁ = 5 x₂ =-2 y₂ = -1
= √(-2-4)² + (-1-5)²
= √(-6)² + (-6)²
= √36 + 36
QS= √72
Area of rhombus = (1/2) x PR x QS
=(1/2) x √32 x √72
= (1/2)4√2 x 6√2
Area of the rhombus = 24 square units.
Area of the rhombus = ½ * diagonals
= ½*PR*QS
= 1/2*√32*√72
= 1/2*48
=24
Area of the rhombus = ½ * diagonals
Hence proved.
Area of rhombus = = 24 square units.
Step-by-step explanation:
Let's take A(3,0) B(4,5) C(-1,4) and D(-2,-1) be the vertices of the rhombus.
Length of diagonal = AC = BD
AC² = (-1-3)² + (4-0)² = (-4)² + (4)² = 16 + 16 = 32
BD² = (-2-4)² + (-1-5)² = (-6)² + (-6)² = 36 + 36 = 72
Area of rhombus = (1/2) * BD * AC
= (1/2) * √32 * √72
= (1/2) * 4√2 * 6√2
= 24 square units.