Question

find the area of rhombus if it's vertices are(3,0),(4,5),(-1,4) and (-2,1) taken in order

Answer 1

HI !

Let A(3,0) , B(4,5) , C(-1,4) and D(-2,1) be the vertices of the rhombus

Area of rhombus = 1/2 d*D

where d and D are the diagonals of the rhombus.

In ABCD, AC and BD are the diagonals.

So , lets find their lengths by distance formula :-

distance formula = $\sqrt{[x _{2} - x_{1} ] ^{2} + [ y_{2} - y_{1} ] ^{2} }$

AC = $\sqrt{[-1-3] ^{2} + [4 -0] ^{2} } = \sqrt{32} = 4 \sqrt{2}$

BD = $\sqrt{[4-(-2)] ^{2} + [5 - 1] ^{2} } = \sqrt{52} = 2 \sqrt{13}$

Area of ABCD = 1/2*AC*BD
                        = 1/2*4√2*2√13
                         = 4√26 sq. units
Area of the rhombus is 4√26 sq. units

Answer 2

Area of rhombus = 1/2 x d1 x d2 

d1 = AC

d2  = BD 

AC = √(-1-3)² + (4 -0)² = √32 

BD = √(4-(-2))² + (5-1)² ) = √52 

Area = 1/2 x √32 x √52 

= 1/2 x 5.65 x 7.21 

= 40.74 /2 = 20.37≈ 20.4 (unit)²