find the area of the Rhombus if its vertices are (3,0) , (4,5) , (-1,4) and (-2,-1) taken in order (area of rhombus = 1/2 product of diagonals
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Let the rhombus have the coordinates as
A(3,0), B(4,5), C(-1,4) and D(-2,-1).
Let us find the lengths of the diagonals AC and BD.
Length of AC = [{3-(-1)}^2 + (0–4)^2}^0.5
= [4^2+4^2]^0.5
= 32^0.5
= 4*2^0.5
Length of BD = [{4-(-2)}^2 +{5-(-1){^2]^0.5
= [6^2+6^2]^0.5
= 72^0.5
= 6*2^0.5
The area of the rhombus ABCD
= AC*BD/2
= 4*2^0.5* 6*2^0.5/2
= 24 sq units.
A(3,0), B(4,5), C(-1,4) and D(-2,-1).
Let us find the lengths of the diagonals AC and BD.
Length of AC = [{3-(-1)}^2 + (0–4)^2}^0.5
= [4^2+4^2]^0.5
= 32^0.5
= 4*2^0.5
Length of BD = [{4-(-2)}^2 +{5-(-1){^2]^0.5
= [6^2+6^2]^0.5
= 72^0.5
= 6*2^0.5
The area of the rhombus ABCD
= AC*BD/2
= 4*2^0.5* 6*2^0.5/2
= 24 sq units.
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Answer:
The diagonal AC is 4 root 2 and diagonal BD is 6root 2. And the answer which is the area is 1/2 * 4 root2 * 6root2 = 24 units
Step-by-step explanation:
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