Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B (2 +√3 , 5) and C(2, 6).
Answers
Thank you for asking this question. Here is your answer:
If there is a diagonal in a parallelogram it bisects into two equal triangles:
which will be:
ar(llgm ABCD) = 2ar (ΔABC) -- (this will be an equation 1)
The area of triagnle with the vertices (x₁,y₁),(x₂,y₂) and (x₃,y₃)
= 1/2 [x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)]
Area of ΔABC = 1/2 [2(5-6) + (2+√3) (6-4) + 2(4-5)]
= 1/2 [-2 + 4 + 2√3-2]
= 1/2 (2√3)
= √3
so ar(llgm ABCD) = 2ar (ΔABC)
= ar(llgm ABCD) = 2 x √3
= 2√3 is the final answer for this question.
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Answer:
We know that the diagonal of a parallelogram divides the parallelogram into 2 equal triangles.
So, Area of parallelogram,
ABCD=2× area of ΔABC ......... (1)
We know that,
Area of ΔABC with vertices (x
1
,y
1
),(x
2
,y
2
),(x
3
,y
3
) is
A=
2
1
×[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
So, Area of ΔABC with vertices [A(2,4),B(2+
3
,5),C(2,6)] is
A=
2
1
×[2(5−6)+(2+
3
)(6−4)+2(4−5)]
A=
2
1
×[−2+2(2+
3
)−2]
A=
2
1
×[−2+4+2
3
−2]
A=
3
Put this value in equation (1), we get,
Area of parallelogram,
ABCD=2× area of ΔABC
ABCD=2
3
sq units
Hence, area of the parallelogram is 2
3
sq units.