Question

Find the area of the Rhombus ABCD, whose vertices are taken in order, are A(– 1,1), B(1, –2),

C(3, 1), D(1, 4).​

Answer 1

Answer:

12

Step-by-step explanation:

The area of rhombus = $\frac{pq}{2}$

where p and q are the diagonals of rhombus

A (-1,1) B ( 1 ,-2 ) C ( 3,1) D(1 , 4 )

The diagonal

AC = $\sqrt{(3 + 1 )^{2} + (1-1)^{2} }$     = 4

BD = $\sqrt{(1 -1 )^{2} + ( 4 + 2 )^{2} }$      =6

Area = $\frac{6 * 4}{2}$   = 12

Answer 2

Answer:

24

Step-by-step explanation:

Let A(3,0),B(4,5), C(−1,4) and D(−2,−1) are the vertices of rhombus ABCD.

AC and BD are the diagonals of rhombus.

∴BD=

(x

1

−x

2

)

2

+(y

1

−y

2

)

2

Here x

1

=4,x

2

=−2 and y

1

=5,y

2

=−1

⇒BD=

(4−(−2))

2

+(5−(−1))

2

⇒BD=

6

2

+6

2

=

36+36

=

72

=6

2

∴AC=

(x

1

−x

2

)

2

+(y

1

−y

2

)

2

Here x

1

=3,x

2

=−1 and y

1

=0,y

2

=4

⇒BD=

(3−(−1))

2

+(0−4)

2

⇒BD=

4

2

+(−4)

2

=

16+16

=

32

=4

2

Area of rhombus =

2

1

×Product of diagonals

=

2

1

×6

2

×4

2

=

2

24×2

=24 Sq.unit