Question

Find the area of a rhombus, if its vertices are \[(3,0)\] , \[(4,5)\],\[( - 1,4)\] and \[( - 2, - 1)\] taken in order.

Answer 1

Hint: The half of the product of the diagonals is the area of rhombus.
Here we are going to use these vertices to find the diagonals and then multiply it, then make half of the product.
Thus the area of rhombus is obtained.

Complete step-by-step answer:
Let the vertices be \[A(3,0)\], \[B(4,5)\], \[C( - 1,4)\], \[D( - 2, - 1)\]
               

We know that
Area of Rhombus = \[\dfrac{1}{2} \times \]product of its diagonals
                                = \[\dfrac{1}{2} \times AC \times BD\]
We need to find \[AC\] and \[BD\] using distance formula
First we have to find \[AC\]
Here the value of \[A(3,0)\], \[C( - 1,4)\]
\[AC\] = \[\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
Here \[{x_1} = 3\], \[{y_1} = 0\]
\[{x_2} = - 1\], \[{y_2} = 4\]
Substitute the above value in the $AC$
\[AC\]= \[\sqrt {{{(- 1 - 3)}^2} + {{(4 - 0)}^2}} \]
On adding the inner terms we get,
\[AC\]= \[\sqrt {{{(- 4)}^2} + {{(4)}^2}} \]
Squaring the terms we get,
\[AC\]= \[\sqrt {16 + 16} \]
Adding the two values we get,
\[AC\]= \[\sqrt {32} \]
We can split it into,
\[AC\]= \[\sqrt {16 \times 2} \]
Square root of $16$ is $4$ and we can write it,
\[AC\]= \[4\sqrt 2 \]
Similarly we can find \[BD\]
Here the values of \[B(4,5)\], \[D( - 2, - 1)\]
Now we use the distance formula we get,
\[BD\]= \[\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
Here \[{x_1} = 4\], \[{y_1} = 5\], \[{x_2} = - 2\], \[{y_2} = - 1\]
Substitute the above values we get,
\[BD\]= \[\sqrt {{{( - 2 - 4)}^2} + {{( - 1 - 5)}^2}} \]
On adding the inner terms we get,
\[BD\]= \[\sqrt {{{( - 6)}^2} + {{( - 6)}^2}} \]
Squaring the terms we get,
\[BD\]= \[\sqrt {36 + 36} \]
Adding the two values we get,
\[BD\]= \[\sqrt {72} \]
We can split it into,
\[BD\]= \[\sqrt {36 \times 2} \]
Square root of $36$ is $6$ and we can write it,
\[BD\] = \[6\sqrt 2 \]
Finally, we have to find
Area of Rhombus = \[\dfrac{1}{2} \times AC \times BD\]
Substitute the value of $AC$ and $BD$ we get,
   = \[\dfrac{1}{2} \times 4\sqrt 2 \times 6\sqrt 2 \]
Multiply terms we can split it into we get,
  = \[\dfrac{1}{2} \times (4 \times 6) \times (\sqrt 2 \times \sqrt 2 )\]
Multiply the terms and also the same value of square root is equal to the same value.
  = \[\dfrac{1}{2} \times (24) \times (2)\]
We multiply the terms we get,
   = \[\dfrac{1}{2} \times (48)\]
Divided the terms we get,
   = \[24\] Square units.
Hence, Area of Rhombus = \[24\] square units

Note: Hence, the area of the Rhombus is half of the product of the diagonals; the rhombus is one of the quadrilaterals. Thus the four sides of the rhombus are equal.
We have to find areas of rhombus in so many ways.
It should give in the diagonals, or base and height.