Question

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

order. [Hint : Area of a rhombus =

½ (product of its diagonals)​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given

vertices of rhombus A(5,4) ,B(-1,-2),C(0,3) & D(-1,4)

To Find

Area of Rhombus

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

step by step explanation

We have given vertices so,we first find diagonals of Rhombus after that we will it's area.

Vertices of Rhombus=A(5,4),B(-1,-2) ,C(0,3) &D(-1,4)

Let AC & BD be two diagonals of Rhombus

we have not given length of diagonals so,we find its length by using distance formula:

$\sf\boxed{ D= \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2 }- y_{1})}^{2} }}$

$\sf AC = \sqrt{ {(0 - 5)}^{2} + {(3 - 4)}^{2} }$

$\sf \: AC = \sqrt{ {(5)}^{2} + {( - 1) }^{2} } = \sqrt{25 + 1} = \sqrt{26} units$

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

$\sf \: BD = \sqrt{ {(0)}^{2} + {(6)}^{2} } = \sqrt{36} = 6units$

Now,we have length of two diagonals .

$\sf \boxed{\: Area \: of \: Rhombus= \dfrac{D_{1 }\times D_{2}}{2}}$

$\sf \: Area \: of \: Rhombus= \dfrac{ \sqrt{26} \times 6 }{2}$

$\sf \: Area \: of \: Rhombus= 3 \sqrt{26} \: sq.units$