Question

Verticles of a rhombus...Find its area..!!​

Answer 1

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Answer 2

GIVEN:

• Points (-4,-7),(-1,2),(8,5),(5,4)

TO FIND:

• To show that given points are vertices of rhombus.
• Area of rhombus

FORMULA USED:

• Distance formula =$\sf{\sqrt{{{x}_{2}-{x}_{1}}^{2} + {{y}_{2} - {y}_{1}}^{2}}}$
• Area of rhombus =$\sf{\dfrac{1}{2} \times Di \times D2 \:\:\: (D1\:and\:D2\:are\:diagonal\:of\:rhombus)}$

PROPERTIES OF RHOMBUS:

• All the sides are equal.
• Diagonals are unequal

SOLUTION:

$\sf{Let\: the\: points \:are \:A(-4,-7),\:B(-1,2),\:C(8,5),\:D(5,4)}$

First, we will find distance between points of vertices of sides of the shape.

$\sf\longmapsto{AB\:=\:\sqrt{{-1-(-4)}^{2} + {2-(-7)}^{2}}}$

$\sf\longmapsto{AB\:=\:\sqrt{{3}^{2} + {9}^{2}}}$

$\sf\longmapsto{AB\:=\:\sqrt{90}=3\sqrt{10}}$

$\sf\longmapsto{BC\:=\:\sqrt{{8-(-1)}^{2} + {5-2}^{2}}}$

$\sf\longmapsto{BC\:=\:\sqrt{{9}^{2} + {3}^{2}}}$

$\sf\longmapsto{BC\:=\:\sqrt{90}=3\sqrt{10}}$

$\sf\longmapsto{CD\:=\:\sqrt{{5-8}^{2} + {-4-5}^{2}}}$

$\sf\longmapsto{CD\:=\:\sqrt{{3}^{2} + {9}^{2}}}$

$\sf\longmapsto{CD\:=\:\sqrt{90}=3\sqrt{10}}$

$\sf\longmapsto{DA\:=\:\sqrt{{-4-5}^{2} + {-7-(-4)}^{2}}}$

$\sf\longmapsto{DA\:=\:\sqrt{{9}^{2} + {3}^{2}}}$

$\sf\longmapsto{DA\:=\:\sqrt{90}=3\sqrt{10}}$

Second, we will find distance between diagonal points.

$\sf\longmapsto{AC\:=\:\sqrt{{8-(-4)}^{2} + {5-(-7)}^{2}}}$

$\sf\longmapsto{AC\:=\:\sqrt{{12}^{2} + {12}^{2}}}$

$\sf\longmapsto{AC\:=\:\sqrt{144}=12\sqrt{2}}$

$\sf\longmapsto{BC\:=\:\sqrt{{5-(-1)}^{2} + {-4-2}^{2}}}$

$\sf\longmapsto{BC\:=\:\sqrt{{6}^{2} + {6}^{2}}}$

$\sf\longmapsto{BC\:=\:\sqrt{72}=6\sqrt{2}}$

Here,

• AB = BC = CD = DA
• AC ≠ BC

Given points satisfies the properties of a rhombus.

hence, given points are vertices of a rhombus.

Area of rhombus = $\sf{\dfrac{1}{2} \times 12\sqrt{2} \times 6\sqrt{2}= 72 {unit}^{2}}$

NOTE:-

• For diagram, refer to attachment.
• Swipe from right towards left to see the complete answer :)

@BrainlicaLDoll