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# Show that the Points (−3, 2), (−5,−5), (2, −3) and (4, 4) Are the Vertices of a Rhombus. Find the Area of this Rhombus. - CBSE Class 10 - Mathematics

ConceptConcepts of Coordinate Geometry

#### Questions

Show that the points (−3, 2), (−5,−5), (2, −3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.

Show that A(-3, 2), B(-5,-5), C(2,-3) and D(4,4) are the vertices of a rhombus.

#### Solution 1

The distance d between two points (x_1,y_1) and (x_2-y_2) is given by the formula.

d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)

In a rhombus, all the sides are equal in length. And the area ‘A’ of a rhombus is given as

A = 1/2(Product of both diagonals)

Here the four points are A(3,2), B(5,5), C(2,3) and D(4,4)

First, let us check if all the four sides are equal.

AB = sqrt((-3+5)^2 + (2 + 5)^2)

=sqrt((2)^2 + (7)^2)

=sqrt(49 + 4)

AB=sqrt(53)

BC =sqrt((-5-2)^2 + (-5+3)^2)

= sqrt((-7)^2 + (-2)^2)

=sqrt(49 + 4)

BC = sqrt(53)

CD = sqrt((2- 4)^2 + (-3 - 4)^2)

sqrt((-2)^2 + (-7)^2)

= sqrt(4 + 49)

CD = sqrt(53)

AD = sqrt((-3-4)^2 + (2 - 4)^2)

= sqrt((-7)^2 + (-2)^2)

= sqrt(49 + 4)

AD = sqrt53

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is a rhombus.

Now let us find out the lengths of the diagonals of the rhombus.

AC = sqrt((-3-2)^2 + (2 + 3))

= sqrt((-5)^2 + (5)^2)

= sqrt(25 + 25)

= sqrt(50)

AC = 5sqrt2

BD = sqrt((-5-4)^2 + (-5-4)^2)

= sqrt((-9)^2 + (-9)^2)

= sqrt(81 + 81)

= sqrt162

BD = 9sqrt2

Now using these values in the formula for the area of a rhombus we have,

A = ((5sqrt2)(9sqrt2))/2

= ((5)(9)(2))/2

A = 45

Thus the area of the given rhombus is 45 square units

#### Solution 2

The given points are A(-3, 2), B(-5,-5), C(2,-3) and D(4,4).

AB = sqrt((-5+3)^2 +(-5-2)^2) = sqrt((-2)^2 +(-7) ^2) = sqrt(4+49) = sqrt(53)   units

BC = sqrt((2+5)^2 +(-3+5)^2 )= sqrt((7)^2 +(2)^2) = sqrt(4+49)= sqrt(53)  units

CD = sqrt((4-2)^2 +(4+3)^2 )= sqrt((2)^2 +(7)^2) = sqrt(4+49)= sqrt(53)  units

DA = sqrt((4+3)^2 +(4-2)^2 )= sqrt((7)^2 +(2)^2) = sqrt(4+49)= sqrt(53)  units

Therefore AB =BC=CD=DA= sqrt(53)  units

Also, AC =- sqrt((2+3)^2 +(-3-2)^2) = sqrt((5)^2 +(-5)^2 ) = sqrt(25+25) = sqrt(50) = sqrt(25xx2) = 5 sqrt(2)  units

BD = sqrt((4+5)^2 +(4+5)^2) = sqrt((9)^2 +(9)^2) = sqrt(81+81) = sqrt(162) = sqrt(81 xx 2) = 9 sqrt(2)  units

Thus, diagonal AC is not equal to diagonal BD.

Therefore ABCD is a quadrilateral with equal sides and unequal diagonals

Hence, ABCD a rhombus
Area of a rhombus = 1/2 xx (product of diagonals)

= 1/2 xx (5 sqrt(2) )xx (9 sqrt(2) )

(45(2))/2

= 45 square units.

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Solution Show that the Points (−3, 2), (−5,−5), (2, −3) and (4, 4) Are the Vertices of a Rhombus. Find the Area of this Rhombus. Concept: Concepts of Coordinate Geometry.
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