Question

If P(2,-1), Q(3,4) R(-2,3) and S(-3,-2) be 4 points in a plane. Show that PQRS is a rhombus but not a square. Find the area of rhombus.

Answer 1

Answer:

The proof is explained strep-wise below :

Step-by-step explanation:

First finding the lengths of all the sides :

$PQ=\sqrt{(2-3)^2+(-1-4)^2}\\\\\implies PQ=\sqrt{26}\\\\QR=\sqrt{(3+2)^2+(4-3)^2}\\\\\implies QR=\sqrt{26}\\\\RS=\sqrt{(-2+3)^2+(3+2)^2}\\\\\implies RS=\sqrt{26}\\\\PS=\sqrt{(2+3)^2+(-1+2)^2}\\\\\implies PS=\sqrt{26}$

Now, length of all the sides are equal. So PQRS may be square or rhombus

So, finding the length of diagonals :

$PR=\sqrt{(2+2)^2+(-1-3)^2}\\\\\implies PR=\sqrt{32}\\\\QS=\sqrt{(3+3)^2+(4+2)^2}\\\\\implies QS =\sqrt{72}$

The length of diagonals are not equal so PQRS cannot be a square.

Therefore, PQRS is a rhombus not a square.

Hence Proved.

Answer 2

$\large{\underline{\bf{\pink{Answer:-}}}}$

✰ The area of the rhombus is 24sq units.

$\large{\underline{\bf{\blue{Explanation:-}}}}$

$\large{\underline{\bf{\green{Given:-}}}}$

Four points are given as :-

✰ A(2 , -1)

✰ B(3 , 4)

✰ C(-2 ,3)

✰ D(-3, -2)

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the area of the Rhombus.

$\huge{\underline{\bf{\red{Solution:-}}}}$

$:\implies\:AB=\sqrt{(3-2)^2+(4+1)^2}$

$:\implies\:AB=\sqrt{1^2+5^2}$

$:\implies\:AB= \sqrt{26}\:\:units$

$:\implies\:BC=\sqrt{(-2-3)^2+(3-4)^2}$

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

$:\implies\:BC =\sqrt{26}\:\:units$

$:\implies\:CD = \sqrt{(-3+2)^2+(-2-3)^2}$

$:\implies\:CD=\sqrt{(-1)^2+(-5)^2}$

$:\implies\:CD =\sqrt{26}\:\:units$

$:\implies\:DA = \sqrt{(-3-2)^2+(-2+1)^2}$

$:\implies\:DA=\sqrt{(-5)^2+(-1)^2}$

$:\implies\:CD =\sqrt{26}\:\:units$

So, AB = BC = CD = DA = $\:\sqrt{26}$

Now ,

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

$:\implies\:AC = \sqrt{(-4)^2+4^2}$

$:\implies\:AC=\sqrt{32}$

$:\implies\:AC=4\sqrt{2}$

$:\implies\:BD=\sqrt{(-3-3)^2+(-2-4)^2}$

$:\implies\:BD = \sqrt{(-6)^2+(-6)^2}$

$:\implies\:BD=\sqrt{72}$

$:\implies\:BD=6\sqrt{2}$

But diagonal AC ≠ diagonal BD.

so, ABCD is a quadrilateral in which all sides are equal but diagonals are not equal.

So we can say that ABCD is a rhombus but not a square.

Now,

Area of rhombus ABCD

${\purple{\bf{=1/2 ×(Product \:of \:diagonals)}}}$

$:\implies\:\frac{1}{2}\times\:AC\times\:BD$

$:\implies\:\frac{1}{2}\times\:4\sqrt{2}\times\:6\sqrt{2}$

$:\implies{\pink{\bf{\:24\:sq\:units.}}}$

Hence,

$\small{\red{\bf{ Area\: of\: rhombus\: ABCD\:<em> </em>is \:24\: sq\: unit's.}}}$

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