Question

p+ q + r + 1. (p =.-3 q =.0 r = 2​

Answer 1

The given points are P(2,-1), Q(3,4), R(-2,3) and S(-3,-2).We have

PQ=

(3−2)

2

+(4+1)

2

=

1

2

+5

2

=

26

units

QR=

(−2−3)

2

+(3−4)

2

=

25+1

=

26

units

RS=

(−3+2)

2

+(−2−3)

2

=

1+25

=

26

units

SP=

(−3−2)

2

+(−2−3)

2

=

26

units

PR=

(−2−2)

2

+(3+1)

2

=

16+16

=4

2

units

and, QS=

(−3−3)

2

+(−2−4)

2

=

36+36

=6

2

units

∴PQ=QR=RS=SP=

26

units

and, PR



=QS

This means that PQRS is quadrilateral whose sides are equal but diagonals are not equal.

Thus, PQRS is a rhombus but not a square.

.Now, Area of rhombus PQRS=

2

1

×(Productoflengthsofdiagonals)

⇒AreaofrhombusPQRS=

2

1

×(PR×QS)

⇒AreaofrhombusPQRS=(

2

1

×4

2

×6

2

)sq.units=24sq.units.

$\huge \fbox \pink{change \: ur \: dp \: into \: full \: face \: its \: so \: beautiful}$

Answer 2

HOPE IT HELPS YOU...

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