Question

quadrilateral with of p(- 3 , 2) q (- 5 , - 5 ) R ( 2 ,- 3) and S ( 4 ,4)is a

Answer 1

Answer:

$\huge{\underline{\mathtt{\red{❥A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)∣

= \frac{1}{2}| (-5)[-6-(-3)]+(-4)[-3-(-3)]+2[-3-(-6)]|=

2

1

∣(−5)[−6−(−3)]+(−4)[−3−(−3)]+2[−3−(−6)]∣

= \frac{1}{2} | (-5)(-6+3)+(-4)(-3+3)+2(-3+6)|=

2

1

∣(−5)(−6+3)+(−4)(−3+3)+2(−3+6)∣

= \frac{1}{2} | (-5)(-3) + 0 + 2\times 3|=

2

1

∣(−5)(−3)+0+2×3∣

= \frac{1}{2} | 15 + 6 |=

2

1

∣15+6∣

= \frac{1}{2} \times 21 = \frac{21}{2}\: --(1)=

2

1

×21=

2

21

−−(1)

\underline { Finding \:Area\:of \:\triangle PRS }

FindingAreaof△PRS

= \frac{1}{2}| (-5)[-3-2]+2[2-(-3)]+1[-3-(-3)]|=

2

1

∣(−5)[−3−2]+2[2−(−3)]+1[−3−(−3)]∣

= \frac{1}{2} | (-5)(-5)+2\times 5+ 1\times 0 |=

2

1

∣(−5)(−5)+2×5+1×0∣

\begin{gathered}= \frac{1}{2} | 25+10|\\= \frac{35}{2}\:---(2)\end{gathered}

=

2

1

∣25+10∣

=

2

35

−−−(2)

Therefore.,

Area \: of \: PQRS = ar(\triangle PQR)+ar(\triangle PRS)AreaofPQRS=ar(△PQR)+ar(△PRS)

\begin{gathered}= \frac{21}{2} + \frac{35}{2}\\= \frac{21+35}{2}\\= \frac{66}{2}\\= 33\: sq\:units\end{gathered}

=

2

21

+ 235

= 221+35= 266=33squnits