Question

if P(-5,-3), Q(-4,-6), R(2,-3) and S(1,2) are the vertices of a quadrilateral PQRS, then find its area.

Answer 1

Answer:

$\red {Area \: of \: PQRS}\green {= 33\:sq\:units }$

Step-by-step explanation:

$Given \: P(-5,-3), Q(-4,-6), R(2,-3) \:and \\S(1,2) \: are \:the \: vertices \: of \: a \: quadrilateral \:PQRS$

$\underline { Finding \:Area\:of \:\triangle PQR }$

$\pink { Area \: of \: triangle \: whose \: vertices\\are \: (x_{1}, y_{1}), \: (x_{2}, y_{2})\:and \: (x_{3}, y_{3})\: is}$

$\pink {= \frac{1}{2} |x_{1}(y_{2 }- y_{3})+x_{2}(y_{3 }- y_{1})+x_{3}(y_{1 }- y_{2})|}$

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

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

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

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

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

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

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

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

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

Therefore.,

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

$= \frac{21}{2} + \frac{35}{2}\\= \frac{21+35}{2}\\= \frac{66}{2}\\= 33\: sq\:units$

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Answer 2

Answer:

Step-by-step explanation: