Question

find the area of quadrilateral pqrs whose vertices are p(-5 -3) q(-4 -6) r(2 -3) s(1 2)
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Answer 1

Answer:

Taking ABP as right angled △. 

BP=25144+x2 (Applying pythogoras) ⟶(1) 

Taking PDC as right angled △. 

PC=2144+(5−x)2                                ⟶(2)

Now,

BP and PC are the two non-hypotenus side of

 another right angled △BPC. 

So, by pythogoras theorem:-

BP2+PC2=BC2=25 ....... from (1)&(2)

⇒25144+x2+25144+(5−x)2=25 

⇒25288+(x2+25+x2−10x)=25

 

⇒25288+2x2−10x=0 

⇒50x2−250x+288=0

hope it's helps