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5

1)

In , we know that  and  Find

2)

In quadrilateral   and  We know that  and  What is

3)

Two diagonals of quadrilateral  are perpendicular to each other at  We know  and  What is

4)

In , we have , and . Point  is on  such that  bisects . If , then what is

5)

Points  and  are on side  of rectangle  such that  and  trisect . If  and , then what is the area of ?

Oct 28, 2017

#1
+99580
+2

1)  Draw altitude  BD.....this will bisect angle ABC

Then angle BCA  = 30°

And since  triangle BDC is a 30 - 60- 90 triangle...then  CD  =  [√3 / 2] [ 6 √3]  = 9

And CD  =  (1/2) AC

So   AC  = 2CD =  18

Oct 28, 2017
#2
+99580
+2

2)  Draw DB

Then DB   is a hypotenuse of triangles DBA, DBC

And its length  is  √  [  (√13)^2  + (√5)^2 ]  =  √18  = 3√2

Then because BC  = CD  then angle CDB  =   angle CBD  = 45°

So because  triangle  DBC  is a 45 - 45 - 90 triangle, BC  =   [ 3√2  ] / √2  =  3

Oct 28, 2017
edited by CPhill  Oct 28, 2017
#3
+99580
+1

3)  AO  = BO   and CO = DO

Then the angles opposite these sides  =  45°

And since triangles ABO and DCO   are 45 - 45 - 90  triangles

Then    DO  =  DC/√2     and  BO  = AB / √2

So   DO  + BO  =    DC/√2 +  AB/√2   =   [ DC + AB ] / √2

And  DO + BO  = BD  =    [ DC + AB ] / √2   =  [ 5 √2 ]  /  √2  =     5

Oct 28, 2017
#4
+99580
+1

4)  Since QX bisects  PQR, then  angles  PQX = RQX = QPX  = 30°

Then angle QXP  =  120°...so angle QXR  = 60°

And triangles  QRP and QRX   are   30 - 60 - 90  triangles

So.........since QR is opposite a 30° angle, it is  (1/2)PQ =  (1/2) 4√3  = 2√3

So.......QX =  [ 2 / √3 ] * [ 2√3 ]  =  4

Oct 28, 2017
#5
+99580
+1

And triangle SAD  is a 30 - 60 - 90  triangle

So.......AD  =  √3SD  = √3 * 1  = √3

And angle TAD  = 60°    so   triangle  ATD  is also a 30 - 60 - 90  triangle

And TD  =  √3 AD  =  √3 √3  =  3

So CD  = CT + TD  = 2√3 - 3  + 3  =  2√3

So......the area of  ABCD  =  CD * AD =  2√3 * √3    =  6  units^2

Oct 28, 2017