+0

# Geometry

0
1
2
+360

Let ABCD be a trapezoid with bases AB and CD.  Let P be a point on side CD, and let X, Y be the feet of the altitudes from P to AD, BC respectively.  Prove that if AD = 5, BC = 7, AB = 6, CD = 12, and CP/PD = 1, then PX = 12*sqrt(6)/5 and PY = 12*sqrt(6)/7.

Jun 3, 2024

#1
+759
+1

Let

A =(0,0)

B =(6,0)

Construct a circle centered at A with a radius of 5   and a circle centered at  B with a radius of 7

If we let the x coordinate of D  = -1  we can  find the y coordinate as

(-1)^2 + y^2  = 25

y^2 = 24

y = 2sqrt (6)

Letting  the x coordinate of C = 11, we can  find the y coordinate as

( 11 - 6)^2 + y^2  = 49

5^2 + y^2   =49

y^2 = 24

y = 2sqrt (6)    ( which we would expect )

Since CP / PD = 1

Let  P  =  ([11 -1) /2, 2sqrt 6)  = (5, 2sqrt 6 )

The slope of the line through AD  = [ 2sqrt 6] / -1  =  -2sqrt 6

And the equation of this  line is  y = (-2sqrt 6)x

In standard form we have (2sqrt6)x + y  =  0        (1)

The slope of the line  through  BC  =  [ 2sqrt 6 - 0] / [ 11 - 6] =  (2/5)sqrt 6

And the equation of the line through BC  is  y  = (2/5)sqrt 6 ( x - 6) =

y =(2/5)sqrt (6)x - (12/5)sqrt 6

In standard from we have   (2/5)sqrt (6)x -y - (12/5)sqrt (6)  = 0     (2)

The  formula for the distance  between a point and a  line is  given  by

abs [ Ax + By + C ]  / sqrt [ A^2 + B^2 ]       where (x,y)  are the coordinates of the point

So   using (1) , the distance from P to the line through AD  = PX  is

abs [ (2sqrt (6) (5) + 2sqrt (6) ] /  sqrt [ (2sqrt (6))^2 + 1^2 ] =  [12sqrt (6)]  / 5

Likewise, the distance between P and the line through BC  = PY   is

abs [ (2/5)sqrt (6) (5) - 2sqrt (6) - (12/5)sqrt (6) ]  / sqrt  [ [(2/5)sqrt (6) ]^2 + 1^2 ]   =

[ (12/5)sqrt 6] / (7/5)  =  [12sqrt (6)]  / 7

Thanks! :)

Jun 3, 2024

#1
+759
+1

Let

A =(0,0)

B =(6,0)

Construct a circle centered at A with a radius of 5   and a circle centered at  B with a radius of 7

If we let the x coordinate of D  = -1  we can  find the y coordinate as

(-1)^2 + y^2  = 25

y^2 = 24

y = 2sqrt (6)

Letting  the x coordinate of C = 11, we can  find the y coordinate as

( 11 - 6)^2 + y^2  = 49

5^2 + y^2   =49

y^2 = 24

y = 2sqrt (6)    ( which we would expect )

Since CP / PD = 1

Let  P  =  ([11 -1) /2, 2sqrt 6)  = (5, 2sqrt 6 )

The slope of the line through AD  = [ 2sqrt 6] / -1  =  -2sqrt 6

And the equation of this  line is  y = (-2sqrt 6)x

In standard form we have (2sqrt6)x + y  =  0        (1)

The slope of the line  through  BC  =  [ 2sqrt 6 - 0] / [ 11 - 6] =  (2/5)sqrt 6

And the equation of the line through BC  is  y  = (2/5)sqrt 6 ( x - 6) =

y =(2/5)sqrt (6)x - (12/5)sqrt 6

In standard from we have   (2/5)sqrt (6)x -y - (12/5)sqrt (6)  = 0     (2)

The  formula for the distance  between a point and a  line is  given  by

abs [ Ax + By + C ]  / sqrt [ A^2 + B^2 ]       where (x,y)  are the coordinates of the point

So   using (1) , the distance from P to the line through AD  = PX  is

abs [ (2sqrt (6) (5) + 2sqrt (6) ] /  sqrt [ (2sqrt (6))^2 + 1^2 ] =  [12sqrt (6)]  / 5

Likewise, the distance between P and the line through BC  = PY   is

abs [ (2/5)sqrt (6) (5) - 2sqrt (6) - (12/5)sqrt (6) ]  / sqrt  [ [(2/5)sqrt (6) ]^2 + 1^2 ]   =

[ (12/5)sqrt 6] / (7/5)  =  [12sqrt (6)]  / 7

Thanks! :)

NotThatSmart Jun 3, 2024
#2
+2
0

Thanks. It is spot on

copa america apuestas

Jun 4, 2024