In the oriented plane, consider a rectangle ABCD such that :
(AB,AD)=pi/2+2kpi , AB=4cm and AD=3cm
Let H be the foot of the perpendicular issued through A to (BD)
Consider the dilation h of center H that transforms D onto B .
1)a- Determine and trace the image of straight line (AD) by h .
b- Deduce the image E of point A by h .
c- Calculate the ratio of h .
d- Construct the point F image of B by h and the point G image of C by h and determine the image of rectangle ABCD under h.
e- Calculate the area of the image of ABCD by h .
This is the first part of the Exercise the other part will be followed after I get the answers of the above questions .
My achievments so far :
For # a) D -----> B by h(H,k) , HB=kHD
but A passes through straight line (AD) so the image of (AD) is a straight line passes by B and parallel to (AD) which is BC .
# b) A ----> E by h(H,k) => HE=kHA => A , H and C are collinear .
Also (AD) ----> (BC) but A belongs to (AD) so image of A which is E belongs to straight line (BC).
Hence E is the point of intersection of (AH) and (BC).
# c) D ----> B by h
A ----> E by h this emplys that [BE]=k[AD] => k = [BE]/[AD] : AD= 3 (given) , So i have to determine [BE] ??
# d) For this part I can't do it without the help of k(ratio).
But I think the image of rectangle ABCD is EFGB .
# e) Area(EFGB) = k^2*Area(ABCD).
Thanks in advance ~
(AB,AD)=pi/2+2kpi , AB=4cm and AD=3cm
Let H be the foot of the perpendicular issued through A to (BD)
Consider the dilation h of center H that transforms D onto B .
1)a- Determine and trace the image of straight line (AD) by h .
b- Deduce the image E of point A by h .
c- Calculate the ratio of h .
d- Construct the point F image of B by h and the point G image of C by h and determine the image of rectangle ABCD under h.
e- Calculate the area of the image of ABCD by h .
This is the first part of the Exercise the other part will be followed after I get the answers of the above questions .
My achievments so far :
For # a) D -----> B by h(H,k) , HB=kHD
but A passes through straight line (AD) so the image of (AD) is a straight line passes by B and parallel to (AD) which is BC .
# b) A ----> E by h(H,k) => HE=kHA => A , H and C are collinear .
Also (AD) ----> (BC) but A belongs to (AD) so image of A which is E belongs to straight line (BC).
Hence E is the point of intersection of (AH) and (BC).
# c) D ----> B by h
A ----> E by h this emplys that [BE]=k[AD] => k = [BE]/[AD] : AD= 3 (given) , So i have to determine [BE] ??
# d) For this part I can't do it without the help of k(ratio).
But I think the image of rectangle ABCD is EFGB .
# e) Area(EFGB) = k^2*Area(ABCD).
Thanks in advance ~