ABCDABCD is a rectangle and point PP is such that PA=5 cm ,PB=42 cm,  and PD=3 cm. PA=5 cm ,PB=42 cm,  and PD=3 cm.  Find the length of PCPC.


Answer:

4 cm 4 cm 

Step by Step Explanation:
  1. We are given that ABCDABCD is a rectangle with a point PP in its interior such that PA=5 cm ,PB=42 cm,  and PD=3 cm. PA=5 cm ,PB=42 cm,  and PD=3 cm. 
  2. Let us assume that AD=x cm AD=x cm  and CD=y cm. CD=y cm. 
    Now, let us plot the perpendiculars at the points S,T,U,S,T,U, and VV on the sides of the rectangle from the point P.P.
    Let, PS=m cm PS=m cm  and PU=n cm PU=n cm 
    Now, PT=(xn) cm PV=(ym) cm 
  3. Now, using pythagoras theorem in PAS, we have PA2=PS2+AS2(1) Similarly, using pythagoras theorem in PCV, PC2=PV2+CV2(2)
  4. Adding (1) and (2), we have, PA2+PC2=PS2+AS2+PV2+CV2=m2+n2+(ym)2+(xn)2(3)
  5. Now, using pythagoras theorem in PBV, we have PB2=PV2+BV2(4) Similarly, using pythagoras theorem in PDS, PD2=PS2+SD2(5)
  6. Adding (4) and (5), we have, PB2+PD2=PV2+BV2+PS2+SD2=(ym)2+n2+m2+(xn)2=m2+n2+(ym)2+(xn)2(6)
  7. On comparing, equation (3) and (6), we have PA2+PC2=PB2+PD2PC2=PB2+PD2PA2PC2=(42)2+3252PC2=32+925PC2=4125PC2=16PC=16PC=4
  8. Thus, the length of PC=4 cm 

You can reuse this answer
Creative Commons License