ABCDABCD is a rectangle and point PP is such that PA=5 cm ,PB=4√2 cm, and PD=3 cm. PA=5 cm ,PB=4√2 cm, and PD=3 cm. Find the length of PCPC.
![](https://www.edugain.com/egdraw/draw.php?num=5&sx=240&sy=140&x0=20&y0=20&A1=shape:polygon;points:0,0,200,0,200,100,0,100;textc:A,D,[0.15]C,[0.15]B&A2=shape:line;x1:50;y1:60;x2:0;y2:0;tc1:[-9.-2]P&A3=shape:line;x1:50;y1:60;x2:200;y2:0&A4=shape:line;x1:50;y1:60;x2:200;y2:100&A5=shape:line;x1:50;y1:60;x2:0;y2:100 )
Answer:
4 cm 4 cm
- We are given that ABCDABCD is a rectangle with a point PP in its interior such that PA=5 cm ,PB=4√2 cm, and PD=3 cm. PA=5 cm ,PB=4√2 cm, and PD=3 cm.
- 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.
Now, PT=(x−n) cm PV=(y−m) cm - Now, using pythagoras theorem in △PAS, we have PA2=PS2+AS2…(1) Similarly, using pythagoras theorem in △PCV, PC2=PV2+CV2…(2)
- Adding (1) and (2), we have, PA2+PC2=PS2+AS2+PV2+CV2=m2+n2+(y−m)2+(x−n)2…(3)
- Now, using pythagoras theorem in △PBV, we have PB2=PV2+BV2…(4) Similarly, using pythagoras theorem in △PDS, PD2=PS2+SD2…(5)
- Adding (4) and (5), we have, PB2+PD2=PV2+BV2+PS2+SD2=(y−m)2+n2+m2+(x−n)2=m2+n2+(y−m)2+(x−n)2…(6)
- On comparing, equation (3) and (6), we have PA2+PC2=PB2+PD2⟹PC2=PB2+PD2−PA2⟹PC2=(4√2)2+32−52⟹PC2=32+9−25⟹PC2=41−25⟹PC2=16⟹PC=√16⟹PC=4
- Thus, the length of PC=4 cm