Question

ABCD is a rectangle in which AB = 10 cms, BC = 8 cms. A point P is taken on AB such that PA = x. Then the minimum value of $P{C}^2+P{D}^2$ is obtained when x =

• A. 10
• B. 5
• C. 8
• D. 4

Answer 1

The correct option is B 5

$P{C}^2=B{C}^2+P{B}^2$
$P{D}^2=A{D}^2+A{P}^2$
$P{C}^2+P{D}^2=8^2+8^2+(10-x{)}^2+x^2$
$(P{C}^2+P{D}^2{)}^1=-2(10-x)+2x$
$20=4x$
$x=5$