Question

ABCD is a square whose vertex A lies on the origin. The coordinates of the mid-point of the diagonal AC are (p/2,1). Find the value of p, the area of square ABCD is 20 sq. units
(1) ±6
(2) ±20
(3) ±10
(4) ±36
(Punjab, Stage-1, 2015-16)​

Answer 1

Step-by-step explanation:

vertex A lies on the origin

so distance between A(0,0) and midpoint of AC (p/2,1) = √{(p/2)^2 + 1^2}

so, the lenth of AC = 2√{(p/2)^2 + 1^2}

Then, the lenth of the side of the squre

= 2√{(p/2)^2 + 1^2}/2

= √[2.{(p/2)^2 + 1^2}]

so, the area of the squre = [√2.{(p/2)^2 + 1^2}]^2

= 2.{(p/2)^2 + 1^2}

Hence, 2.{(p/2)^2 + 1^2} = 20

or, (p/2)^2 + 1 = 10

or, p^2 /4 = 9

or, p^2 = 36

or, p = ±√36

or, p = ±6

Answer 2

Good luck for your NTSE.......