Explanation:
let the point be (h,k)
for a given line, ax+by+c
d=∣∣∣ah+bk+ca2+b2−−−−−−√∣∣∣
so, 2≤d1p+d2p≤4
2≤∣∣∣h−k2–√∣∣∣+∣∣∣h+k2–√∣∣∣≤4
22–√≤|h−k|+|h+k|
ifh>k
22–√≤h−k+h+k≤42–√
2–√≤h≤22–√
ifh<k
22–√≤−h+k+h+k≤42–√
2–√≤k≤22–√
area of shaded region
area = (22–√)2−2–√2
=6unit2