JEE Advanced 2020 Concept Recapitulation Test 1 Paper 2

(Q. 53−54)
It is given that one of the diagonals of the square is parallel to the line y=x.
Also, the length of the diagonal of the square is 4√2.
Hence, the equation of one of the diagonal is ‌

x−3

1 √2

=‌

y−4

1 √2

=r=±2√2
Hence, x−3=y−4=±2
⇒‌‌x=5,1 and y=6,2
Hence, two vertices are (1,2) and (5,6).
The other diagonal is parallel to the line y=−x, so that its equation is ‌

x−3

−‌ 1 √2

=‌

y−4

1 √2

=r=±2√2
Hence, the two verticeson this diagonal are (1,6) and (5,2).
AB=4,AC=4√2 ⇒AE=2√2
In figure (i) EF+FA=AE
⇒r+√2r=2√2‌‌⇒r=‌

2√2

√2+1

=2√2(√2−1)
In figure (ii), EG+GF=EF
⇒‌‌√2r+r=2‌‌⇒r=‌

2

√2+1

=2(√2−1)