From the given graph it can be observed that the coordinate of PointAAis(3,4)(3,4), coordinate of PointBBis(6,7)(6,7), coordinate of PointCCis(9,4)(9,4)and coordinate of PointDDis(6,1)(6,1).

The distance between the two pointsP(x1,y1)P(x1,y1)andQ(x2,y2)Q(x2,y2)is given by,

PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√PQ=(x2−x1)2+(y2−y1)2

The distance between the points A(3,4)(3,4)and B(6,7)(6,7)is,

AB=(6−3)2+(7−4)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitAB=(6−3)2+(7−4)2=9+9=32unit

The distance between the points B(6,7)(6,7)and C(9,4)(9,4)is,

BC=(9−6)2+(4−7)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitBC=(9−6)2+(4−7)2=9+9=32unit

The distance between the points C(9,4)(9,4)and D(6,1)(6,1)is,

CD=(6−9)2+(1−4)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitCD=(6−9)2+(1−4)2=9+9=32unit

The distance between the points A(3,4)(3,4)and D(6,1)(6,1)is,

AD=(3−6)2+(4−1)2−−−−−−−−−−−−−−−√=9+9−−−−√=32–√unitAD=(3−6)2+(4−1)2=9+9=32unit

Therefore, the distance between AB, BC, CD and AD is same. All the distances are equal. So it may be a square or a rhombus.

We need to find the distance between the diagonals.

Now, the distance between the points A(3,4)(3,4)and C(9,4)(9,4)is,

AC=(9−3)2+(4−4)2−−−−−−−−−−−−−−−√=36+0−−−−−√=6unitAC=(9−3)2+(4−4)2=36+0=6unit

The distance between the points B(6,7)(6,7)and D(6,1)(6,1)is,

BD=(6−6)2+(1−7)2−−−−−−−−−−−−−−−√=0+36−−−−−√=6unitBD=(6−6)2+(1−7)2=0+36=6unit

The length of the diagonals is equal. So, the quadrilateral is a square.

Thus, ABCD is a square and Champa is correct.