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Given, ABCD is a rectangle. If side AB is 8 units long and side BC is 6 units long, which are possible lengths of line DX?

I. 5 -> when x and y coincides then dx = sqrt (6^2+8^2) /2 = 5
II. 6 -> To increase length of DX, we need to move point x towards c. And similarly, point y towards A. Now, y will be close to point A, not point x. Then whole figure will be changed. Not possible.
III. 7 -> similar to DX = 6

So, I think A. :)
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Bunuel

ABCD is a rectangle. If side AB is 8 units long and side BC is 6 units long, which are possible lengths of line DX?

I. 5
II. 6
III. 7


A. I only
B. III only
C. I and II only
D. I and III only
E. I, II and III


Attachment:
1.png

As there is no fixed point X, we will have to find the extremes for values of DX.

Let us take \(\triangle ACD\)
AC=\(\sqrt{8^2+6^2}=10\)

Least value : Least value of DX will be when it is perpendicular to AC
The area is \(\frac{1}{2}*8*6=\frac{1}{2}*10*DX........DX=4.8\)

Maximum Value: When X coincides with the point C or is just short of C. Here DX will be just less than CD, so DX~8.

\(4.8\leq DX<8\)

All three values fall in the range.

E
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It's such a trap that we all thought that DX is a leg of the triangle 😬

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