Join QT and say this side as x.

Now say ST as y

we have right angled triangle QTS = ( \(\sqrt{5}\))^2 = x^2+ y^2.

Trick here is that x and y have to be only positive can't be negative.

x^2+ y^2 = 5.

Let's take the combination value.

when x = 2 and y = 1 , this satisfies the equation.

when y = 2 and x = 1 , this satisfies the equation and I don't think any values are possible apart from this.

To get area of QSTP = 1/2 ( xy ) + 1/2 ( xy) = xy.

What ever the value of xy from above we finally get as 2. This value is greater than 1.5

When we consider x and y as \(\sqrt{2}\) and \(\sqrt{3}\) or vice versa,

xy value is greater than 1.5



So answer A.

Hope this is clear.

Hi can you explain how you jump to the above conclusion?
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You can refer the diagram and we are clearly given that figure is a square.

QSTP is a parallelogram i.e. QST is one triangle and QPT is another triangle as highlighted in the diagram as T1 and T2.

QSTP = T1 + T2 = two triangles = 1/2 ( xy ) + 1/2 ( xy) = xy.

Hope this clears.

Your graph is fascinating in its reasoning. However, the best and easier way to solve this question is, as almost always, an evaluation upfront and observation.



Notice that we do have the triplet \(x:x:x \sqrt2\)

So, \(x \sqrt{2} = \sqrt{5} = \frac{\sqrt5}{sqrt2} = 1.5\)

\(x =\sqrt{5} = 2.2\)

Area of the rectangle \(QS * ST= b*h = 1.5 * 2.2 = 3.3\) this is quantity A

Quantity B, which of course, is 1.5.

A is the answer
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Hi,

I doubt if you can use y=1 and x= 2 satisfies PRSU as square or either taking as \(\sqrt{2}\)and \(\sqrt{3}\). Moreover QSTP is a rectangle and we can easily find out the area as = Length * Breadth

However I agree with Caracass with approach
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