Why do two different methods to calculate the path difference yield different results?

HCV Concepts(Physics) 17.26

Note that $Delta$ means triangle

Consider a point source of light S, which sends two rays of light of wavelength $lambda$ (order of magnitude $10^{-7}$ m) – one to point O (directly in front of it) and another to point P, which is at a distance $a$ (order of magnitude $10^{-4}$ m) from O. Points O and P lie on a screen and the ray SO is perpendicular (normal) to the screen, while ray SP makes an angle $theta$ with ray SO. SO = $D$ (order of magnitude $10^{0}$ m), such that $D$ >> $a$. A diagram of this is given below:

It is given in the question that the path difference between the two rays is $frac{lambda}{4}$ (i.e. SP $= D+ frac{lambda$}{4} $ $implies $ SP – SO = $ frac{lambda}{4}) $ and we are told to express $a$ in terms of $D$ and $lambda$. Using two different approaches, I get two different results.

Approach 1:
Because $Delta$SOP is a right angled triangle, and SP is the hypotenuse,

SP = $sqrt{D^2+a^2}$
$ implies frac{lambda}{4} = sqrt{D^2+a^2} – D
frac{lambda}{4} = Dsqrt{1+frac{a^2}{D^2}} – D
frac{lambda}{4} = D(1+frac{a^2}{D^2})^{frac{1}{2}} – D $

Because $a <<D$, $frac{a^2}{D^2}$ is very small, meaning that $(1+frac{a^2}{D^2})^{frac{1} {2}} approx 1+frac{a^2}{D^2} times frac{1} {2} = 1+frac{a^2}{2D^2}$

Using this, $ frac{lambda}{4} = D(1+frac{a^2}{2D^2}) – D
frac{lambda}{4} = D+ Dfrac{a^2}{2D^2} – D
frac{lambda}{4} = frac{a^2}{2D}
a^2 = frac{ lambda times 2D} {4}
a^2 = frac{ D lambda} {2}
a = sqrt{ frac{ D lambda} {2}} $

This ($sqrt{ frac{ D lambda} {2}} $) is also the answer given in the textbook.

But, using an approach often used in diffraction and interference questions, I got a different answer…

Approach 2: $ tan theta = frac{a}{D} $
Because $a<<D implies tan theta $ is very small. Because $ tan theta $ is very small, $ tan theta approx theta $

Now, another right triangle can be constructed by drawing a line segment (OQ) which is perpendicular to SP and passes through O:

From the diagram, it can be seen that $angle$ POQ is also equal to $theta$. As $theta$ is so small, $angle$ SOQ is nearly $frac{pi}{2}$ ($90^circ$), making $Delta$ SOQ nearly an isosceles triangle and making SQ $approx $ SO = $D$. Given SQ = SO, the path difference, SP-SO, is equal to QP, i.e. QP = $frac{lambda}{4}$

In $Delta $ POQ $ implies sin theta = frac{left(frac{lambda}{4} right)}{a}
sin theta = frac{lambda}{4a}$

Now, because $ theta $ is small, $ sin theta approx theta $ (I could have just done $ sin theta approx tan theta $ instead of $ tan theta approx theta $ and $ sin theta approx theta $)

$ implies frac{lambda}{4a} = frac{a}{D}
frac{lambda times D}{4} = a^2
therefore a = sqrt{ frac{D lambda}{4} } $

As I have already mentioned, this is not the answer which is given in my textbook. Although I make a few more approximations in approach 2 than in approach 1, in which I only make 1 assumption, it is worth noting that many of the formulae for single slit and double slit interference have been derived using similar methods in my textbook.

As the two answers vary quite significantly – by a factor of $ sqrt 2 $, both answers cannot be simultaneously correct, and I don’t think that simply approximating more times in approach would bring about such an error. I don’t know for sure which of the two approaches is correct even though the textbook’s answer matches that of approach 1. I would greatly appreciate an answer which clarifies which answer is correct, and why the approach that lead to the other answer is wrong. Thank you for your time.