2. ABCD is a square and P, Q are the midpoints of BC, CD respectively. If AP = a and AQ = b, find in terms of a and b, the directed line segments (i) AB, (ii) AD, (iii) BD and (iv) AC.

It looks like you're wanting vectors. So,

AB + BP = a = b - PQ
AD + DQ = b = a + PQ
play around with those and see what you get.

To find the directed line segments in terms of a and b, we need to analyze the given information about the square ABCD and the midpoints P and Q.

(i) AB:
Since ABCD is a square, AB is one of its sides. Let's call the length of AB as s.
Since P is the midpoint of BC, we can say that AP = PB = a. Similarly, since Q is the midpoint of CD, we can say that AQ = QD = b.
Therefore, AB = AP + PB = a + s + PB.
But since ABCD is a square, each side has the same length.
So, AB = a + s + PB = a + s + QB.
Now, QB is the distance from Q to B, which is equivalent to AD, which we'll determine in the next step.
Hence, AB = a + s + AD.

(ii) AD:
Since ABCD is a square, AD is one of its sides, and we can label the length of AD as s. Additionally, we know that AQ = QD = b.
Therefore, AD = AQ + QD = b + s.

(iii) BD:
BD is the diagonal of the square ABCD. Let's call the length of BD as d.
Since ABCD is a square, each side has the same length, so we can say that AB = BC = CD = DA = s.
We also know that P is the midpoint of BC, so BP = PC = (1/2)BC = (1/2)s.
Using Pythagoras' theorem, we can find the length of BD in terms of s. Since BD is the hypotenuse of the right-angled triangle PBD with PB and PD as its sides, we have:
BD^2 = BP^2 + PD^2
d^2 = (1/2s)^2 + s^2
d^2 = (1/4)s^2 + s^2
d^2 = (1/4 + 1)s^2
d^2 = (5/4)s^2

(iv) AC:
AC is the other diagonal of the square ABCD, and it has the same length as BD.
So, AC = BD = sqrt((5/4)s^2)
= (s/2)sqrt(5)

To summarize:
(i) AB = a + s + AD
(ii) AD = b + s
(iii) BD = sqrt((5/4)s^2)
(iv) AC = (s/2)sqrt(5)

To find the directed line segments in terms of a and b, let's analyze the given information step by step:

(i) AB:

Since ABCD is a square, AB is one of its sides. Let's call the side length of the square "s".

Since P is the midpoint of BC, we can conclude that BP = PC = s/2.
Similarly, since Q is the midpoint of CD, we can conclude that CQ = QD = s/2.

Since AP = a, we can express AB in terms of a and s.
AB = AP + PB
AB = a + s/2

Hence, the directed line segment AB is equal to a + s/2.

(ii) AD:

To find AD, we can use the fact that AD is a diagonal of the square.

By using the Pythagorean theorem, we can calculate AD.

AD^2 = AP^2 + PD^2
AD^2 = a^2 + (QD + PQ)^2
AD^2 = a^2 + (s/2 + b)^2

Taking the square root of both sides, we get:

AD = √(a^2 + (s/2 + b)^2)

Therefore, the directed line segment AD is equal to √(a^2 + (s/2 + b)^2).

(iii) BD:

To find BD, we can use the fact that BD is another diagonal of the square.

Similarly, using the Pythagorean theorem, we can calculate BD.

BD^2 = BP^2 + PD^2
BD^2 = (BC/2)^2 + (CQ + QD)^2
BD^2 = (s/2)^2 + (s/2 + b)^2

Taking the square root of both sides, we get:

BD = √((s/2)^2 + (s/2 + b)^2)

Therefore, the directed line segment BD is equal to √((s/2)^2 + (s/2 + b)^2).

(iv) AC:

AC is a diagonal of the square.

Since the diagonals in a square are equal, the length of AC is equal to AD.

Therefore, the directed line segment AC is equal to √(a^2 + (s/2 + b)^2), which is the same as AD.

In summary:

(i) AB = a + s/2
(ii) AD = √(a^2 + (s/2 + b)^2)
(iii) BD = √((s/2)^2 + (s/2 + b)^2)
(iv) AC = √(a^2 + (s/2 + b)^2)

Note: The lengths are expressed in terms of a, b, and s, where s is the side length of the square.