(i) On combining two identical set-squares which have angles of measure 30° ,60° and 90°, we get the following shapes.

Here, rectangle has 2 lines of symmetry and kite has 1 line of symmetry which are shown by dotted linesin above figures.
(ii) On combining two identical set-squares which have angles of measure 45°,45° and 90°, we get the following shapes.

Here, right angled isosceles triangle has 1 line of symmetry, square has 4 lines of symmetry, which are shown by dotted lines in figures.
(iii)

On combining three identical set-squares which have angles of measure 45°, 90° and 45°, we get the adjoining shape. Here, isosceles trapezium has 1 line of symmetry which is shown by dotted line in the adjoining figure.

The given word is 'SYMMETRY' The letter S has no line of symmetry. Letter Y has one line of symmetry.

The letter M has one line of symmetry

The letter E has one line of symmetry

The letter T has one line of symmetry

The letter R has no line of symmetry.

Given, in figure, the image of the point A is C, in the line l.

(a) Yes, the image of P in line is the point P itself.
(b) Yes, PA = PC
(c) Yes, PA + PB = PC + PB because the distance PA=PC.
(d) Yes, from the point P in the line I, the sum of the distances of points A and B is minimum.

The images of the point P, Q and R in the line n are P',Q' and R', respectively.

It is clear that ㄥPQR = ㄥP'Q'R'

The complete figures are shown below:

4 lines of symmetry

I, M, A

Three

We can use reflectional symmetry here. Let A' be the image of A in the mirror line which is the street here. Then the point P is the ideal place to park the cycle (where the mirror line and A'B meet?
Using reflectional symmetry,
AP = A'P
AP' = A'P
\(\therefore\)  AP' + BP' = A'P' + BP' > A'B
\(\Rightarrow\)  AP' + BP' > A'P + PB
\(\Rightarrow\)  AP' + BP' > AP + PB