Class 7
Mathematics
Federal Board of Intermediate and Secondary Education (FBISE) (FBISE)
📚 Model Textbook of Mathematics Grade 7 (FBISE / NBF)
Class 7 Mathematics Ch 7 Mastery Guide: Transformations, Cartesian Plane, Translations, Reflections & Rotational Symmetry (FBISE)
📖 Chapter 7: Transformations
📅 Updated: Sep 12, 2026
Teacher & Parent Roadmap
Grade 7 Mathematics • FBISE / National Curriculum (NBF)
Instructional Guide: Unit 07 Transformations
Target Learning Outcomes (SLOs)
- Understand and use the Cartesian coordinate system (axes, origin, ordered pairs $(x, y)$, abscissa, and ordinate).
- Identify and plot points across all four quadrants and along coordinate axes.
- Connect plotted points to identify and classify 2D geometric shapes (triangles, rectangles, squares, polygons).
- Define and perform translations on points, lines, and shapes using algebraic mapping $(x, y) \to (x \pm a, y \pm b)$.
- Understand reflections and draw lines of symmetry for various geometric and real-world figures.
- Determine the order of rotational symmetry ($n$) and calculate the angle of rotation ($\theta = \frac{360^\circ}{n}$).
Prerequisites & Bridge Concepts
- Number Line: Positive numbers lie to the right/up; negative numbers lie to the left/down.
- Basic Geometry: Properties of triangles, rectangles, squares, kites, parallelograms, and regular polygons.
- Basic Arithmetic: Adding and subtracting integers on horizontal and vertical number lines.
Common Misconceptions & Pitfalls
- Ordered Pair Reversal: Confusing $(x, y)$ with $(y, x)$. Emphasize that $(5, 2)$ (5 right, 2 up) is completely different from $(2, 5)$ (2 right, 5 up).
- Points on Axes: Thinking $(0, 4)$ or $(-3, 0)$ belongs to a quadrant. Points on axes lie on the boundaries and belong to no quadrant.
- Rectangle Diagonals: Mistakenly believing diagonals of a non-square rectangle are lines of symmetry. (A rectangle only has 2 lines of symmetry through side midpoints).
- Parallelogram Symmetry: Thinking a general parallelogram has reflection lines. It has zero lines of symmetry, but rotational symmetry of order 2.
Pedagogical Tips & Real-World Connections
Connect transformations to video game animation (sprites moving across screens), GPS navigation maps (latitude/longitude coordinates), kaleidoscope patterns (reflections and rotations), and architectural tile mosaics.
Quick Jump Navigation (Unit 07 Transformations)
1. The Cartesian Coordinate Plane & Ordered Pairs
Just as a street map uses names and numbers to pinpoint houses, the Cartesian Coordinate Plane allows mathematicians and scientists to pinpoint any exact location in a flat 2D space.
Core Components of the Coordinate System:
- x-axis: The horizontal number line extending infinitely left (-) and right (+).
- y-axis: The vertical number line extending infinitely down (-) and up (+), perpendicular ($90^\circ$) to the x-axis.
- Origin $O(0, 0)$: The central intersection point where both number lines cross at zero.
- Ordered Pair $(x, y)$: A pair of numbers giving the exact horizontal distance ($x$-coordinate or abscissa) and vertical distance ($y$-coordinate or ordinate).
The Four Quadrants:
| Quadrant | x-coordinate (Abscissa) | y-coordinate (Ordinate) | Sign Pattern | Example Point |
|---|---|---|---|---|
| Quadrant I (Top Right) | Positive ($x > 0$) | Positive ($y > 0$) | (+, +) | $(4, 5)$ |
| Quadrant II (Top Left) | Negative ($x < 0$) | Positive ($y > 0$) | (-, +) | $(-3, 6)$ |
| Quadrant III (Bottom Left) | Negative ($x < 0$) | Negative ($y < 0$) | (-, -) | $(-5, -2)$ |
| Quadrant IV (Bottom Right) | Positive ($x > 0$) | Negative ($y < 0$) | (+, -) | $(2, -7)$ |
| Coordinate Axes & Origin | Points on the x-axis have $y = 0$ (e.g. $(6, 0)$). Points on the y-axis have $x = 0$ (e.g. $(0, -4)$). They lie on boundaries and belong to no quadrant. | |||
2. Exercise 7.1 (100% Solved Step-by-Step)
Question 1: Plot each point in coordinate plane and state its location:
(a) $P_1(1, 5)$ • (b) $P_2(-2, 1)$ • (c) $P_3(-2, -2)$ • (d) $P_4(2, -4)$ • (e) $P_5(6, -2)$ • (f) $P_6(3, 3)$ • (g) $P_7(5, 0)$ • (h) $P_8(5, 2)$ • (i) $P_9(0, -4)$ • (j) $P_{10}(-6, -2)$
Solution:
• (a) $P_1(1, 5)$: $x = 1 > 0, y = 5 > 0 \implies$ Quadrant I
• (b) $P_2(-2, 1)$: $x = -2 < 0, y = 1 > 0 \implies$ Quadrant II
• (c) $P_3(-2, -2)$: $x = -2 < 0, y = -2 < 0 \implies$ Quadrant III
• (d) $P_4(2, -4)$: $x = 2 > 0, y = -4 < 0 \implies$ Quadrant IV
• (e) $P_5(6, -2)$: $x = 6 > 0, y = -2 < 0 \implies$ Quadrant IV
• (f) $P_6(3, 3)$: $x = 3 > 0, y = 3 > 0 \implies$ Quadrant I
• (g) $P_7(5, 0)$: $y = 0, x = 5 \implies$ On the positive x-axis (not in any quadrant)
• (h) $P_8(5, 2)$: $x = 5 > 0, y = 2 > 0 \implies$ Quadrant I
• (i) $P_9(0, -4)$: $x = 0, y = -4 \implies$ On the negative y-axis (not in any quadrant)
• (j) $P_{10}(-6, -2)$: $x = -6 < 0, y = -2 < 0 \implies$ Quadrant III
• (a) $P_1(1, 5)$: $x = 1 > 0, y = 5 > 0 \implies$ Quadrant I
• (b) $P_2(-2, 1)$: $x = -2 < 0, y = 1 > 0 \implies$ Quadrant II
• (c) $P_3(-2, -2)$: $x = -2 < 0, y = -2 < 0 \implies$ Quadrant III
• (d) $P_4(2, -4)$: $x = 2 > 0, y = -4 < 0 \implies$ Quadrant IV
• (e) $P_5(6, -2)$: $x = 6 > 0, y = -2 < 0 \implies$ Quadrant IV
• (f) $P_6(3, 3)$: $x = 3 > 0, y = 3 > 0 \implies$ Quadrant I
• (g) $P_7(5, 0)$: $y = 0, x = 5 \implies$ On the positive x-axis (not in any quadrant)
• (h) $P_8(5, 2)$: $x = 5 > 0, y = 2 > 0 \implies$ Quadrant I
• (i) $P_9(0, -4)$: $x = 0, y = -4 \implies$ On the negative y-axis (not in any quadrant)
• (j) $P_{10}(-6, -2)$: $x = -6 < 0, y = -2 < 0 \implies$ Quadrant III
Question 2: Determine the ordered pair $(x, y)$ of the points in the given graphs:
Solution:
Graph (a):
• $A = (-2, 4)$ (2 units left, 4 units up)
• $B = (3, 3)$ (3 units right, 3 units up)
• $C = (4, 2)$ (4 units right, 2 units up)
• $D = (3, -1)$ (3 units right, 1 unit down)
• $E = (4, -3)$ (4 units right, 3 units down)
• $F = (3, -4)$ (3 units right, 4 units down)
• $G = (-2, -4)$ (2 units left, 4 units down)
• $H = (0, -3)$ (On y-axis at -3)
• $I = (-2, 0)$ (On x-axis at -2)
Graph (b):
• $S = (-3, 4)$ • $W = (-2, 0)$ • $U = (-2, -2)$ • $T = (0, -3)$ • $R = (2, -4)$ • $V = (2, -2)$ • $X = (3, 0)$ • $Y = (1, 2)$ • $Z = (2, 4)$
Graph (a):
• $A = (-2, 4)$ (2 units left, 4 units up)
• $B = (3, 3)$ (3 units right, 3 units up)
• $C = (4, 2)$ (4 units right, 2 units up)
• $D = (3, -1)$ (3 units right, 1 unit down)
• $E = (4, -3)$ (4 units right, 3 units down)
• $F = (3, -4)$ (3 units right, 4 units down)
• $G = (-2, -4)$ (2 units left, 4 units down)
• $H = (0, -3)$ (On y-axis at -3)
• $I = (-2, 0)$ (On x-axis at -2)
Graph (b):
• $S = (-3, 4)$ • $W = (-2, 0)$ • $U = (-2, -2)$ • $T = (0, -3)$ • $R = (2, -4)$ • $V = (2, -2)$ • $X = (3, 0)$ • $Y = (1, 2)$ • $Z = (2, 4)$
Question 3: Without plotting the given points, name the quadrants in which the points lie:
Solution:
• (a) $A(2, 4) \to$ Quadrant I (+, +)
• (b) $B(-6, 2) \to$ Quadrant II (-, +)
• (c) $D(-6, -7) \to$ Quadrant III (-, -)
• (d) $F(-4, -5) \to$ Quadrant III (-, -)
• (e) $P(1, 3) \to$ Quadrant I (+, +)
• (f) $K(4, -3) \to$ Quadrant IV (+, -)
• (g) $R(0, 7) \to$ On positive y-axis (not in any quadrant)
• (h) $H(6, -2) \to$ Quadrant IV (+, -)
• (i) $S(-5, 0) \to$ On negative x-axis (not in any quadrant)
• (j) $T(-3, 5) \to$ Quadrant II (-, +)
• (a) $A(2, 4) \to$ Quadrant I (+, +)
• (b) $B(-6, 2) \to$ Quadrant II (-, +)
• (c) $D(-6, -7) \to$ Quadrant III (-, -)
• (d) $F(-4, -5) \to$ Quadrant III (-, -)
• (e) $P(1, 3) \to$ Quadrant I (+, +)
• (f) $K(4, -3) \to$ Quadrant IV (+, -)
• (g) $R(0, 7) \to$ On positive y-axis (not in any quadrant)
• (h) $H(6, -2) \to$ Quadrant IV (+, -)
• (i) $S(-5, 0) \to$ On negative x-axis (not in any quadrant)
• (j) $T(-3, 5) \to$ Quadrant II (-, +)
Question 4: Answer the following conceptual questions:
(a) Why do we consider $(5, 2)$ and $(2, 5)$ to be different pairs?
Answer: In an ordered pair $(x, y)$, order is essential. The first value is the horizontal position along the x-axis and the second value is the vertical position along the y-axis. The point $(5, 2)$ is located $5$ units right and $2$ units up, while $(2, 5)$ is located $2$ units right and $5$ units up. Because they locate two entirely different physical positions on the Cartesian plane, $(5, 2) \ne (2, 5)$.
(b) Under what circumstances is a point not in any quadrant?
Answer: A point is not in any quadrant when at least one of its coordinates is zero ($x = 0$ or $y = 0$). If $y = 0$, the point lies on the x-axis; if $x = 0$, the point lies on the y-axis; if both $x = 0$ and $y = 0$, it is the origin $(0, 0)$. The axes are the boundary lines dividing the plane and do not belong to any quadrant.
Answer: In an ordered pair $(x, y)$, order is essential. The first value is the horizontal position along the x-axis and the second value is the vertical position along the y-axis. The point $(5, 2)$ is located $5$ units right and $2$ units up, while $(2, 5)$ is located $2$ units right and $5$ units up. Because they locate two entirely different physical positions on the Cartesian plane, $(5, 2) \ne (2, 5)$.
(b) Under what circumstances is a point not in any quadrant?
Answer: A point is not in any quadrant when at least one of its coordinates is zero ($x = 0$ or $y = 0$). If $y = 0$, the point lies on the x-axis; if $x = 0$, the point lies on the y-axis; if both $x = 0$ and $y = 0$, it is the origin $(0, 0)$. The axes are the boundary lines dividing the plane and do not belong to any quadrant.
Question 5: Draw and identify the geometric figure by joining the given points:
• (a) $(2, 0), (6, 5), (1, 3)$: Joining 3 non-collinear points forms a Scalene Triangle.
• (b) $(-3, -2), (4, -1), (-2, 4)$: 3 non-collinear points form a Triangle.
• (c) $(1, 2), (1, -2), (5, 2), (5, -2)$: Vertical side length $= |2 - (-2)| = 4$ units; Horizontal side length $= |5 - 1| = 4$ units. All sides are 4 units and perpendicular, forming a Square (Area $= 16\text{ sq units}$).
• (d) $(-4, 1), (-4, 6), (-1, 1), (-1, 6)$: Vertical length $= |6 - 1| = 5$ units; Horizontal width $= |-1 - (-4)| = 3$ units. Opposite sides are equal with $90^\circ$ angles, forming a Rectangle (Area $= 15\text{ sq units}$).
• (e) $(0, -6), (2, -2), (5, -4)$: Joining the 3 points forms a Triangle.
• (b) $(-3, -2), (4, -1), (-2, 4)$: 3 non-collinear points form a Triangle.
• (c) $(1, 2), (1, -2), (5, 2), (5, -2)$: Vertical side length $= |2 - (-2)| = 4$ units; Horizontal side length $= |5 - 1| = 4$ units. All sides are 4 units and perpendicular, forming a Square (Area $= 16\text{ sq units}$).
• (d) $(-4, 1), (-4, 6), (-1, 1), (-1, 6)$: Vertical length $= |6 - 1| = 5$ units; Horizontal width $= |-1 - (-4)| = 3$ units. Opposite sides are equal with $90^\circ$ angles, forming a Rectangle (Area $= 15\text{ sq units}$).
• (e) $(0, -6), (2, -2), (5, -4)$: Joining the 3 points forms a Triangle.
3. Geometric Transformations: Translations
A transformation is an operation that maps an original geometric figure (called the pre-image) onto a new figure (called the image). A translation (or slide) moves every point of a figure by the same distance in a specified direction without rotating, flipping, or resizing it.
Coordinate Translation Rules:
| Shift $a$ units Right: | $(x, y) \to (x + a, y)$ |
| Shift $a$ units Left: | $(x, y) \to (x - a, y)$ |
| Shift $b$ units Up: | $(x, y) \to (x, y + b)$ |
| Shift $b$ units Down: | $(x, y) \to (x, y - b)$ |
| Combined Translation: | $(x, y) \to (x \pm a, y \pm b)$ |
Key Property: Translation is a rigid isometry. The original shape and its image are strictly congruent ($\Delta ABC \cong \Delta A'B'C'$). Side lengths, angles, perimeter, and area are completely preserved.
4. Exercise 7.2 (100% Solved Step-by-Step)
Question 1: Write vertices of figures in Figure A & B and find image coordinates under: (a) 5 right, 3 up • (b) 2 left, 4 down
Figure A Solutions:
• Shape 1 (Polygon): Original: $(4,0), (5,0), (6,2), (6,3), (3,3), (3,2), (4,2)$
→ (a) $(x+5, y+3): (9,3), (10,3), (11,5), (11,6), (8,6), (8,5), (9,5)$
→ (b) $(x-2, y-4): (2,-4), (3,-4), (4,-2), (4,-1), (1,-1), (1,-2), (2,-2)$
• Shape 2 (Trapezium): Original: $(-3,2), (-2,4), (-5,6), (-6,4)$
→ (a) $(x+5, y+3): (2,5), (3,7), (0,9), (-1,7)$
→ (b) $(x-2, y-4): (-5,-2), (-4,0), (-7,2), (-8,0)$
• Shape 3 (Triangle): Original: $(2,-1), (1,2), (-2,0)$
→ (a) $(x+5, y+3): (7,2), (6,5), (3,3)$
→ (b) $(x-2, y-4): (0,-5), (-1,-2), (-4,-4)$
• Shape 4 (Square): Original: $(2,-2), (0,-2), (0,-4), (2,-4)$
→ (a) $(x+5, y+3): (7,1), (5,1), (5,-1), (7,-1)$
→ (b) $(x-2, y-4): (0,-6), (-2,-6), (-2,-8), (0,-8)$
• Shape 1 (Polygon): Original: $(4,0), (5,0), (6,2), (6,3), (3,3), (3,2), (4,2)$
→ (a) $(x+5, y+3): (9,3), (10,3), (11,5), (11,6), (8,6), (8,5), (9,5)$
→ (b) $(x-2, y-4): (2,-4), (3,-4), (4,-2), (4,-1), (1,-1), (1,-2), (2,-2)$
• Shape 2 (Trapezium): Original: $(-3,2), (-2,4), (-5,6), (-6,4)$
→ (a) $(x+5, y+3): (2,5), (3,7), (0,9), (-1,7)$
→ (b) $(x-2, y-4): (-5,-2), (-4,0), (-7,2), (-8,0)$
• Shape 3 (Triangle): Original: $(2,-1), (1,2), (-2,0)$
→ (a) $(x+5, y+3): (7,2), (6,5), (3,3)$
→ (b) $(x-2, y-4): (0,-5), (-1,-2), (-4,-4)$
• Shape 4 (Square): Original: $(2,-2), (0,-2), (0,-4), (2,-4)$
→ (a) $(x+5, y+3): (7,1), (5,1), (5,-1), (7,-1)$
→ (b) $(x-2, y-4): (0,-6), (-2,-6), (-2,-8), (0,-8)$
Question 2: Translate points $(4,-5), (-2,0), (-6,-1), (0,-4), (-3,-8)$ under the four translations:
• (a) 3 units right, 4 units up $(x+3, y+4)$:
$(4+3, -5+4) = (7, -1)$; $(-2+3, 0+4) = (1, 4)$; $(-6+3, -1+4) = (-3, 3)$; $(0+3, -4+4) = (3, 0)$; $(-3+3, -8+4) = (0, -4)$
• (b) 0 units right, 4 units down $(x, y-4)$:
$(4, -9), (-2, -4), (-6, -5), (0, -8), (-3, -12)$
• (c) 1 unit left, 0 units up $(x-1, y)$:
$(3, -5), (-3, 0), (-7, -1), (-1, -4), (-4, -8)$
• (d) 5 units left, 3 units down $(x-5, y-3)$:
$(-1, -8), (-7, -3), (-11, -4), (-5, -7), (-8, -11)$
$(4+3, -5+4) = (7, -1)$; $(-2+3, 0+4) = (1, 4)$; $(-6+3, -1+4) = (-3, 3)$; $(0+3, -4+4) = (3, 0)$; $(-3+3, -8+4) = (0, -4)$
• (b) 0 units right, 4 units down $(x, y-4)$:
$(4, -9), (-2, -4), (-6, -5), (0, -8), (-3, -12)$
• (c) 1 unit left, 0 units up $(x-1, y)$:
$(3, -5), (-3, 0), (-7, -1), (-1, -4), (-4, -8)$
• (d) 5 units left, 3 units down $(x-5, y-3)$:
$(-1, -8), (-7, -3), (-11, -4), (-5, -7), (-8, -11)$
Question 3: Rectangle ABCD with vertices $A(1,1), B(5,1), C(5,5), D(1,5)$:
(a) Translation rule: $(x, y) \to (x + 3, y)$ (3 units right, 0 units up)
(b) Vertices of image:
• $A'(1+3, 1) = A'(4, 1)$
• $B'(5+3, 1) = B'(8, 1)$
• $C'(5+3, 5) = C'(8, 5)$
• $D'(1+3, 5) = D'(4, 5)$
(b) Vertices of image:
• $A'(1+3, 1) = A'(4, 1)$
• $B'(5+3, 1) = B'(8, 1)$
• $C'(5+3, 5) = C'(8, 5)$
• $D'(1+3, 5) = D'(4, 5)$
Question 4: Graph ABCDA with $A(-10,-10), B(10,-10), C(10,10), D(-10,10)$:
(i) Identify Figure: The distance between adjacent vertices is $|10 - (-10)| = 20$ units. All 4 sides equal 20 units and meet at $90^\circ$ angles. Thus, ABCDA is a Square (Area $= 20 \times 20 = 400\text{ sq units}$).
(ii) Translation 0 units left and 3 units down $(x, y - 3)$:
• $A'(-10, -10 - 3) = A'(-10, -13)$
• $B'(10, -10 - 3) = B'(10, -13)$
• $C'(10, 10 - 3) = C'(10, 7)$
• $D'(-10, 10 - 3) = D'(-10, 7)$
(ii) Translation 0 units left and 3 units down $(x, y - 3)$:
• $A'(-10, -10 - 3) = A'(-10, -13)$
• $B'(10, -10 - 3) = B'(10, -13)$
• $C'(10, 10 - 3) = C'(10, 7)$
• $D'(-10, 10 - 3) = D'(-10, 7)$
Question 5: Rectangle with $A(0,-2), B(2,-2), C(2,-4), D(0,-4)$:
(i) Translation 0 right, 5 up $(x, y + 5)$:
• $A'(0, -2 + 5) = A'(0, 3)$
• $B'(2, -2 + 5) = B'(2, 3)$
• $C'(2, -4 + 5) = C'(2, 1)$
• $D'(0, -4 + 5) = D'(0, 1)$
(ii) Area of rectangle:
$\text{Length} = |2 - 0| = 2\text{ units}$, $\text{Width} = |-2 - (-4)| = 2\text{ units}$.
$\text{Area} = \text{Length} \times \text{Width} = 2 \times 2 = 4\text{ square units}$.
• $A'(0, -2 + 5) = A'(0, 3)$
• $B'(2, -2 + 5) = B'(2, 3)$
• $C'(2, -4 + 5) = C'(2, 1)$
• $D'(0, -4 + 5) = D'(0, 1)$
(ii) Area of rectangle:
$\text{Length} = |2 - 0| = 2\text{ units}$, $\text{Width} = |-2 - (-4)| = 2\text{ units}$.
$\text{Area} = \text{Length} \times \text{Width} = 2 \times 2 = 4\text{ square units}$.
5. Reflections & Line of Symmetry
A reflection is a flip of a figure across a mirror line called the line of symmetry (or line of reflection). A figure has reflective symmetry (line symmetry) if it can be folded along a straight line such that both halves match exactly.
Summary of Lines of Symmetry for Standard Figures:
| Geometric Shape | Number of Lines of Symmetry | Description of Symmetry Lines |
|---|---|---|
| Equilateral Triangle | 3 | From each vertex perpendicular to the opposite midpoint |
| Isosceles Triangle | 1 | Vertical line through apex to base midpoint |
| Scalene Triangle | 0 | No matching sides or angles |
| Square | 4 | 2 joining opposite midpoints + 2 along diagonals |
| Rectangle (non-square) | 2 | 1 horizontal + 1 vertical through side midpoints (diagonals are NOT symmetry lines) |
| Rhombus | 2 | Along both perpendicular diagonals |
| Parallelogram (general) | 0 | No reflection lines (cannot be folded onto itself) |
| Regular Pentagon | 5 | From each vertex to opposite side midpoint |
| Regular Hexagon | 6 | 3 joining opposite vertices + 3 joining opposite midpoints |
| Circle | Infinite ($\infty$) | Every straight line passing through the center (diameter) is a symmetry line |
6. Rotations & Rotational Symmetry
A rotation is a circular turning motion of a figure around a fixed central point called the centre of rotation. It can turn in a clockwise or anticlockwise direction.
Order of Rotational Symmetry ($n$)
The number of times a figure matches its starting appearance during one complete $360^\circ$ rotation about its centre.
Angle of Rotation ($\theta$)
The smallest positive angle required to turn the shape onto itself:
$\text{Angle of Rotation} = \frac{360^\circ}{\text{Order } (n)}$
7. Exercise 7.3 (100% Solved Step-by-Step)
Question 1: Trace the figures and draw all lines of symmetry:
• (i) Rectangle: 2 lines of symmetry (1 horizontal and 1 vertical through midpoints of opposite sides).
• (ii) Isosceles Trapezium: 1 line of symmetry (vertical line through the midpoints of the parallel top and bottom sides). (If general trapezium: 0 lines).
• (iii) Symmetrical Rabbit: 1 line of symmetry (vertical line down the middle).
• (iv) Kite: 1 line of symmetry (along the main diagonal).
• (v) Arched Window / Tomb Shape: 1 line of symmetry (vertical line down the middle).
• (ii) Isosceles Trapezium: 1 line of symmetry (vertical line through the midpoints of the parallel top and bottom sides). (If general trapezium: 0 lines).
• (iii) Symmetrical Rabbit: 1 line of symmetry (vertical line down the middle).
• (iv) Kite: 1 line of symmetry (along the main diagonal).
• (v) Arched Window / Tomb Shape: 1 line of symmetry (vertical line down the middle).
Questions 2 & 3: Write Order of Rotational Symmetry & Measure of Angle of Rotation:
| Figure | Order of Rotational Symmetry ($n$) | Angle of Rotation ($\theta = 360^\circ/n$) |
|---|---|---|
| (i) Rectangle | 2 | $360^\circ / 2 = \mathbf{180^\circ}$ |
| (ii) Regular Pentagon | 5 | $360^\circ / 5 = \mathbf{72^\circ}$ |
| (iii) Circle | Infinite ($\infty$) | Any angle / Infinitely small |
| (iv) 5-pointed Star | 5 | $360^\circ / 5 = \mathbf{72^\circ}$ |
| (v) Regular Hexagon | 6 | $360^\circ / 6 = \mathbf{60^\circ}$ |
Question 4: An animated movie takes 24 still drawings for 1 second of motion. About how many still drawings are needed for a 2 hours animated movie?
Step 1: Convert 2 hours into seconds:
$\text{Time} = 2 \times 60 \text{ min} \times 60 \text{ sec} = 7,200 \text{ seconds}$.
Step 2: Multiply by drawing frame rate:
$\text{Total Drawings} = 7,200 \times 24 = \mathbf{172,800 \text{ drawings}}$.
$\text{Time} = 2 \times 60 \text{ min} \times 60 \text{ sec} = 7,200 \text{ seconds}$.
Step 2: Multiply by drawing frame rate:
$\text{Total Drawings} = 7,200 \times 24 = \mathbf{172,800 \text{ drawings}}$.
Question 5: Sketch a car at the top of a hill sliding down the hill. Is your drawing an example of translation or rotation?
Answer: It is an example of Translation (a diagonal translation along the downward slope). The car slides along a straight path without changing its orientation, shape, or size.
8. Review Exercise 7 (100% Solved Master Guide)
Q1: Multiple Choice Questions (MCQs):
(i) Location of $(-1, 0)$ is on:
Correct: (a) x-axis ($y = 0$).
Correct: (a) x-axis ($y = 0$).
(ii) Point $(-3, -1)$ lies on:
Correct: (c) third quadrant (-, -).
Correct: (c) third quadrant (-, -).
(iii) Point $(-x, y)$ for positive $x, y$ lies on:
Correct: (b) second quadrant (-, +).
Correct: (b) second quadrant (-, +).
(iv) Points on x-axis have y-coordinate:
Correct: (a) 0.
Correct: (a) 0.
(v) Point in third quadrant has coordinates:
Correct: (b) $(-x, -y)$.
Correct: (b) $(-x, -y)$.
(vi) Point $(-2, -3)$ translated 3 up becomes:
Correct: (c) $(-2, 0)$ ($y = -3 + 3 = 0$).
Correct: (c) $(-2, 0)$ ($y = -3 + 3 = 0$).
(vii) $(0, 0)$ translated 3 left and 5 down becomes:
Correct: (a) $(-3, -5)$.
Correct: (a) $(-3, -5)$.
(viii) $(x, y)$ translated $h$ left and $k$ up:
Correct: (b) $(x - h, y + k)$.
Correct: (b) $(x - h, y + k)$.
(ix) $(x, y)$ translated $h$ right and $k$ below:
Correct: (d) $(x + h, y - k)$.
Correct: (d) $(x + h, y - k)$.
(x) $(0, y)$ translated 2 left and 1 up:
Correct: (c) $(-2, y + 1)$.
Correct: (c) $(-2, y + 1)$.
(xi) $(x, 0)$ translated 2 left and 1 up:
Correct: (d) $(x - 2, 1)$.
Correct: (d) $(x - 2, 1)$.
Q2 to Q10 Detailed Solutions:
• Q2 & Q3 (Points & Figures): Given $A(1,5), B(4,2), C(0,2), D(0,-2), E(-1,5), F(4,-2), G(-1,-5)$:
- $BCDB$ forms a Right-Angled Triangle.
- $ACBA$ forms a Triangle.
- $ABEFGA$ forms a Polygon.
• Q4 (Line Segment Translation): $A(0, -1)$ and $B(0, 1)$ translated 3 left and 4 up $(x-3, y+4)$:
$A' = (0 - 3, -1 + 4) = \mathbf{(-3, 3)}$; $B' = (0 - 3, 1 + 4) = \mathbf{(-3, 5)}$.
• Q5 (Translation Rules):
(a) 0 right, 1 down $\to \mathbf{(x, y - 1)}$
(b) 1 left, 0 up $\to \mathbf{(x - 1, y)}$
(c) $h$ right, $k$ down $\to \mathbf{(x + h, y - k)}$
(d) $h$ left, $k$ down $\to \mathbf{(x - h, y - k)}$
• Q6 (Geometric Figures Identification):
(i) $(-3, -2), (-3, 3), (4, 3), (4, -2) \to$ Rectangle (Width 7, Height 5).
(ii) $(4, -1), (4, 1), (0, 1), (0, -1), (4, -2) \to$ Pentagon / Polygon.
(iii) $(0, 6), (4, 3), (0, 1) \to$ Isosceles Triangle.
• Q7 (School Coordinates): 3 units left and 4 units up from origin $\to \mathbf{(-3, 4)}$.
• Q8 (House & School): Ahmad's house $= \mathbf{(2, 5)}$; School $= \mathbf{(-3, -7)}$.
• Q9 (Square Quadrant): First coordinate positive ($x > 0$), second negative ($y < 0$) $\implies$ Quadrant IV (Fourth Quadrant).
• Q10 (Point Translation): 2 units right of $(-4, 2) \to (-4 + 2, 2) = \mathbf{(-2, 2)}$.
- $BCDB$ forms a Right-Angled Triangle.
- $ACBA$ forms a Triangle.
- $ABEFGA$ forms a Polygon.
• Q4 (Line Segment Translation): $A(0, -1)$ and $B(0, 1)$ translated 3 left and 4 up $(x-3, y+4)$:
$A' = (0 - 3, -1 + 4) = \mathbf{(-3, 3)}$; $B' = (0 - 3, 1 + 4) = \mathbf{(-3, 5)}$.
• Q5 (Translation Rules):
(a) 0 right, 1 down $\to \mathbf{(x, y - 1)}$
(b) 1 left, 0 up $\to \mathbf{(x - 1, y)}$
(c) $h$ right, $k$ down $\to \mathbf{(x + h, y - k)}$
(d) $h$ left, $k$ down $\to \mathbf{(x - h, y - k)}$
• Q6 (Geometric Figures Identification):
(i) $(-3, -2), (-3, 3), (4, 3), (4, -2) \to$ Rectangle (Width 7, Height 5).
(ii) $(4, -1), (4, 1), (0, 1), (0, -1), (4, -2) \to$ Pentagon / Polygon.
(iii) $(0, 6), (4, 3), (0, 1) \to$ Isosceles Triangle.
• Q7 (School Coordinates): 3 units left and 4 units up from origin $\to \mathbf{(-3, 4)}$.
• Q8 (House & School): Ahmad's house $= \mathbf{(2, 5)}$; School $= \mathbf{(-3, -7)}$.
• Q9 (Square Quadrant): First coordinate positive ($x > 0$), second negative ($y < 0$) $\implies$ Quadrant IV (Fourth Quadrant).
• Q10 (Point Translation): 2 units right of $(-4, 2) \to (-4 + 2, 2) = \mathbf{(-2, 2)}$.
9. Chapter 7 Formula Card & Active Recall
Quadrant Signs:
QI: $(+, +)$ • QII: $(-, +)$
QIII: $(-, -)$ • QIV: $(+, -)$
QI: $(+, +)$ • QII: $(-, +)$
QIII: $(-, -)$ • QIV: $(+, -)$
Translation Rule:
$(x, y) \to (x \pm a, y \pm b)$
Preserves shape, size & area.
$(x, y) \to (x \pm a, y \pm b)$
Preserves shape, size & area.
Angle of Rotation:
$\theta = \frac{360^\circ}{n}$
where $n$ is the order of rotation.
$\theta = \frac{360^\circ}{n}$
where $n$ is the order of rotation.
Regular $n$-gon Symmetry:
Lines of symmetry $= n$
Order of rotation $= n$
Lines of symmetry $= n$
Order of rotation $= n$
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