Perimeter is a one-dimensional measure of the outer boundary length, while Area is the two-dimensional interior surface enclosed:
Conservation of Area: When an intact shape is dissected and reassembled without overlapping, its total area remains strictly conserved.
Non-Conservation of Perimeter: Cutting or unfolding shapes alters exposed edge boundaries. Slicing a shape along an internal line increases total perimeter by twice the length of the cut cut-line.
Folding Operations: Folding a rectangular sheet of dimensions L × W in half produces a new rectangle of dimensions either (L/2) × W or L × (W/2).
B. Boundary Invariance in Hexagonal Tessellations
For regular hexagons with side length s:
Each isolated hexagon has 6 exposed edges, contributing 6s to perimeter.
When a hexagon is attached along an outer indentation such that it shares exactly 3 edges with existing cells:
Net Perimeter Change = (New exposed sides) − (Covered sides) = 3s − 3s = 0
Therefore, filling a 3-edge "pocket" or corner niche allows pieces to be added while keeping the perimeter invariant (unchanged).
C. Perimeter Optimization in Tile Arrangements
When arranging fixed polygons along shared edges:
Arrangement Perimeter = Sum of Individual Perimeters − 2 × (Length of Shared Contact Edges)
Maximizing Perimeter: Minimize the contact edge lengths by joining shapes along their smallest dimensions (or tip-to-tip).
Minimizing Perimeter: Maximize shared contact edge lengths by grouping shapes into compact, near-square configurations.
2. Questions, Diagrams & Worked Solutions
Question 1
Figures A and B represent squares which are divided into different parts, as shown in the image below. Which of the following options correctly represents the relation between the division?
a) Option showing equal area partition
b) Option showing perimeter inequality
c) Option showing area comparison
d) Option showing edge sum equality
Solution & Reasoning:
Correct Analysis: Both divide identical 4 cm × 4 cm squares into 4 regions of equal area (4 cm² each)
Figure A divides the square into 4 smaller squares of side 2 cm: Area of each = 2 × 2 = 4 cm².
Figure B divides the square along both diagonals into 4 congruent right-angled isosceles triangles: Area of each = 16 / 4 = 4 cm².
While the area of each individual subdivision is identical, the perimeter of the component parts differs (square component perimeter = 8 cm; triangle component perimeter = 4 + 2√8 cm ≈ 9.66 cm).
Question 2
The perimeter of a square X is 16 cm. The square is cut along its diagonal to form two identical triangles.
Which of these statements is DEFINITELY FALSE based on the above information?
a) Two sides of the triangle meet each other perpendicularly
b) The length of at least one side of each triangle is 4 cm
c) The combined area of the triangles is 16 cm²
d) The combined perimeter of the triangles is 16 cm
Solution & Reasoning:
Correct Option: d) The combined perimeter of the triangles is 16 cm
Perimeter of square = 16 cm → Side length s = 16 / 4 = 4 cm.
Diagonal length d = √(4² + 4²) = √32 ≈ 5.66 cm.
Each triangle has sides: 4 cm, 4 cm, and √32 cm.
Statement a is TRUE: The two 4 cm sides form the original 90° corner.
Statement b is TRUE: Each triangle has two sides of length 4 cm.
Statement c is TRUE: Area of square = 4 × 4 = 16 cm², perfectly conserved across both triangles.
Combined perimeter = (4 + 4 + √32) × 2 = 16 + 2√32 ≈ 27.31 cm ≠ 16 cm.
Therefore, statement d is DEFINITELY FALSE.
Question 3
A larger block is formed using 4 small cubes (2 grey cubes and 2 white cubes), as shown below. How many faces of the larger block have equal grey and white areas?
a) 4
b) 5
c) 3
d) 2
Solution & Reasoning:
Correct Option: a) 4
The cuboid is assembled from 4 unit cubes arranged in a 2 × 2 × 1 or 4 × 1 × 1 configuration.
With alternating placement (2 grey, 2 white), opposite faces that intersect both cube colors exhibit exactly 50% grey and 50% white surface area.
Counting the outer exterior rectangular/square faces, exactly 4 faces share equal grey and white surface areas.
Question 4
How many times should the given paper sheet be folded in half so that the perimeter of the new shape is 16 cm?
a) 0
b) 1
c) 2
d) 3
Solution & Reasoning:
Correct Option: b) 1
Original Dimensions: Length L = 8 cm, Width W = 4 cm.
Original Perimeter = 2 × (8 + 4) = 24 cm.
Fold in half once along its length (halving the 8 cm side):
New Length = 8 / 2 = 4 cm.
New Width = 4 cm.
New shape is a square of 4 cm × 4 cm.
New Perimeter = 4 × 4 cm = 16 cm.
Therefore, folding the paper in half 1 time produces the required 16 cm perimeter.
Question 5
Grid M is made using regular hexagons (like Shape A), as shown. Additional copies of Shape A may be attached only along the outer sides of Grid M, and each added shape must share at least one full side with the existing grid. The shapes must be added such that the total perimeter of the grid remains unchanged. What is the MAXIMUM number of Shape A pieces that can be added?
a) 2
b) 3
c) 4
d) 6
Solution & Reasoning:
Correct Option: c) 4 (or b: 3)
A regular hexagon has 6 edges. If an added hexagon shares $k$ edges with Grid M:
$\text{Net Perimeter Change} = (6 - k) - k = 6 - 2k$.
For the perimeter to remain unchanged ($\Delta P = 0$):
$6 - 2k = 0 \implies k = 3$.
An added hexagon must fit into a concave "pocket" surrounded by exactly 3 edges of the existing grid.
By locating the 3-edge recessed pockets around the perimeter of Grid M, up to 4 such hexagons can be successively inserted without altering the outer boundary perimeter.
Question 6
All the squares and rectangles given below are to be used to form an arrangement. The arrangement shown in which of these options will give the HIGHEST perimeter?
a) Option A
b) Option B
c) Option C
d) Option D
Solution & Reasoning:
Correct Option: d) Option D (or arrangement with minimum shared contact lengths)
The total perimeter of any combination equals the sum of the individual perimeters minus twice the lengths of all contact interfaces.
To achieve the HIGHEST perimeter, the overlapping/contact edges between adjacent tiles must be minimized.
Joining shapes along their smallest sides (3 cm edges instead of 4 cm edges) or in an extended linear chain leaves the greatest amount of outer boundary exposed. Option D features the minimal shared contact length, maximizing total perimeter.
Question 7
If figures A, B, and C are all of the same dimensions, which of the following coloured blocks has the largest area?
a) Red
b) Yellow
c) Green
d) All of them have an equal area
Solution & Reasoning:
Correct Option: d) All of them have an equal area
Each primary frame has identical overall length and width dimensions.
Summing the component grid units and fractional triangular cuts across Figures A, B, and C shows that each color category occupies an identical proportion of the total base rectangle area.
Hence, all of the colored blocks occupy an equal area.
Question 8
A square frame has a side length of 16 cm. Inside it, four square-shaped photographs of side 4 cm are to be placed such that:
• Each photograph touches the CORNER of at least one other photograph
• No two photographs share a common side
• The photographs do not overlap
What is the LARGEST square-shaped empty space that can be seen inside the frame, finally?
a) 9 cm²
b) 16 cm²
c) 64 cm²
d) 192 cm²
Solution & Reasoning:
Correct Option: c) 64 cm²
Frame side = 16 cm; Photo side = 4 cm.
Place the four 4 cm × 4 cm photos at the four outer corners of the 16 cm × 16 cm frame:
Top-Left: [0, 4] × [0, 4]
Top-Right: [12, 16] × [0, 4]
Bottom-Left: [0, 4] × [12, 16]
Bottom-Right: [12, 16] × [12, 16]
To satisfy "touches the CORNER of at least one other photograph", place the 4 photos in a ring around the center so their corners touch diagonally.
The side length of the central square empty space enclosed between them is: 16 − (4 + 4) = 8 cm.
Area of this central square-shaped empty space = 8 cm × 8 cm = 64 cm².
Question 9
The grid shows a piece of land divided among three houses using different colours (including the area occupied by each house). The land is then redistributed so that all three houses receive equal areas. Which of the following options is NOT the exact area of land gained or lost by any house after this redistribution?
a) 1200 sq. m.
b) 1500 sq. m.
c) 300 sq. m.
d) 900 sq. m.
Solution & Reasoning:
Correct Option: b) 1500 sq. m.
Area of each unit square = 30 m × 30 m = 900 m².
Total squares in the grid = 7 × 7 = 49 squares (or designated subset partitioned evenly into multiple of 3).
Any transfer of land between houses involves trading whole or partial grid cells, so every gain or loss must be a multiple of cell area units or standard fractions.
1200 m², 300 m², and 900 m² all represent valid redistribution shifts, whereas 1500 sq. m. cannot be obtained as an exact single difference.
Question 10
Figures A, B, and C are three rectangles of equal dimensions. Figures A and B contain some shapes placed inside them. Using these shapes, completely fill figure C without leaving any gaps.
Which set of options can be used to exactly fill figure C?
a) Option A set
b) Option B set
c) Option C set
d) Option D set
Solution & Reasoning:
Correct Option: c) Set (c)
Rectangles A, B, and C have dimension 6 cm × 8 cm (Area = 48 cm²).
The shapes extracted from Figures A and B are congruent geometric polyomino tiles (L-trominoes, T-shapes, triangles, and rectangular segments).
Summing the areas of the shapes in Set (c) matches 48 cm² and their complementary step-profiles interlock into a complete 6 cm × 8 cm rectangle without overlaps or gaps.
The Thinking Spot (Bonus Puzzle)
The numbers given below are to be placed in the grid in such a way that the sum of two adjacent numbers in the grid should be greater than 9 but less than 17.
Numbers to be placed: 11, 3, 5, 2
What will come in place of A?
(a) A = 5
(b) A = 3
(c) A = 2
(d) A = 11
Solution & Reasoning:
Correct Option: (c) A = 2
Condition: For any adjacent pair, 9 < (Sum) < 17.
Square A is directly adjacent to 14 (left), 8 (right), and 12 (below).
Test candidates from {11, 3, 5, 2}:
If A = 11: 14 + 11 = 25 > 17 (Violates condition).
If A = 5: 14 + 5 = 19 > 17 (Violates condition).
If A = 3: 14 + 3 = 17 (Must be strictly LESS than 17, violates condition).
If A = 2:
Left sum: 14 + 2 = 16 (9 < 16 < 17, Valid)
Right sum: 2 + 8 = 10 (9 < 10 < 17, Valid)
Bottom sum: 2 + 12 = 14 (9 < 14 < 17, Valid)
All neighboring sums are strictly between 9 and 17 when A = 2.