14 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 148
How many different ways can you divide a square into 4 parts of equal area? Describe several genuinely different ways, and explain the book's trick of compressing a part along one edge while expanding it along another.
Hint. You do not have to keep the four parts congruent — only their *areas* have to match.
The three obvious cuts. Take a square of side s, so its area is s².
| Cut | The four parts | Area of each |
|---|
| Two perpendicular lines through the centre | 4 small squares of side s/2 | s²/4 |
| The two diagonals | 4 congruent triangles, base s, height s/2 | ½ · s · s/2 = s²/4 |
| Three cuts parallel to one side | 4 strips of size s × s/4 | s²/4 |
The book's compress-and-expand idea. Start from any of these and redraw one boundary as a wiggle instead of a straight line. Wherever the wiggle bulges into part A it takes some area away from A and hands exactly that much to part B; wherever it bulges back it returns exactly as much. If every bulge is matched by an equal dent, both parts keep the area they started with, so all four parts stay equal. Since a wiggle can be drawn in endlessly many shapes, this alone gives infinitely many divisions.
The cleanest infinite family — the pinwheel. Draw any curve from the centre O of the square to a point on its boundary. Now rotate that curve about O through 90°, 180° and 270°. The four copies cut the square into four pieces that are exact rotations of one another, so they are congruent and therefore equal in area. Every different starting curve gives a different division.
✦ Answer: Infinitely many. Four squares, four triangles and four strips are the simplest; any curve from the centre plus its three 90° rotations gives a fresh one; and any boundary can be wiggled so long as every bulge is matched by an equal dent.
22 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 149
Two rangoli rectangles are to be filled evenly with powder — one measures 7 cm by 4 cm and the other 8 cm by 3 cm. Which one needs more powder? What do you notice about their perimeters?
Hint. Count the 1 cm unit squares that pack into each rectangle.
Count unit squares. The rectangle 7 cm by 4 cm holds 7 × 4 = 28 unit squares of side 1 cm, so its area is 28 cm². The rectangle 8 cm by 3 cm holds 8 × 3 = 24 such squares, so its area is 24 cm².
Powder spread evenly means the powder used is proportional to area, so the first rectangle needs more — 4 cm² more, which is one-sixth extra.
The surprise. Look at the boundaries:
- Perimeter of the 7 × 4 rectangle = 2(7 + 4) = 22 cm
- Perimeter of the 8 × 3 rectangle = 2(8 + 3) = 22 cm
The two rectangles have exactly the same perimeter but different areas. That is the first warning that the length of a boundary tells you nothing reliable about the space inside it.
✦ Answer: The 7 cm × 4 cm rectangle needs more powder — 28 cm² against 24 cm² — even though both rectangles have the same perimeter, 22 cm.
32 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 150
A diagonal is drawn in the 7 cm by 4 cm rectangle. What is the area of each of the two triangles it creates?
Hint. What does a diagonal do to a rectangle?
A diagonal cuts a rectangle into two triangles that are congruent — each has the rectangle's length and width as two of its sides and the diagonal as the third, so one is a half-turn of the other. Congruent figures have equal areas, and the two together make up the whole rectangle, so each is exactly half of it.
Area of each triangle = ½ × 28 = 14 cm²
In terms of unit squares: the 28 unit squares of the rectangle get shared out so that each triangle covers the equivalent of 14 of them — some squares are cut by the diagonal, but each such square is split into two halves that go one to each triangle.
Written with the sides directly: ½ × 7 × 4 = 14 cm². This is the first appearance of the formula ½ × base × height.
✦ Answer: 14 cm² each.
45 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 150 (Math Talk)
Find two rectangles, Region 1 and Region 2, with Perimeter of Region 1 greater than Perimeter of Region 2 but Area of Region 1 less than Area of Region 2. Then give an example using shapes that are not rectangles, where the same reversal is obvious just by looking.
Hint. Long and thin is the way to buy a lot of boundary with very little area.
Two rectangles. Take
- Region 1 = a 1 cm × 10 cm strip: perimeter 2(1 + 10) = 22 cm, area 1 × 10 = 10 cm²
- Region 2 = a 5 cm × 5 cm square: perimeter 2(5 + 5) = 20 cm, area 5 × 5 = 25 cm²
So Perimeter 1 (22) > Perimeter 2 (20) while Area 1 (10) < Area 2 (25). The strip has more boundary and less than half the space.
The reason is easy to see: stretching a rectangle thin adds two long sides to the perimeter while squeezing the area towards zero. Push it further — 1 cm × 100 cm has perimeter 202 cm and area 100 cm², while a 20 cm × 20 cm square has perimeter 80 cm and area 400 cm².
Shapes that are not rectangles. Cut a comb: start with a 10 cm × 1 cm strip and slit nine 0.8 cm deep notches into it. Every notch adds about 1.6 cm of boundary and removes area, so the comb ends up with a perimeter well over 30 cm and an area under 3 cm² — against a 2 cm × 2 cm square with perimeter 8 cm and area 4 cm². The comb looks spindly and its area really is smaller, yet its outline is nearly four times as long. A spiral or a star does the same job.
What this settles. Perimeter measures the length of the edge; area measures the space inside. Neither controls the other, which is why area has to be defined by counting unit squares rather than by measuring the boundary.
✦ Answer: e.g. a 1 cm × 10 cm rectangle (perimeter 22 cm, area 10 cm²) against a 5 cm × 5 cm square (perimeter 20 cm, area 25 cm²); among other shapes, a notched comb beats a small square on perimeter while losing badly on area.
Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp207.pdf), Chapter 7 'Area', pages 148-171. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every answer was derived from first principles and independently recomputed in Python. FIGURES READ OFF HIGH-DPI RENDERS AND RE-SOLVED: the p.150 pinwheel, whose four rectangles all turn out to have a side of 7 in (areas 14, 21, 28, 35), giving the missing width 2 in; the p.150 step figure, where the BOLD outline encloses the dotted region together with the region below it - 50 sq m in all - so the widths are 29/4 = 7.25 m and 11/4 = 2.75 m (summing to exactly 10 m) and the missing height is (50-29)/7.25 = 84/29 = 2.90 m, which matches the 2.74 m measured off the printed drawing; the p.152 spiral tube, whose nine arms total 120 and whose eight corners are each double-counted once, giving 112 sq units - confirmed by rasterising the nine arms at 20 cells per unit and counting - with the hint's L-tube (arms 5 and 5) coming to 5+5-1 = 9; the p.152 square whose regions are s^2/4, s^2/4 and s^2/2, so doubling the side raises each by 3 times its own area; the p.157 triangles, whose '4 cm' label is centred on BC (not EC) giving areas 6, 8 and 6 sq cm; the p.158 obtuse triangle giving BY = 3 units; the p.158 three-square figure, where the line from D to H crosses the top of the first square at its midpoint, making the red region exactly s^2 and the blue exactly s^2/4, hence 12.25 sq units and 144 sq units; the p.160 quadrilateral (66 sq cm) and shaded region (180-30-40 = 110 sq cm); the p.160 blue 'bowtie', whose two triangles share the full width and whose heights sum to the rectangle's, so it is exactly half - measured as 0.4987 of the printed rectangle; the seven p.162 parallelograms, measured by connected-component analysis to have identical filled areas to within 0.02% (all base 5, height 3, area 15) with leans of 0.4, 1.2, 2.5, 0.6, 0.9, 2.0 and 2.8 grid units, so (g) has the maximum perimeter and (a) the minimum; the p.163 parallelograms (28, 15, 24 and 8.8 sq cm) and QN = 72/7.6 = 180/19 = 9.47 cm; the p.169 trapezia (136 sq ft, 420 sq m, 100 sq in and 120 sq ft); the p.170 hexagon, whose long diagonal and radius cut it into 3, 1 and 2 of the six unit equilateral triangles, giving the ratio 3:1:2; and the p.170 trapezium ZYXW, where ASA gives triangle ZYA congruent to triangle BXA. Unit work checked: A4 = 21 x 29.7 = 623.7 sq cm; 1 sq in = 2.54^2 = 6.4516 sq cm; 161.29 / 6.4516 = 25 sq in exactly; 5 in = 12.7 cm; 7.4 in = 18.796 cm; 5.08 cm = 2 in; 11.43 cm = 4.5 in; 1 sq ft = 144 sq in; 1 sq km = 1,000,000 sq m; 1 acre = 43,560 sq ft.. Questions are referenced from the NCERT textbook for identification.
