1 (i)3 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 162, Q1(i)
Seven parallelograms (a) to (g) are drawn on the same grid, each standing on a base of 5 grid units between the same pair of horizontal gridlines 3 units apart. What can we say about their areas?
Hint. Base and height are the only two things the formula asks for.
Read the two numbers off the grid. Every one of the seven parallelograms has
- base = 5 grid units (count along the bottom edge)
- height = 3 grid units (the gap between the horizontal line the base sits on and the line the opposite side sits on)
The figures lean by wildly different amounts, from almost upright in (a) to strongly slanted in (g), but leaning does not change either number.
Area = base × height = 5 × 3 = 15 square units, for every one of them
Why leaning cannot matter. Shearing a parallelogram — sliding its top edge along its own line — is exactly the situation of triangles between parallel lines: the base stays put and the height is the fixed distance between two parallel lines. So the area is locked.
(Measuring the seven printed figures confirms it: the shaded areas agree with one another to within 0.02%.)
✦ Answer: all seven have the same area, 15 square units — the slant is irrelevant because base and height are the same in every case.
1 (ii)4 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 162, Q1(ii)
What can we say about the perimeters of those same seven parallelograms? Which appears to have the maximum perimeter and which the minimum?
Hint. The two horizontal sides never change. Only the slanted sides do.
Equal areas, unequal perimeters. Each parallelogram has two sides of length 5 (the base and the side opposite it) and two slanted sides. If the top edge is shifted sideways by k units relative to the bottom edge, then each slanted side is the hypotenuse of a right triangle with legs 3 (the height) and k, so
slant = √(9 + k²) and Perimeter = 2(5 + √(9 + k²))
The perimeter therefore grows steadily as the lean k grows — and it is never less than 2(5 + 3) = 16, the value for a perfectly upright rectangle.
Measuring the seven figures.
| Figure | lean k | slant | perimeter |
|---|
| (a) | 0.4 | 3.03 | 16.1 |
| (d) | 0.6 | 3.05 | 16.1 |
| (e) | 0.9 | 3.14 | 16.3 |
| (b) | 1.2 | 3.22 | 16.4 |
| (f) | 2.0 | 3.63 | 17.3 |
| (c) | 2.5 | 3.93 | 17.9 |
| (g) | 2.8 | 4.11 | 18.2 |
Answering the question. Figure (g) leans the most, so it has the largest perimeter, about 18.2 units. Figure (a) is the most nearly upright, so it has the smallest, about 16.1 units — with (d) so close behind that the two are hard to separate by eye.
The moral. Equal areas say nothing about perimeters, just as equal perimeters said nothing about areas back on page 149. The two measurements are independent.
✦ Answer: the perimeters differ even though the areas do not; (g) has the maximum (about 18.2 units) and (a) the minimum (about 16.1 units), with (d) a very close second.
2 (i)1 markGanita Prakash Cl-8 Part 2, Figure it Out, page 163, Q2(i)
Find the area of a parallelogram with base 7 cm and the perpendicular height marked as 4 cm.
Hint. Straight substitution.
The dashed segment marked 4 cm meets the base at a right angle, so it is the height belonging to the 7 cm base.
Area = base × height = 7 × 4 = 28 cm²
The slanted sides are not marked, and they are not needed — only the base and its perpendicular height enter the formula.
✦ Answer: 28 cm².
2 (ii)1 markGanita Prakash Cl-8 Part 2, Figure it Out, page 163, Q2(ii)
Find the area of a parallelogram with base 5 cm and perpendicular height 3 cm.
Hint. The diagonal drawn across the figure is a distraction.
Using the marked pair:
Area = base × height = 5 × 3 = 15 cm²
The diagonal drawn inside this figure splits it into two triangles of ½ × 5 × 3 = 7.5 cm² each, which add back to 15 cm² — a quick way to check the answer, and a reminder that a diagonal always halves a parallelogram, since the two halves are congruent.
✦ Answer: 15 cm².
2 (iii)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 163, Q2(iii)
Find the area of a parallelogram in which the side marked 5 cm carries a perpendicular of length 4.8 cm drawn to it from the opposite vertex.
Hint. The 5 cm side is the base here, not the horizontal one.
This figure is labelled the other way round from the first two: the 5 cm measurement runs down the right-hand side, and the dashed 4.8 cm segment meets that side at a right angle. So the base–height pair to use is 5 cm with 4.8 cm.
Area = 5 × 4.8 = 24 cm²
Any side of a parallelogram may serve as the base as long as the height used is the one perpendicular to it, which is exactly why the figure has been drawn tipped over — to make you match the pair rather than reach for whatever is at the bottom of the page.
✦ Answer: 24 cm².
2 (iv)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 163, Q2(iv)
Find the area of a parallelogram in which the side marked 2 cm carries a perpendicular of 4.4 cm drawn to it.
Hint. Again, pair the marked perpendicular with the side it meets.
The dashed 4.4 cm segment ends on the 2 cm side at a right angle, so those two are the matching pair.
Area = 2 × 4.4 = 8.8 cm²
It may look strange that the height (4.4 cm) is more than twice the base (2 cm), but that is perfectly normal for a very flat, strongly leaning parallelogram — the height belonging to a short side is large, precisely so that the product stays equal to the area computed from the long side.
✦ Answer: 8.8 cm².
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.
