Figure it Out — Rectangles, Paths and Composite Regions — Area

1. 1 (i)4 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 150, Q1(i)

   Four rectangles are arranged in a pinwheel. Their areas are 28 in², 21 in², 35 in² and 14 in². The figure marks 4 in (the part of the 28 in² rectangle sticking above the 21 in² one), 7 in (the length of the 21 in² rectangle), 3 in (the part of the 35 in² rectangle sticking out to the left) and 2 in (the part of the 14 in² rectangle hanging below the 35 in² one). Find the missing width marked ?.

   Hint. Work round the pinwheel. Each rectangle hands you one side of the next one.

   Go round the pinwheel one rectangle at a time; each answer unlocks the next.

   Step 1 — the 21 in² rectangle. Its length is marked 7 in, so its width is 21 ÷ 7 = 3 in.

   Step 2 — the 28 in² rectangle. Its bottom edge lines up with the bottom of the 21 in² rectangle, and it sticks 4 in above the top of it. So its height is 4 + 3 = 7 in, which makes its width 28 ÷ 7 = 4 in.

   Step 3 — the 35 in² rectangle. Its right edge lines up with the right edge of the 28 in² rectangle and it sticks out 3 in to the left of it, so its width is 3 + 4 = 7 in. Its height is therefore 35 ÷ 7 = 5 in.

   Step 4 — the 14 in² rectangle. Its top lines up with the top of the 35 in² rectangle and it hangs 2 in below it, so its height is 5 + 2 = 7 in. That gives the missing width:

   ? = 14 ÷ 7 = 2 in

   The pattern worth spotting. Every one of the four rectangles turned out to have a side of 7 in — that is what makes a pinwheel close up. Their areas 14, 21, 28, 35 are 7 × 2, 7 × 3, 7 × 4, 7 × 5.

   ✦ Answer: ? = 2 in.

2. 1 (ii)4 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 150, Q1(ii)

   A step-shaped figure is made of three rectangles. The top band is 4 m tall; its left part (dotted) has area 29 m² and its right part (hatched) has area 11 m². Below the dotted rectangle sits a third rectangle of the same width. The bold outline — the dotted rectangle together with the one below it — encloses 50 m². Find the two missing widths and the missing height.

   Hint. Every rectangle in the top band has height 4 m, so dividing an area by 4 gives a width.

   The two widths. Both parts of the top band are 4 m tall, so:

   width of the dotted rectangle = 29 ÷ 4 = 7.25 m width of the hatched rectangle = 11 ÷ 4 = 2.75 m

   A useful check: the whole top band is 29 + 11 = 40 m² and 4 m tall, so it must be 40 ÷ 4 = 10 m wide — and 7.25 + 2.75 = 10 ✓.

   The missing height. The bold outline encloses the dotted rectangle and the rectangle below it, together 50 m². The dotted one accounts for 29 m², so the lower rectangle has area 50 − 29 = 21 m². It is the same width as the dotted one, 7.25 m, which gives

   height = 21 ÷ 7.25 = 2100 ÷ 725 = 84/29 ≈ 2.9 m

   A note on the answer. This one does not come out round, and it is not meant to: 7.25 does not divide 21 neatly. Leave it as the exact fraction 84/29 m and quote 2.9 m as the rounded value. (Measuring the printed figure gives about 2.7 m, so the drawing is to scale and the answer is in the right range.)

   ✦ Answer: widths 7.25 m and 2.75 m (which add to exactly 10 m), and height 84/29 m ≈ 2.9 m.

3. 25 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 151, Q2

   A path is laid all round a rectangular park EFGH; the outer boundary of the path is the rectangle ABCD. (i) What measurements do you need to find the area of the path? Assign values of your own and give a formula. (ii) If instead you are told the width of the path along each side, can you find its area? What else do you need? (iii) Does the area of the path change if the outer rectangle is slid about while the park stays inside it?

   Hint. The path is what is left when you remove the park from the big rectangle.

   (i) Two rectangles, one subtraction. The path is exactly the big rectangle with the park removed, so

   Area of path = Area(ABCD) − Area(EFGH)

   The measurements needed are the length and width of each rectangle: AB and BC for the outside, EF and FG for the park. Taking AB = 20 m, BC = 15 m, EF = 16 m, FG = 11 m:

   Area of path = (20 × 15) − (16 × 11) = 300 − 176 = 124 m²

   Formula: Area of path = L × W − ℓ × w

   (ii) The widths alone are not enough. Knowing that the path is, say, 2 m wide all round tells you how much wider the outside is than the park, but not how big either rectangle is — a 2 m path round a tiny park and a 2 m path round a huge park have very different areas. You still need the park's length and width (or the outer ones).

   With park ℓ = 16 m, w = 11 m and a uniform width t = 2 m, break the path into pieces:

   Piece	Count	Each	Total
   Strips along the long sides	2	16 × 2 = 32 m²	64 m²
   Strips along the short sides	2	11 × 2 = 22 m²	44 m²
   Corner squares	4	2 × 2 = 4 m²	16 m²
   			124 m²

   which agrees with part (i). In symbols, for a uniform width t:

   Area of path = 2t(ℓ + w) + 4t² = 2t(ℓ + w + 2t)

   Check: 2(2)(16 + 11 + 4) = 4 × 31 = 124 ✓. If the four widths differ — p and q at the sides, r and s at top and bottom — then the outer rectangle is (ℓ + p + q) by (w + r + s) and the path is (ℓ + p + q)(w + r + s) − ℓw.

   (iii) No, the area does not change. Sliding the outer rectangle changes neither rectangle's size, and the path is always the difference of the two areas. What changes is the shape of the path: as one strip widens, the strip opposite it narrows by exactly the same amount. So long as the park stays completely inside, the path keeps its 124 m².

   ✦ Answer: (i) the two rectangles' lengths and widths, giving Area = LW − ℓw = 124 m² for 20 × 15 and 16 × 11; (ii) no — you also need the park's dimensions, after which Area = 2t(ℓ + w + 2t) = 124 m²; (iii) no change — only the shape of the path moves, not its area.

4. 34 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 151, Q3

   A plot measures 14 m by 12 m and has a crosspath running right across it. What further measurements do you need to find the area of the crosspath? Choose values and give a formula.

   Hint. The two arms overlap in the middle — do not pay for that patch twice.

   What is missing. The plot's sides are given, but a path has a width; you need the width of each arm. Let a be the width of the arm running the 12 m way (top to bottom) and b the width of the arm running the 14 m way (left to right).

   Building the formula. Each arm crosses the whole plot, so

   vertical arm = 12 × a and horizontal arm = 14 × b

   Adding those two counts the small rectangle where the arms cross twice, once in each, so subtract one copy of it:

   Area of crosspath = 12a + 14b − ab

   With numbers. Take a = 2 m and b = 2 m:

   Area = 12(2) + 14(2) − (2)(2) = 24 + 28 − 4 = 48 m²

   The arms need not be equally wide. With a = 2 m and b = 1.5 m: 24 + 21 − 3 = 42 m².

   A point worth noticing. Where the arms are placed makes no difference. Whether the cross sits dead centre or off to one side, each arm still spans the full plot and the overlap is still a × b, so the answer is the same — the same principle as sliding the outer rectangle in the previous question.

   ✦ Answer: you need the width of each arm; then Area = 12a + 14b − ab, which is 48 m² when both arms are 2 m wide.

5. 45 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 152, Q4

   Find the area of the spiral tube, which has width 1 throughout. Its arms measure 20, 20, 20, 15, 15, 10, 10, 5 and 5. The hint shows a bent tube with two arms of length 5 — what length of straight tube has the same area as that bent tube?

   Hint. Straighten the tube. Each bend costs you exactly one 1 × 1 square.

   The hint first. The bent tube has arms of length 5 and 5 and width 1, so the two arms cover 5 + 5 = 10 unit squares — except that the 1 × 1 square at the corner belongs to both arms and has been counted twice. Removing that one square:

   Area of the bent tube = 5 + 5 − 1 = 9 sq units

   A straight tube of width 1 and length L has area L, so the straight tube must be 9 units long.

   Now the spiral. It is made of nine arms with eight bends between them:

   20 + 20 + 20 + 15 + 15 + 10 + 10 + 5 + 5 = 120

   Each of the eight bends double-counts one 1 × 1 corner square, exactly as in the hint, so subtract eight:

   Area of the spiral = 120 − 8 = 112 sq units

   Check by straightening. Unroll the spiral into one straight tube of width 1. Every bend that is straightened out gives back one corner square, so the straight tube is 120 − 8 = 112 units long and its area is 112 × 1 = 112 sq units — the same number, which is what you would expect since unrolling neither creates nor destroys material.

   (I also checked this by drawing the nine arms on a fine grid and counting the squares covered: 112 exactly.)

   ✦ Answer: the spiral tube has area 112 sq units; the straight tube matching the hint's bent tube is 9 units long.

6. 54 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 152, Q5

   A square is cut by one diagonal, and by a segment from another corner to the midpoint of that diagonal, giving regions 1, 2 and 3. If the sidelength of the square is doubled, what is the increase in the areas of regions 1, 2 and 3? Give reasons.

   Hint. First say what fraction of the square each region is; then ask what doubling the side does to the square itself.

   What the three regions are. Let the square have side s, so its area is s². The diagonal cuts it into two equal halves, so region 3 — the half on one side of the diagonal — has area s²/2. The other half is a triangle, and the segment drawn to the midpoint of the diagonal is a median of that triangle (the tick marks in the figure mark the two equal halves of the diagonal). A median splits a triangle into two triangles of equal area, so regions 1 and 2 are each half of that half:

   Region 1 = s²/4 Region 2 = s²/4 Region 3 = s²/2

   Doubling the side. Replacing s by 2s makes the square's area (2s)² = 4s², that is four times as big. Each region is a fixed fraction of the square — one quarter, one quarter, one half — and those fractions do not depend on s, so every region also becomes four times as big.

   Region	Before	After	Increase
   1	s²/4	s²	3s²/4
   2	s²/4	s²	3s²/4
   3	s²/2	2s²	3s²/2

   Reading the table. In every row the increase is exactly three times the original area — a 300% rise — because going from one unit to four units means adding three. The three regions gain different amounts (3s²/4, 3s²/4, 3s²/2) but the same proportion.

   With s = 4, for instance: regions 4, 4, 8 become 16, 16, 32, increases of 12, 12, 24.

   ✦ Answer: each region becomes 4 times its old area, so the increase is 3 times the original in every case — region 1 rises by 3s²/4, region 2 by 3s²/4 and region 3 by 3s²/2.

7. 65 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 152, Q6

   Divide a square into 4 parts by drawing two perpendicular lines inside it, as in the figure. Rearrange the pieces to get a larger square with a hole inside. Explain why it works and what the hole's size is.

   Hint. Two perpendicular cuts each running from one side of a square to the opposite side always have the same length. That common length becomes the new square's side.

   The cut. Draw two lines inside the square that are perpendicular to each other, each running from one side across to the opposite side, crossing at the centre. They slice the square into four pieces.

   Why the two cuts are equal in length. Rotating the whole square 90° about its centre carries one cut onto the direction of the other and carries the square onto itself, so the two cuts must be the same length. Call it L. If the square has side s and one cut lands k units to the side of the perpendicular through its start, then by the Baudhāyana–Pythagoras relation

   L² = s² + k²

   The rearrangement. The four pieces are congruent — each is the next one turned 90° about the centre. Now re-glue them in a pinwheel, matching each piece's cut edge of length L against a different piece's cut edge. The outside closes up into a square of side L, and because the pieces no longer nest as tightly, a square hole opens in the middle.

   Size of the hole. The four pieces still carry all of the original s² of material, and they now sit inside a square of area L², so

   hole = L² − s² = (s² + k²) − s² = k²

   — a square hole of side k, exactly the offset of the cut. Choosing k = 0 puts the cuts along the diagonals and the hole disappears; the further the cuts lean, the bigger the hole.

   Where you have met this before. This is Perigal's dissection, and the picture — four pieces round a small square filling a bigger square — is the Baudhāyana–Pythagoras figure from Chapter 3 read backwards. Cutting card or thick paper and sliding the pieces makes it obvious in a way no amount of algebra does.

   ✦ Answer: the four pieces re-form as a square of side L = √(s² + k²) with a square hole of side k in the middle, where s is the original side and k the offset of the cuts; the areas balance because L² − k² = s².

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.