Figure it Out — Parallelograms, Triangles and Sulba-Sutra Constructions — Area

1. 34 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 163, Q3

   In parallelogram PQRS, QM is perpendicular to SR with QM = 6 cm and SR = 12 cm, and QN is perpendicular to PS with PS = 7.6 cm. Find QN.

   Hint. Compute the area once from the pair you know, then read QN off the other pair.

   Area from the pair you are given. Base SR with its height QM:

   Area(PQRS) = SR × QM = 12 × 6 = 72 cm²

   The same area from the other pair. Base PS with its height QN:

   Area(PQRS) = PS × QN = 7.6 × QN

   Equating.

   7.6 × QN = 72 QN = 72 ÷ 7.6 = 720/76 = 180/19 ≈ 9.47 cm

   Does that make sense? PS = 7.6 cm is much shorter than SR = 12 cm, so its height has to be correspondingly larger — and 9.47 > 6, as expected. The height also has to be no more than the other side length, 12 cm, and 9.47 comfortably clears that test.

   On the untidy answer. 72 ÷ 7.6 is not a whole number and does not terminate. Leave it as the exact fraction 180/19 and quote 9.47 cm as the rounded value rather than pretending it is 9.5.

   ✦ Answer: QN = 180/19 ≈ 9.47 cm.

2. 43 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 163, Q4

   A rectangle and a parallelogram both have sidelengths 5 cm and 4 cm. Which has the greater area?

   Hint. Build them on the same 5 cm base and compare their heights.

   Put them on the same base. Stand both figures on a base of 5 cm. The other side is 4 cm in each case.

   • For the rectangle, that 4 cm side is upright, so it is the height: Area = 5 × 4 = 20 cm².
   • For the parallelogram, the 4 cm side leans over. Its height h is the perpendicular from the top edge down to the base, and it is a leg of a right triangle whose hypotenuse is that 4 cm side. A leg is always shorter than the hypotenuse, so h < 4 and Area = 5 × h < 20 cm².

   With numbers. A parallelogram of sides 5 and 4 leaning by 3 cm has height √(16 − 9) = √7 ≈ 2.65, giving area ≈ 13.2 cm² — well under 20.

   The general statement. Among all parallelograms with two given sidelengths, the rectangle has the greatest area, because it is the one whose height reaches the full sidelength. Lean it at all and the area falls; lean it flat and the area tends to zero.

   ✦ Answer: the rectangle, with area 20 cm², beats every parallelogram with the same two sidelengths, since a leaning side gives a height of less than 4 cm.

3. 54 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 163, Q5

   Give a method to obtain a rectangle whose area is twice that of a given triangle. What different methods can you think of?

   Hint. The ½ in the triangle formula is exactly what you are being asked to cancel.

   A triangle of base b and height h has area ½bh, so a rectangle of twice that area must measure bh — which is the rectangle on the triangle's own base and its own height.

   Method 1 — box the triangle in. Draw the line through the apex parallel to the base and drop perpendiculars from the two base ends to it. The rectangle formed has base b and height h, so its area is bh = twice the triangle. The triangle sits inside it and covers exactly half. This is the picture the chapter opened with.

   Method 2 — copy and half-turn. Take a second copy of the triangle and give it a half turn about the midpoint of one side. The copy joins the original to make a parallelogram of base b and height h, whose area is bh. Then square that parallelogram up into a rectangle by the cut-and-slide dissection.

   Method 3 — stretch a rectangle. First build the rectangle of the same area as the triangle (base b, height h/2, from the midline dissection), then simply double one of its sides.

   All three end at a rectangle of area bh. Method 1 needs only a set square, method 2 needs only scissors, and method 3 reuses work already done.

   With numbers. A triangle of base 8 cm and height 5 cm has area 20 cm²; the rectangle 8 cm × 5 cm has area 40 cm² = 2 × 20 ✓

   ✦ Answer: the rectangle on the triangle's own base and own height has exactly twice the area — obtainable by boxing the triangle in, by half-turning a second copy, or by doubling a side of the equal-area rectangle.

4. 63 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 164, Q6 [Sulba-Sutras]

   [Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle.

   Hint. Halve the height rather than keeping it.

   What is wanted. ½bh has to be matched by a rectangle, and the cleanest choice of dimensions is b by h/2.

   Construction. Given ∆ABC with base BC = b and height h:

   1. Bisect the height — mark the midpoints M of AB and N of AC, so MN lies at height h/2 and is parallel to BC.
   2. Cut along MN.
   3. Half-turn the top piece ∆AMN about N; the vertex A lands exactly on C, so the piece slots in beside the bottom piece and the two form a parallelogram on base BC of height h/2.
   4. Square that parallelogram up by dropping a perpendicular at one end and sliding the corner triangle across.

   Result. A rectangle b by h/2, of area bh/2 = the triangle's area.

   Why this is the Śulba-Sūtra problem. Vedic altar builders had to reshape a prescribed area into a prescribed form without gaining or losing an inch of ground, so every one of their transformations is a dissection like this — physical, checkable with cord and pegs, and exact.

   Numerical check. Base 9 cm, height 6 cm: triangle 27 cm²; rectangle 9 × 3 = 27 cm² ✓

   ✦ Answer: the rectangle of the triangle's base and half its height, built by cutting along the midline and half-turning the top piece.

5. 74 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 164, Q7 [Sulba-Sutras]

   [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Find how. [Hint: show that ∆ADB and ∆ADC can be made into halves of a rectangle.]

   Hint. One cut along the altitude, then one half-turn.

   The cut. Let ∆ABC be isosceles with AB = AC, and let AD be the altitude from A to BC. Because the triangle is isosceles, D is the midpoint of BC, and ∆ABD and ∆ACD are congruent right triangles.

   The move. Cut along AD. Keep ∆ACD where it is, and give ∆ABD a half turn about the midpoint of AD.

   Tracking the corners. Put D = (0, 0), B = (−m, 0), C = (m, 0) and A = (0, h), so the midpoint of AD is (0, h/2). The half turn sends

   D(0, 0) → (0, h) = A B(−m, 0) → (m, h) A(0, h) → (0, 0) = D

   so ∆ABD lands as the triangle with corners (0, h), (m, h), (0, 0). Together with ∆ACD, whose corners are (0, 0), (m, 0), (0, h), the two pieces fill the rectangle with corners (0, 0), (m, 0), (m, h), (0, h) — no gap and no overlap.

   The rectangle. Its width is m = BC/2 and its height is h = AD, so its area is ½ × BC × AD, which is the triangle's area ✓

   This is simpler than the general midline method because it needs only one cut and one turn, which is why the Śulba-Sūtras single out the isosceles case.

   ✦ Answer: cut along the altitude AD and half-turn one half about the midpoint of AD — the two right triangles then form a rectangle of width BC/2 and height AD.

6. 83 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 164, Q8 [Sulba-Sutras]

   [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection.

   Hint. Run the previous construction backwards.

   Reverse the moves of the last question. Given a rectangle of width m and height h, place it with corners (0, 0), (m, 0), (m, h), (0, h).

   1. Cut along the diagonal from (0, h) to (m, 0).
   2. Keep the lower triangle (0, 0), (m, 0), (0, h) fixed.
   3. Half-turn the upper triangle (0, h), (m, h), (0, 0) about the midpoint of the left edge, the point (0, h/2). It lands as the triangle (0, 0), (−m, 0), (0, h).

   The result. The two pieces together form the triangle with corners (−m, 0), (m, 0) and (0, h) — base 2m, height h, and isosceles because the apex sits directly above the midpoint of the base.

   Area = ½ × 2m × h = mh = the rectangle ✓

   So the rule is: the isosceles triangle on twice the rectangle's width, with the rectangle's height, matches it exactly.

   With numbers. A 6 cm × 5 cm rectangle (30 cm²) becomes the isosceles triangle of base 12 cm and height 5 cm, area ½ × 12 × 5 = 30 cm² ✓

   ✦ Answer: cut along a diagonal and half-turn one piece about the midpoint of a side — this yields an isosceles triangle of base twice the rectangle's width and the same height.

7. 94 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 164, Q9

   Which has the greater area — an equilateral triangle or a square of the same sidelength? And which is greater — two identical equilateral triangles together, or a square of that same sidelength? Give reasons.

   Hint. Compare the triangle's height with its own side, without reaching for a calculator.

   The key fact. In an equilateral triangle of side s, the altitude from a vertex splits the base in half and is a leg of a right triangle whose hypotenuse is the side s. A leg is always shorter than the hypotenuse, so the height h satisfies

   h < s (exactly, h = (√3/2)s ≈ 0.87s)

   One triangle against the square.

   triangle = ½ × s × h < ½ × s × s = ½s²

   so the triangle is less than half the square s². Numerically, 0.433s² against s². The square wins easily. You can even see it without arithmetic: stand the triangle on the bottom side of the square and its apex, at height 0.87s, stays inside — the triangle fits within the square with room to spare.

   Two triangles against the square. Two of them together give

   2 × ½ × s × h = s × h < s × s = s²

   so even the pair falls short — 0.866s² against s². The square still wins, though now only by about 13%.

   How many would it take? Since each triangle is about 0.433s², you would need just over two of them — about 2.31 — to cover the square.

   ✦ Answer: the square is larger in both comparisons, because the equilateral triangle's height (≈ 0.87s) is less than its side, making one triangle ≈ 0.433s² and two of them ≈ 0.866s², both under 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.