Figure it Out — Rhombus and Trapezium — Area

1. 11 markGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q1

   Find the area of a rhombus whose diagonals are 20 cm and 15 cm.

   Hint. Half the product of the diagonals.

   Area = ½ × d₁ × d₂ = ½ × 20 × 15 = 150 cm²

   The sidelength was not needed and was not given, which is the whole convenience of the diagonal formula. (For the record, the diagonals halve each other at right angles, so the side is √(10² + 7.5²) = 12.5 cm; checking with base × height, the height would be 150 ÷ 12.5 = 12 cm.)

   ✦ Answer: 150 cm².

2. 3 (i)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q3(i)

   Find the area of a trapezium whose parallel sides are 10 ft and 7 ft, with the perpendicular distance between them marked 16 ft.

   Hint. The dashed segment with the right-angle mark is the height, however the figure is tilted.

   This trapezium is drawn tipped over, so the parallel sides are the slanted-looking ones, 10 ft and 7 ft, and the dashed 16 ft segment — marked with a right angle — is the perpendicular distance between them.

   Area = ½ × height × (sum of the parallel sides) = ½ × 16 × (10 + 7) = 8 × 17 = 136 ft²

   Reading the figure as though the longest number were the base would give ½ × 7 × 26 = 91 ft², which is wrong — the height must be the segment carrying the right-angle mark.

   ✦ Answer: 136 ft².

3. 3 (ii)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q3(ii)

   Find the area of a trapezium with parallel sides 24 m and 36 m and height 14 m.

   Hint. Add the parallel sides first — it keeps the arithmetic clean.

   Area = ½ × 14 × (24 + 36) = ½ × 14 × 60 = 7 × 60 = 420 m²

   Taking half of the 14 before multiplying is the quickest route: 7 × 60 = 420. The average of the two parallel sides is (24 + 36)/2 = 30 m, so this trapezium covers the same ground as a 30 m by 14 m rectangle ✓

   ✦ Answer: 420 m².

4. 3 (iii)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q3(iii)

   Find the area of a trapezium whose parallel sides are 14 in and 6 in, with 10 in marked as the perpendicular distance between them.

   Hint. Here the parallel sides are the vertical ones.

   The two vertical sides, 14 in on the left and 6 in on the right, are the parallel pair, and the dashed 10 in segment runs perpendicular between them.

   Area = ½ × 10 × (14 + 6) = 5 × 20 = 100 in²

   The figure is drawn on its side compared with the previous one, which is why it pays to identify the parallel pair from the shape rather than from where the numbers happen to sit on the page.

   ✦ Answer: 100 in².

5. 3 (iv)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q3(iv)

   Find the area of a trapezium with parallel sides 12 ft and 18 ft and height 8 ft.

   Hint. Straightforward substitution.

   Area = ½ × 8 × (12 + 18) = 4 × 30 = 120 ft²

   As a check, this trapezium matches a rectangle of width 15 ft (the average of 12 and 18) and height 8 ft, and 15 × 8 = 120 ✓

   ✦ Answer: 120 ft².

6. 24 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q2

   Give a method to convert a rectangle into a rhombus of equal area using dissection.

   Hint. Two isosceles triangles glued base to base make a rhombus.

   Aim. A rectangle L by w has area Lw. A rhombus of diagonals d₁ and d₂ has area ½d₁d₂, so taking d₁ = L and d₂ = 2w gives ½ × L × 2w = Lw — a match.

   Construction.

   1. Cut the rectangle in half across its length, into two rectangles of size (L/2) by w.
   2. Turn each of those into an isosceles triangle by the earlier move (cut along a diagonal, half-turn one piece about the midpoint of a side). Each becomes an isosceles triangle of base L and height w.
   3. Place the two triangles base to base, one the mirror image of the other.

   Why the result is a rhombus. The common base is the diagonal of length L, and the two apexes lie w above and w below it, directly over its midpoint — so the other diagonal is 2w and the two diagonals bisect each other at right angles. Each of the four sides is one of the equal sides of an isosceles triangle, and the triangles are congruent, so all four sides are equal: it is a rhombus.

   Area = ½ × L × 2w = Lw ✓ — the rectangle's area, exactly.

   With numbers. A 12 cm × 5 cm rectangle (60 cm²) becomes a rhombus with diagonals 12 cm and 10 cm, area ½ × 12 × 10 = 60 cm² ✓

   ✦ Answer: halve the rectangle, turn each half into an isosceles triangle, and join the two along their bases — the rhombus that results has diagonals L and 2w and the same area Lw.

7. 44 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q4 [Sulba-Sutras]

   [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.

   Hint. Cut vertically through the midpoints of the two slanted sides.

   Construction. Let the isosceles trapezium have parallel sides a (top) and b (bottom) and height h. Mark M and N, the midpoints of the two slanted sides, and drop perpendiculars from M and N to the bottom side, meeting it at P and Q. Cut along MP and NQ, then half-turn each cut-off triangle about M and N respectively.

   Where the pieces land. Put the bottom side from (0, 0) to (b, 0) and the top from (t, h) to (t + a, h), with t = (b − a)/2 for an isosceles trapezium. Then M = (t/2, h/2), and the half turn about M sends

   (0, 0) → (t, h) and (t/2, 0) → (t/2, h)

   so the left triangle flips up and fills the notch above exactly. The right-hand triangle behaves the same way by symmetry.

   The rectangle. What is left spans from x = t/2 to x = (t + a + b)/2, a width of

   (t + a + b)/2 − t/2 = (a + b)/2

   with height h. So the rectangle measures (a + b)/2 by h, giving area ½h(a + b) ✓ — the trapezium formula, obtained with scissors rather than algebra.

   With numbers. An isosceles trapezium with a = 6 cm, b = 10 cm, h = 4 cm has area 32 cm²; the rectangle is 8 cm × 4 cm = 32 cm² ✓

   ✦ Answer: cut vertically through the midpoints of the two slanted sides and half-turn each corner piece upward — the result is a rectangle of width (a + b)/2 and height h.

8. 54 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 169, Q5

   Trapezium ABCD is converted into rectangle EFGH of equal area, with ∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ. Given the trapezium, how do you find the vertices of the rectangle?

   Hint. The congruences tell you exactly where I and J must be.

   What the congruences force. In ∆AHI and ∆DGI the angles at I are vertically opposite and the angles at H and G are both right angles, so the only way the two triangles can be congruent is if AI = DI — that is, I is the midpoint of the side AD. The same reasoning on the other side makes J the midpoint of BC.

   The construction.

   1. Mark I, the midpoint of the slanted side AD, and J, the midpoint of BC.
   2. Through I draw a line perpendicular to DC; it meets the line AB at H and DC at G.
   3. Through J draw a line perpendicular to DC; it meets AB at E and DC at F.
   4. EFGH is the rectangle.

   Why the areas match. Going from trapezium to rectangle removes ∆AHI and adds ∆DGI, which are congruent, so nothing is lost or gained; the same happens on the right with ∆BEJ and ∆CFJ. Hence area(EFGH) = area(ABCD).

   What the rectangle measures. With D = (0, 0), C = (b, 0), A = (p, h), B = (p + a, h), the midpoints give HG at x = p/2 and EF at x = (p + a + b)/2, so the width is (a + b)/2 and the height is h — the trapezium formula once again.

   ✦ Answer: take I and J to be the midpoints of the two non-parallel sides and erect perpendiculars to DC through them — those perpendiculars cut out the rectangle EFGH, of width (a + b)/2 and height h.

9. 63 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 170, Q6

   Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm².

   Hint. Start from a rectangle of area 144 and read the trapezium off it.

   Work backwards from a rectangle. A trapezium of height h and parallel sides a and b has the same area as a rectangle of height h and width (a + b)/2. So pick any rectangle of area 144 cm² and split its width into two numbers averaging that width.

   Choice 1. Take the rectangle 12 cm × 12 cm. Then h = 12 and (a + b)/2 = 12, so a + b = 24. Choose a = 10 cm and b = 14 cm.

   Check: ½ × 12 × (10 + 14) = 6 × 24 = 144 cm² ✓

   Choice 2. Take the rectangle 18 cm × 8 cm. Then h = 8 and a + b = 36; choose a = 16 cm and b = 20 cm.

   Check: ½ × 8 × (16 + 20) = 4 × 36 = 144 cm² ✓

   Drawing it. Rule the base b, erect the height h at one end, draw a parallel to the base at that height, mark off a on it, and join up. The offset you choose — whether the trapezium is isosceles or leaning — makes no difference to the area, so there are infinitely many correct answers.

   ✦ Answer: any trapezium with ½h(a + b) = 144, for example h = 12 cm with parallel sides 10 cm and 14 cm, or h = 8 cm with parallel sides 16 cm and 20 cm.

10. 74 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 170, Q7

    A regular hexagon is divided by a long diagonal and a radius into a trapezium, an equilateral triangle and a rhombus. Find the ratio of their areas.

    Hint. Measure everything in units of the small equilateral triangle of side a.

    Set up. Let the hexagon have side a and centre O. Joining O to all six vertices cuts the hexagon into six equilateral triangles of side a, each of area T = (√3/4)a². The whole hexagon is 6T.

    The trapezium is the half of the hexagon cut off by the long diagonal, so it is three of those triangles:

    trapezium = 3T

    Checking with the formula: its parallel sides are the diagonal 2a and the opposite side a, and its height is (√3/2)a, so ½ × (√3/2)a × 3a = (3√3/4)a² = 3T ✓

    The equilateral triangle has O, one vertex and the next vertex as its corners. All three of those distances are a, so it is one of the six:

    triangle = T

    The rhombus is what remains of that half:

    rhombus = 3T − T = 2T

    Checking directly: its diagonals are a and √3a, so ½ × a × √3a = (√3/2)a² = 2T ✓

    The ratio.

    trapezium : triangle : rhombus = 3T : T : 2T = 3 : 1 : 2

    and the three add to 6T, the whole hexagon ✓

    ✦ Answer: 3 : 1 : 2.

11. 84 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 170, Q8

    ZYXW is a trapezium with ZY parallel to WX. A is the midpoint of XY, and ZA extended meets WX extended at B. Show that the area of trapezium ZYXW equals the area of ∆ZWB.

    Hint. Find a pair of congruent triangles at A — one inside the trapezium, one outside it.

    The two triangles to compare. Look at ∆ZYA and ∆BXA, which meet at A.

    • YA = AX, since A is the midpoint of XY
    • ∠ZYA = ∠BXA, alternate angles, because ZY ∥ WX with YX as transversal
    • ∠ZAY = ∠BAX, vertically opposite angles at A

    Two angles and the included side match, so by ASA, ∆ZYA ≅ ∆BXA, and congruent triangles have equal areas.

    Building both figures from the same parts. Let Q denote the quadrilateral ZAXW — the part of the trapezium below ZA. Then

    trapezium ZYXW = Q + ∆ZYA ∆ZWB = Q + ∆BXA

    because ∆ZWB is that same quadrilateral together with the triangle sticking out beyond X.

    Concluding. The two added triangles are equal in area, so the two totals are equal:

    Area(ZYXW) = Area(∆ZWB) ∎

    What has been achieved. A four-sided figure has been replaced by a triangle of exactly the same area — the trapezium version of the Śulba-Sūtra reshaping problems, and the reason ∆ZWB has base WB = WX + XB = WX + ZY, the sum of the parallel sides, with the same height. That is the area formula ½h(a + b) appearing yet again.

    ✦ Answer: ∆ZYA ≅ ∆BXA by ASA, so swapping one for the other turns the trapezium into ∆ZWB without changing the area — the two are equal.

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.