In-text — Trapezium — Area

1. 13 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 166

   Find the areas of the three given trapeziums by breaking them into figures whose areas can already be computed. Describe the break-up in each case.

   Hint. Drop perpendiculars from the ends of the shorter parallel side.

   The general recipe. Drop a perpendicular from each end of the shorter parallel side down to the longer one. That splits any trapezium into a middle rectangle with a right triangle on each side, and all three are shapes whose areas you already know.

   Trapezium ABCD, with one perpendicular DM drawn. Only one perpendicular is needed here because one of the slanted sides is already upright — this is a right trapezium. It breaks into one rectangle plus one right triangle:

   Area = (rectangle) + ½ × (base of triangle) × (height)

   Trapezium PQRS, with nothing drawn. Either drop the two perpendiculars as above, giving rectangle + two triangles, or simply join a diagonal, which gives two triangles standing on the two parallel sides with the same height:

   Area = ½ × a × h + ½ × b × h

   Trapezium WXYZ, with perpendiculars WM and XN drawn. This is the full case: rectangle WXNM in the middle, with ∆WMZ on the left and ∆XNY on the right.

   Area = ½ × MZ × WM + WX × WM + ½ × NY × XN

   Every one of these break-ups leads to the same formula once you tidy it up, which is the point of the next question.

   ✦ Answer: rectangle + triangle for a right trapezium, rectangle + two triangles (or two triangles via a diagonal) in general — every piece being a shape whose area is already known.

2. 25 marksGanita Prakash Cl-8 Part 2, Section 7.1, pages 166-167

   Take trapezium WXYZ with WX parallel to ZY. Construct WM and XN perpendicular to ZY, put MZ = x, WM = XN = h, WX = a, NY = y, ZY = b. Show that the area works out to ½h(a + b).

   Hint. Write the total in x, y, a, h first — then get rid of x and y.

   Is WXNM really a rectangle? WX ∥ ZY, and WM, XN are both perpendicular to ZY, so ∠MWX = ∠NXW = 90° (interior angles on the same side of a transversal add to 180°). With four right angles it is a rectangle, so MN = WX = a and WM = XN = h.

   Add the three pieces.

   Area WXYZ = Area(∆WMZ) + Area(WXNM) + Area(∆XNY) = ½ × x × h + a × h + ½ × y × h

   Factor out h.

      = h(½x + a + ½y)
      = ½h(x + y + 2a)
   

   Now remove x and y. Walking along the long parallel side, ZY is made of MZ, MN and NY, so

   b = x + a + y which gives x + y = b − a

   Substitute.

   Area WXYZ = ½h(b − a + 2a) = ½h(a + b)

   Area of a trapezium = ½ × height × sum of the parallel sides

   Reading it. The bracket (a + b)/2 is the average of the two parallel sides, so a trapezium has the same area as a rectangle of that average width and the same height. Setting a = b turns it back into a parallelogram: ½h(2a) = ah ✓

   ✦ Answer: the three pieces give ½h(x + y + 2a), and x + y = b − a turns that into ½h(a + b).

3. 35 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 167

   Will the formula ½h(a + b) still hold for a trapezium that leans so far that the perpendicular from one end of the short side falls outside the long side? Complete the two approaches sketched in the book.

   Hint. Approach 1 subtracts a triangle instead of adding one; Approach 2 avoids the problem altogether.

   Approach 1 — rectangle and triangles. Put D = (0, 0) and C = (b, 0) along the long side, with A = (−p, h) leaning out to the left, so its foot F = (−p, 0) lies outside DC, and B = (a − p, h) with foot E inside. Then ABEF is a rectangle of area a × h, and the book's two lines say

   Area ABED = Area ABEF − Area(∆AFD) = ah − ½ph Area ABCD = Area ABED + Area(∆BEC) = ah − ½ph + ½(b − a + p)h

   since EC = b − (a − p) = b − a + p. Expanding the bracket,

   = h[a − p/2 + b/2 − a/2 + p/2] = h[a/2 + b/2] = ½h(a + b) ✓

   The two p terms cancel, which is exactly why the lean does not matter.

   Approach 2 — parallelogram and triangle. Draw BG parallel to AD, with G on DC. Then ABGD has both pairs of opposite sides parallel, so it is a parallelogram with base DG = AB = a and height h:

   Area(ABGD) = ah Area(∆BGC) = ½ × GC × h = ½(b − a)h total = ah + ½(b − a)h = ½h(2a + b − a) = ½h(a + b) ✓

   Does Approach 2 work for any trapezium? It needs G to land on the segment DC, which requires a ≤ b. That is no restriction, because you can always call the shorter parallel side a. If a = b the shape is a parallelogram, ∆BGC shrinks to nothing, and the formula still returns ½h(2a) = ah. So yes — Approach 2 covers every trapezium, and it is the tidier of the two since it never has to subtract.

   ✦ Answer: the formula holds in every case — Approach 1 works because the two terms in p cancel, and Approach 2 works for any trapezium once the shorter parallel side is named a.

4. 44 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 168

   Two copies of a trapezium ABCD (with AB parallel to CD) are joined along BC, the second copy rotated. Why must the result be a quadrilateral rather than a six-sided figure, what quadrilateral is it, and what area formula does this give?

   Hint. Add up the angles that meet where the two copies join.

   Ruling out the six-sided figure. At the join, the angles x and y of the two copies meet along BC. Since AB ∥ CD and BC is a transversal, x and y are interior angles on the same side, so

   x + y = 180°

   An angle of 180° is a straight line, so ABD′ and A′CD are straight, not bent. The outline therefore has four corners, not six — the result is a quadrilateral.

   Identifying it. Look at the other two angles, u and v. They too are co-interior angles, so u + v = 180°, which makes AD ∥ D′A′. Combined with AD′ ∥ A′D (already established), both pairs of opposite sides are parallel, so AD′A′D is a parallelogram.

   Reading off the area. The parallelogram's base is a + b, since the two parallel sides of the trapezium now lie end to end, and its height is h. So

   Area(parallelogram) = (a + b) × h

   and the trapezium is exactly half of it:

   Area(trapezium) = ½ × (a + b) × h = ½h(a + b) ✓

   Why this proof is worth knowing. It never once splits the trapezium into pieces, so it needs no separate argument for a leaning trapezium — the angle sums that drive it hold whatever the lean. It is the same trick as turning two copies of a triangle into a parallelogram.

   ✦ Answer: a parallelogram of base (a + b) and height h, because x + y = 180° and u + v = 180°; halving it gives Area of a trapezium = ½h(a + b).

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.