14 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 161
Give a method to convert a parallelogram into a rectangle of equal area, and justify that the pieces really do fit.
Hint. Cut off the slanted end and carry it round to the other side.
The cut. Given parallelogram ABCD with AB ∥ DC, construct AX perpendicular to DC (X on DC). This cuts the parallelogram into ∆AXD and the trapezium ABCX.
The move. Lift ∆AXD off the left end and slide it round to the right end. To see where it lands, extend XC to the right and drop a perpendicular to it through B, meeting it at Y. The triangle ∆BYC is precisely the piece missing from ABCX before it becomes the rectangle ABYX.
Why the two triangles match. Compare ∆AXD and ∆BYC:
- BY = AX, since ABYX is a rectangle (opposite sides equal)
- ∠BYC = ∠AXD = 90°, by construction
- BC = AD, since ABCD is a parallelogram
Two right triangles with equal hypotenuses and one pair of equal legs are congruent by the RHS criterion, so ∆BYC ≅ ∆AXD. The cut-off piece therefore covers the gap exactly — no overlap, no space left over.
The name for it. Cutting a figure into pieces and reassembling them into a different figure of the same area is called dissection. It is the technique that produces every formula in the rest of this chapter.
✦ Answer: drop a perpendicular from one top vertex to the base, cut off the resulting triangle and slide it to the other end — it fits exactly, by RHS congruence, turning the parallelogram into a rectangle of the same area.
24 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 162
After the dissection, is there a relation between XY and DC? Use it to derive the formula for the area of a parallelogram.
Hint. DX and CY are the two matching legs of the congruent triangles.
Comparing XY with DC. From the congruence ∆AXD ≅ ∆BYC we get DX = CY. Now add the common stretch XC to each of them:
DX + XC = CY + XC
DC = XY
So the rectangle's length equals the parallelogram's base.
The formula. Dissection does not change area, so
Area(parallelogram ABCD) = Area(rectangle ABYX) = AX × XY
and substituting XY = DC together with AX = the height:
Area of a parallelogram = base × height
Reading it correctly. The height here is AX, the perpendicular distance between the two parallel sides — never the slanted side AD. That is the single most common slip in this topic: a parallelogram with sides 5 cm and 4 cm does not have area 20 cm², because its height is less than 4 cm.
With numbers. Base 7 cm, height 4 cm gives 28 cm², whatever the slant.
✦ Answer: XY = DC, so Area = AX × XY = base × height, with the height measured perpendicular to the base.
33 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 162
Can the area of a parallelogram be found by taking another side as the base and using its corresponding height? Can the parallelogram be cut along CZ and rearranged into a rectangle?
Hint. Turn the page sideways — nothing in the argument depended on which side was at the bottom.
Yes to both. Construct CZ perpendicular to AD (Z on AD), which is the same move as before but with AD playing the role of the base. Cutting along CZ and sliding the cut-off triangle round to the opposite end assembles a rectangle in exactly the same way — the congruence argument is unchanged, since it never used which pair of sides was called the base.
What this gives. The parallelogram has two different base–height pairs, and both must produce the same area:
base₁ × height₁ = base₂ × height₂
A worked consequence. Take a parallelogram with sides 12 cm and 7.6 cm whose height on the 12 cm side is 6 cm. Then the area is 12 × 6 = 72 cm², and the height on the 7.6 cm side must be 72 ÷ 7.6 ≈ 9.47 cm. The longer side carries the shorter height, because the product has to stay at 72.
Careful, though. Each height must be paired with its own base. Mixing the 12 cm base with the 9.47 cm height gives 113.6, which is not the area of anything in the figure.
✦ Answer: Yes — any side may be the base provided you use the height perpendicular to that side, and the two pairings always give the same area.
45 marksGanita Prakash Cl-8 Part 2, Section 7.1, pages 164-165
A rhombus ABCD is dissected into a rectangle WXYZ. What are the sidelengths of WXYZ, and what formula for the area of a rhombus does this give?
Hint. Split the rhombus along a diagonal into two isosceles triangles and rebuild each as a rectangle.
Setting up. In a rhombus all four sides are equal and the diagonals are perpendicular bisectors of each other. Cutting along the diagonal BD gives ∆ABD and ∆CBD, and each is isosceles, since AB = AD and CB = CD.
The dissection. Each isosceles triangle can be turned into a rectangle by the Śulba-Sūtra move — cut along the altitude from the apex and half-turn one piece. Doing that to both triangles and joining the two rectangles side by side gives one rectangle WXYZ.
Its sidelengths. Tracking where the pieces go:
XW = the full diagonal AC
WZ = half the other diagonal, BD/2
The formula.
Area(rhombus ABCD) = Area(rectangle WXYZ) = XW × WZ = AC × BD/2
Area of a rhombus = ½ × product of the diagonals
Check against the parallelogram rule. A rhombus is a parallelogram, so base × height must give the same number. A rhombus of side 5 cm with diagonals 6 cm and 8 cm has area ½ × 6 × 8 = 24 cm²; its height is then 24 ÷ 5 = 4.8 cm, and 5 × 4.8 = 24 ✓ Both formulas agree, and you use whichever data you are given.
✦ Answer: WXYZ has sides AC and BD/2, giving Area = ½ × d₁ × d₂.
53 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 166
The area of rhombus ABCD can also be found by adding the areas of ∆ADB and ∆CDB. Simplify that expression and show it gives the same formula.
Hint. The diagonals are perpendicular, so each half-diagonal is a height.
Set up the two triangles. Let the diagonals meet at O. Because AC ⊥ BD, the segment AO is the perpendicular height of ∆ADB above the base BD, and CO is the height of ∆CDB above the same base. So
Area(∆ADB) = ½ × AO × BD
Area(∆CDB) = ½ × CO × BD
Add and factorise.
Area(ABCD) = ½ × AO × BD + ½ × CO × BD
= ½ × BD × (AO + CO)
= ½ × BD × AC
since AO + CO is the whole diagonal AC.
Area of a rhombus = ½ × d₁ × d₂ ✓ — the same formula the dissection gave.
Worth noticing. The only property used was that the diagonals are perpendicular; the equal sides were never needed. So this formula holds for any quadrilateral with perpendicular diagonals — a kite, for instance — and not only for a rhombus.
✦ Answer: ½·AO·BD + ½·CO·BD = ½·BD·(AO + CO) = ½ × AC × BD, the same rule, and it holds for every quadrilateral with perpendicular diagonals.
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.
