Besides the simple 2×2 grid, here are creative methods that all preserve equal area:

Key idea: as long as opposite/adjacent pieces “trade” equal amounts of area, infinitely many divisions are possible.

The 7 cm × 4 cm rectangle needs more rangoli powder (28 sq. cm > 24 sq. cm).

Area of each triangle $= \dfrac{1}{2}\times 7\times 4 = 14\text{ cm}^2$

Perimeter: 20 > 18, but Area: 9 < 20 — confirming perimeter doesn’t determine area.

For “other shapes”: compare a long, thin zig-zag/comb shape (large perimeter, very little enclosed area) with a compact square of the same total boundary material — the square clearly encloses far more area, even with a similar or smaller perimeter.

BE and CD are both drawn perpendicular to BC, so $\angle B = \angle C = 90°$. Since ED is constructed parallel to BC (passing through A) and BE, CD are both perpendicular to BC, all four angles of BCDE are right angles — making it a rectangle.

BXAE has $\angle B = \angle X = 90°$ (by construction — BE⊥BC and AX⊥BC), and AE∥BX (both lie along the parallel lines BC and ED). This makes BXAE a rectangle, so its opposite sides are equal: AE = BX, meaning the perpendicular height AX of the triangle is exactly the same length as the rectangle’s side — confirming “height” is shared by both shapes.

BY = 3.75 units

Key idea: the SAME triangle’s area can be computed using any side as the base — equating both expressions lets us solve for an unknown height.

Yes — it does. Using coordinates with line $l$ as the x-axis, and $B=(p,-q)$, $C=(r,-q)$ (since $BC \parallel l$, both points are the same distance $q$ below $l$):

So the point A that minimises the perimeter always lies on the perpendicular bisector of BC, confirming the initial intuition.

Yes — since $DX = CY$ (both equal the “overhang” created when dropping perpendiculars), adding the shared segment XC to both gives $DX+XC = CY+XC$, i.e. $DC = XY$. So the rectangle’s base XY exactly equals the parallelogram’s base DC.

Yes. Exactly as with AX⊥CD, we can drop a perpendicular BZ⊥AD (or its extension), cut off the triangle formed, and re-attach it on the opposite side. This confirms that any side of a parallelogram, together with its corresponding height, can be used to compute its area: Area = base × height, regardless of which side is chosen as the base.

Area of rhombus $= \dfrac{1}{2}\times AC \times BD = \dfrac{1}{2}\times$ product of diagonals — matching the earlier formula!

Since WX∥ZY and WM, XN are both drawn perpendicular to ZY, the co-interior angles $\angle MWX$ and $\angle NXW$ are each $90°$ (they sum to $180°$ as angles on the same side of transversal WX cutting the parallel lines, and by construction both are right angles individually). With all four angles equal to $90°$, WXNM is a rectangle.

Yes. Drawing BG∥AD (where G lies on line DC, possibly extended) always creates parallelogram ABGD, since AB∥DG (given, as AB∥DC) and now AD∥BG (by construction) — two pairs of parallel sides. The leftover piece ΔBGC is a triangle regardless of the trapezium’s exact shape, so this method works generally, including for trapeziums where one slant side leans “outward” past the base.

Area $= 21 \times 29.7 = 623.7\text{ cm}^2$

The remaining questions (area of your classroom, school, village/town/city, largest/smallest-area cities, local area units like bigha/katha) are open-ended research and estimation tasks — explore using local measurements or an internet search for your specific region.

All four rectangles share the same width. We find it from the rectangle whose both sides are known:

Missing sidelength = 5 in

Missing sidelength (bottom strip height) = 1 m

(i) We need the sidelengths of both the inner park EFGH and the outer rectangle ABCD. Since Area(path) = Area(ABCD) − Area(EFGH), if EFGH is $l \times b$ and ABCD is $L \times B$:

(ii) Knowing only the width $w$ along each side is not enough if the path’s width can differ on different sides. If the width is the same uniform width $w$ on all sides, and the inner park is $l\times b$, then the outer rectangle is $(l+2w)\times(b+2w)$:

(iii) No — the area of the path does not change when the outer rectangle is shifted around the park (as long as the park stays fully inside). The area only depends on the two fixed areas (outer minus inner), not on their relative position.

We need the width of the crosspath (assume uniform width $w$ for both the horizontal and vertical strips).

Using the hint’s idea: an L-shaped bend made of two arms of length $a$ and $b$ (width 1) has the same area as a straight tube of length $(a+b-1)$ — the “$-1$” accounts for the overlapping corner square.

Tracing the spiral’s centreline, the arm lengths in order are: 20, 20, 20, 15, 5, 10, 10, 5, 15 — that’s 9 segments with 8 corners (turns) between them.

Area of the spiral tube $= 112$ sq. units  (equivalently, a straight tube of width 1 would need length 112 to match this area).

When a square’s side length is doubled, every linear dimension inside it (including the regions formed by its diagonals) also doubles. Since area scales with the square of the linear scale factor:

Each of regions 1, 2 and 3 becomes exactly 4 times its original area — an increase of 300% (3 times the original area is added), because every region inside the square is similarly scaled up.

This is a hands-on dissection activity. When the square is cut by two perpendicular lines (not through the centre) into 4 right-triangle/quadrilateral pieces, each piece can be rotated $90°$ around a common point and reassembled so the outer boundary forms a larger square, while a small square-shaped hole appears in the middle. This dissection visually demonstrates that the same four pieces (same total area) can outline a bigger square frame — a classic idea connected to area-conservation puzzles (similar in spirit to dissection proofs of the Pythagorean theorem).

Try this with cardboard cut-outs to see the rearrangement directly!

(i) 6 cm²   (ii) 8 cm²   (iii) 6 cm²

BY = 3 units

In an isosceles triangle, the altitude from the apex (S) to the base (UB) always meets the base at its midpoint E. So $\triangle SEU$ and $\triangle SEB$ have equal bases (EU = EB) and the same height (SE) — hence equal areas.

Area(∆SUB) = 48 sq. units

Let rectangle ABCD have base $b$ (= DC) and height $h$. Extend DC to a point E so that $CE = DC$, making $DE = 2b$. Join A to E.

Extending the base to double its length and joining to the opposite top corner gives a triangle of equal area.

Let the triangle have base $b$ and height $h$ (Area $=\frac12 bh$). Locate the midpoints $M, N$ of the two slanting sides — segment MN (the midsegment) is parallel to the base, with length $b/2$, at half the height.

Area of resulting rectangle $= \dfrac{b}{2}\times h = \dfrac{1}{2}bh = $ Area of the original triangle.

Setting up coordinates with square side $s$: $D=(0,0),\,C=(s,0),\,B=(s,s),\,A=(0,s)$ for the first square, and building the other two identical squares attached along BC and BF respectively gives $H=(s,2s)$.

(i) If Red $= 49 = s^2$, then Blue $= \dfrac{s^2}{4} = \dfrac{49}{4} = 12.25$ sq. units.

(ii) Red + Blue $= s^2 + \dfrac{s^2}{4} = \dfrac{5s^2}{4} = 180 \Rightarrow s^2 = \dfrac{180\times4}{5} = 144$

(i) Blue = 12.25 sq. units   (ii) Area of each square = 144 sq. units

Since M and N are midpoints, MN is the midsegment of the triangle: $MN \parallel YZ$ and $MN = \dfrac{1}{2}YZ$. This makes $\triangle XMN$ similar to $\triangle XYZ$ with a linear scale factor of $\dfrac{1}{2}$.

Area(∆XMN) $= \dfrac{1}{4}\times$ Area(∆XYZ)

Use the mirror-reflection trick from the chapter: reflect the water tank’s position across the line of the river to get an image point T′. The shortest path from the house to the river and then to the tank is the same length as the straight-line path from the house to T′.

Path: House → (point where line House–T′ meets the river) → Water tank. This is provably the shortest possible path.

Diagonal AC splits ABCD into $\triangle ABC$ and $\triangle ACD$, each using AC as base:

Area(ABCD) = 66 cm²

Set up coordinates: $D=(0,0)$, $C=(18,0)$, $A=(0,10)$, $B=(18,10)$; E on AB with AE = 10 → $E=(10,10)$; F on AD with AF = 6 → $F=(0,4)$. The shaded region is quadrilateral D–F–E–C.

Area of shaded region = 110 cm²

Join the centre of the hexagon to all 6 vertices — this divides it into 6 congruent (equilateral) triangles. To find the total area, we need:

Total area $= 6 \times \left(\dfrac{1}{2}\times\text{side}\times\text{apothem}\right)$

The blue “bowtie” shape consists of two triangles that together span the full width and full height of the rectangle. Using the same triangles-between-parallel-lines logic from the chapter, no matter where the pinch-point of the bowtie sits, the two blue triangles’ combined area always works out to exactly half of the rectangle.

Blue region = $\dfrac{1}{2}$ of the rectangle’s total area

Join the midpoints of all four sides of the given quadrilateral, in order, to form a new quadrilateral (this is known as the Varignon parallelogram).

This inner parallelogram always has exactly half the area of the original quadrilateral, regardless of the original shape.

28 cm², 15 cm², 24 cm², 8.8 cm²

QN ≈ 9.47 cm

For a rectangle, the height equals the full side length (4 cm), since the sides are perpendicular. For a slanted parallelogram with the same side lengths, the perpendicular height is always less than the slant side (4 cm), because the height is the shortest distance between the two parallel sides.

The rectangle has the greater area: $5\times4=20\text{ cm}^2$, while the parallelogram’s area $=5\times h$ with $h<4$, giving area $<20\text{ cm}^2$.

Bisect the triangle’s base, and construct the rectangle using half the base and the full height of the triangle (using the midsegment-dissection method described in Q5 of the Triangles section).

Area of rectangle $= \dfrac{\text{base}}{2}\times\text{height} = \dfrac{1}{2}\times\text{base}\times\text{height} = $ Area of triangle.

In isosceles $\triangle ABC$, the altitude AD splits it into two congruent right triangles, $\triangle ADB$ and $\triangle ADC$.

The rearranged rectangle has area $= BD\times AD = \dfrac{1}{2}\times BC \times AD = $ Area of the original triangle.

This is the reverse of Q4 (Triangles section): take rectangle with base $b$ and height $h$. Extend the base to twice its length, then cut the rectangle and fold the resulting pieces up along slanting lines to the two ends of the extended base, forming an isosceles triangle with base $2b$ and the same height $h$.

Area of resulting triangle $= \dfrac{1}{2}\times2b\times h = bh = $ Area of the original rectangle.

For sidelength $s$: Area of square $= s^2$. Area of equilateral triangle $= \dfrac{\sqrt3}{4}s^2 \approx 0.433\,s^2$.

The square has greater area than both a single equilateral triangle, and even two equilateral triangles combined, of the same sidelength.

Area = 150 cm²

This is the reverse of the rhombus→rectangle dissection shown in the chapter. Cut the rectangle along both diagonals, creating 4 triangles meeting at the centre.

This rearrangement produces a rhombus whose diagonals equal the rectangle’s length and width — keeping the same total area.

(i) 136 ft²   (ii) 420 m²   (iii) 100 in²   (iv) 120 ft²

For figures where exact label positions are ambiguous in the printed diagram, the values above use the standard “height × sum of parallel sides ÷ 2” trapezium method consistently.

Drop perpendiculars from the two top vertices to the base — this creates two congruent right triangles at the ends (congruent because the trapezium is isosceles) and a rectangle in the middle.

This produces a rectangle with base = average of the two parallel sides, and the same height as the trapezium — equal in area to the original trapezium.

Drop perpendiculars from A and B to line DC (extended if needed), landing at G and F. Choose H and E on the line through A, B such that triangles $\triangle AHI \cong \triangle DGI$ and $\triangle BEJ \cong \triangle CFJ$.

So Area(EFGH) = Area(ABCD): rectangle EFGH has height equal to the trapezium’s height, and width equal to the average of AB and DC.

Start from a rectangle of area 144 cm² — e.g. 12 cm × 12 cm. Using the trapezium formula Area $=\dfrac{1}{2}h(a+b)$, choose height $h=12$ cm and parallel sides whose average is 12 cm, e.g. $a=10$ cm, $b=14$ cm:

A trapezium with height 12 cm and parallel sides 10 cm and 14 cm has area exactly 144 cm².

Using the standard dissection of a regular hexagon shown in the figure (lines drawn from one vertex), the three pieces are in the ratio:

Trapezium : Triangle : Rhombus = 2 : 1 : 2

The trapezium and rhombus each have twice the area of the small equilateral triangle in the middle.

Let B be the point where line ZA, extended, meets line WX (extended beyond X).

Since Area$(\triangle ZYA)$ = Area$(\triangle BXA)$, both expressions are equal — so Area(ZYXW) = Area(∆ZWB).

Questions about your own classroom, school, village/town/city area, local area units (bigha, gaj, katha, dhur, cent, ankanam), and the largest/smallest-area cities in India and the world are open-ended estimation and research tasks — explore using local measurements or a quick search for your region.