Part II, Chapter 7Area

Infinitely many! Start with the obvious division into 4 equal quarter-squares:

Now alter each part: compress its area along one edge (push a notch inward) and expand it by the exact same amount along an adjacent edge (a matching tab outward). Since what's removed from one edge exactly equals what's added on the other, each piece keeps precisely the same area, even though the boundary is now a jagged, creative shape:

Count the non-overlapping unit squares (1 cm × 1 cm) that pack into each rectangle: the 7 cm × 4 cm rectangle contains \(7\times4=28\) unit squares, and the 8 cm × 3 cm rectangle contains \(8\times3=24\) unit squares.

The diagonal of a rectangle always splits it into two congruent triangles, so each triangle's area is exactly half the rectangle's area. In terms of unit squares, half the total area fills exactly half the number of unit squares.

Two rectangles: Take a 1 cm × 20 cm rectangle (a long thin strip) versus a 6 cm × 6 cm square.

Rectangle 1: perimeter \(=2(1+20)=42\) cm, area \(=1\times20=20\) cm². Rectangle 2: perimeter \(=2(6+6)=24\) cm, area \(=6\times6=36\) cm².

Here Perimeter(Rectangle 1) = 42 cm > Perimeter(Rectangle 2) = 24 cm, but Area(Rectangle 1) = 20 cm² < Area(Rectangle 2) = 36 cm² — exactly the property we wanted.

Two other shapes: compare a long, thin "star" or zig-zag outline (large perimeter from all its jagged edges, but a small enclosed area) against a plain circle or square of modest size (a much smaller perimeter, but a visibly larger enclosed area). Any shape that "wiggles" a lot along its boundary can have an enormous perimeter while enclosing very little area — this should be visually obvious once drawn.

You need the width of each strip of the crosspath (they need not be equal). Let the vertical strip have width \(w_1\) and the horizontal strip have width \(w_2\).

Area of crosspath = Area(vertical strip) + Area(horizontal strip) − Area(their overlapping square, counted twice):

For example, with \(w_1=w_2=2\) m: Area \(=12(2)+14(2)-(2)(2)=24+28-4=48\) m².

Dropping the altitude from X (or Y) down to DC, both triangles share the exact same base DC and the exact same height (the full height of the rectangle, since X and Y both lie on AB, directly opposite DC).

Just as before, dropping the altitude from X or from Y to line BC (or DC) shows both triangles share the same base (BC, a side of the rectangle) and the same height (the rectangle's other side), so both triangles again work out to exactly half the rectangle's area.

Area(△XDC) \(=\dfrac12\times\)base\(\times\)height\(=\dfrac12\times5\times4=10\).

Yes. The area of △ABC is the difference of the areas of △ADC and △ADB, each of which can be enclosed in a rectangle (since D, the foot of the altitude, now falls on the extension of CB):

First find the area of △ABC using the known altitude AX and base BC: Area\((\triangle ABC)=\dfrac12\times AX\times BC=\dfrac12\times3\times5=\dfrac{15}{2}\).

The same area can also be written using AC as the base and BY as its corresponding height: Area\((\triangle ABC)=\dfrac12\times BY\times AC=\dfrac12\times4\times BY=2\,BY\).

Setting the two expressions equal: \(2\,BY=\dfrac{15}{2} \Rightarrow BY=\dfrac{15}{4}=3.75\).

The 4 triangles are clearly not congruent, so we compare their areas by picking suitable base-height pairs. For adjacent triangles 1 and 2, use OD and OB as their respective bases — both triangles then share the same altitude (the perpendicular distance from A down to line BD). Since a rectangle's diagonals bisect each other, \(OB=OD\), so triangles 1 and 2 have equal areas. The same reasoning applies all the way around.

(i) Area: since every such triangle shares the same base BC and the same height (the fixed perpendicular distance between the parallel lines \(l\) and BC, regardless of where the third vertex sits on \(l\)), every triangle in this family has exactly the same area — there's no unique maximum or minimum.

(ii) Perimeter: the base BC is common to all of them, so only the sum of the other two sides varies. As we'll show below using a mirror-reflection argument, the triangle with the minimum perimeter is the isosceles one obtained by placing the third vertex where the perpendicular bisector of BC meets \(l\). As the vertex moves further from this point in either direction along \(l\), the perimeter grows without bound, so there is no maximum perimeter.

Imagine line \(l\) as a mirror, reflecting B and C to B′ and C′ below it. Since reflection preserves distance, \(AB=AB'\) and \(AC=AC'\) for any point A on \(l\) — so the path length \(B\to A\to C\) always equals \(B\to A\to C'\).

Finding the point A that minimises path \(B\to A\to C\) is therefore the same as finding the A that minimises path \(B\to A\to C'\). But the shortest path between two fixed points B and C′ is simply the straight line between them — so the minimising A is exactly where line \(BC'\) crosses \(l\).

This particular triangle (with A chosen this way) therefore has the minimum perimeter, since BC itself is fixed and we've minimised \(AB+AC\).

Yes. Since B′ is the mirror-image of B, and C is at the same "mirror distance" as C′, the situation is symmetric: the straight line \(BC'\) crosses the mirror line \(l\) at a point equidistant from B and C, by the reflective symmetry of the whole picture (reflecting the entire configuration left-to-right swaps B with C and A stays fixed as its own mirror image, only if A is equidistant from both). This equidistant point is exactly where the perpendicular bisector of BC meets \(l\).

Let the side of each square be \(s\). Using coordinate geometry on this layout, the smaller ("blue") region works out to exactly \(\frac{s^2}{4}\), and the larger ("red") region works out to exactly \(\frac{7s^2}{4}\) — so blue : red = 1 : 7 always, regardless of the actual square size. Also, blue + red \(=\frac{s^2}{4}+\frac{7s^2}{4}=2s^2\), i.e. exactly twice the area of one square.

(i) If red \(=49\), then blue \(=\dfrac{49}{7}=7\) sq. units.

(ii) If blue + red \(=180\), then since blue + red \(=\frac{s^2}{4}+\frac{7s^2}{4}=2s^2\): \(2s^2=180 \Rightarrow s^2=90\).

By the midsegment theorem, MN is parallel to YZ and exactly half its length, so △XMN is similar to △XYZ with a linear scale factor of \(\frac12\). Since area scales with the square of the linear factor: Area ratio \(=\left(\frac12\right)^2=\frac14\).

This is the same mirror-reflection idea used earlier for minimum perimeter: reflect the water tank across the river (to the other side of the river-band) to get an image point. Since reflecting preserves distance, the length of any path House → (point on near riverbank) → Water tank equals House → (same point) → (reflected water tank).

The shortest path to the reflected point is simply the straight line from House to the reflected Water tank — so the optimal crossing point is exactly where this straight line crosses the near riverbank.

Joining diagonal BD splits quadrilateral ABCD into two triangles, △ABD and △BCD. Finding the area of each (needing a base and corresponding height in each triangle) and adding them gives the area of the whole quadrilateral.

Pick any one vertex and draw diagonals from it to every other non-adjacent vertex — this splits the pentagon into 3 triangles (in general, an \(n\)-sided polygon splits into \(n-2\) triangles this way). Adding up the 3 triangles' areas gives the pentagon's total area.

Yes, any polygon can be divided into triangles this way. So, knowing how to compute a triangle's area lets us find the area of any polygon, however many sides it has.

Splitting the regular hexagon from its centre into 6 identical equilateral triangles, you'd only need the hexagon's sidelength — since in a regular hexagon, each of these 6 triangles is equilateral with side equal to the hexagon's own side. Knowing the sidelength lets you find each equilateral triangle's area (using the formula \(\frac{\sqrt3}{4}s^2\), or via base-and-height), then multiply by 6.

General method: a "bowtie" shape like this is typically formed by two triangles that share a single crossing point, each triangle having one full side of the rectangle as its base and the opposite side (or a point on it) as its apex. Since each such triangle's height equals the rectangle's own height (or width), each triangle's area works out to exactly half the rectangle's area, by the "same base, same height as half the rectangle" idea used throughout this chapter. Where the two triangles overlap in the middle, that overlapping sliver gets subtracted once (since it would otherwise be double-counted).

Draw one diagonal of the quadrilateral (say AC), locate its midpoint, and draw a line through this midpoint to any point on one of the other sides such that the new quadrilateral's vertices include this midpoint. More simply: connect the midpoints of two opposite sides of the quadrilateral with a straight line passing through the intersection point of the diagonals — this line divides the original quadrilateral into two regions of equal area, by the same "vertex-to-midpoint bisects area" idea seen earlier for triangles (applied here to each of the two triangles that make up the quadrilateral via a diagonal).

Construct AX perpendicular to CD (this is called a height of the parallelogram). Cutting along AX splits the parallelogram into △AXD and trapezium ABCX. If you slide △AXD around to the right side, it fits exactly onto the "missing" triangular notch at the right end of ABCX (completing it into a rectangle).

Since \(\angle X=90^\circ\) and AB∥XC, we also have \(\angle A=90^\circ\). Extending XC to the right and constructing a perpendicular to it through B creates point Y, forming △BYC. Then:

\(BY=AX\) (since ABYX is a rectangle), \(\angle BYC=\angle AXD=90^\circ\), and \(BC=AD\) (opposite sides of the original parallelogram ABCD).

By the RHS congruency criterion, \(\triangle BYC\cong\triangle AXD\). So △AXD fits exactly over the region occupied by △BYC, converting the parallelogram into rectangle ABYX.

Yes — taking side AD (or BC) as the base instead, with the perpendicular distance from that side to its opposite side as the corresponding height, gives exactly the same area, since it's the same dissection idea applied to a different pair of parallel sides. Cutting along CZ (the perpendicular from C to line AD, or its extension) and rearranging the two pieces this way also produces a rectangle of equal area.

Since ABCD is a rhombus, all sides have equal length, and the diagonals are perpendicular bisectors of each other. So △ABD and △CBD are both isosceles triangles (each with two equal rhombus-sides as its slanted sides). Each isosceles triangle can be dissected into a rectangle (as shown in the earlier isosceles-triangle method), and the two resulting rectangles can then be joined together to form one single rectangle WXYZ of the same total area as rhombus ABCD.

Since the diagonals are perpendicular: Area\((\triangle ADB)=\frac12\times AO\times BD\), and Area\((\triangle CDB)=\frac12\times CO\times BD\).

Area(ABCD) \(=\frac12\times AO\times BD+\frac12\times CO\times BD=\frac12\times BD\times(AO+CO)=\frac12\times BD\times AC\)

(using \(AO+CO=AC\), since O lies on diagonal AC between A and C).

Construct \(WM\perp ZY\) and \(XN\perp ZY\). Since \(WX\parallel ZY\) makes \(\angle MWX=\angle NXW=90^\circ\) (co-interior angles along the transversals WM, XN summing to 180°), WXNM is a rectangle.

Let \(MZ=x\), \(WM=XN=h\), \(WX=a\), \(NY=y\). Then:

Let \(b=ZY\) (the other parallel side), so \(b=x+y+a\), i.e. \(x+y=b-a\). Substituting:

Approach 1 (Rectangle and Triangles): here, only one perpendicular (from A) falls within the extended base, while the other (from B, at E on line DC) falls outside segment DC on the far side. Area(ABCD) = Area(ABED) + Area(△BEC), where Area(ABED) itself is found as Area(ABEF) − Area(△AFD) (subtracting the small triangle that sticks out where the rectangle overshoots past D).

Approach 2 (Parallelogram and Triangle): draw BG parallel to AD (G on line DC, chosen so ABGD is a parallelogram). Then Area(ABCD) = Area(parallelogram ABGD) + Area(△BGC). Since ABGD is a parallelogram with base DG and height \(h\), and △BGC is a triangle with base GC and the same height \(h\) (as BG∥AD and both G,C sit on the same line DC), this reduces to the same base×height and ½×base×height pieces as before, and simplifying gives exactly \(\frac12h(a+b)\) again.

Rotating the second copy 180° and joining it to the first along BC produces a parallelogram (not a 6-sided figure) — this follows since the two angles meeting at B and at C along the shared edge each sum to 180° (co-interior angles from AB∥DC), forcing the boundary to run straight through, and a similar argument at the other pair of angles shows both pairs of opposite sides end up parallel.

This parallelogram has base \((a+b)\) (the two trapezium bases laid end-to-end) and height \(h\) (same as the trapezium's height), so its area is \(h(a+b)\) — exactly twice the original trapezium's area, since the parallelogram is built from two copies of it.

Drop perpendiculars from A and B down to the base line (extended if needed), meeting it at H (below/aligned with A) and E (aligned with B) — these define the rectangle's top corners' projections, giving H and E their positions, with G and F directly below on the base line.

Since \(\triangle AHI\cong\triangle DGI\) (matching up the small triangle cut off near D with the matching gap near H) and \(\triangle BEJ\cong\triangle CFJ\) (similarly near C and F), the material trimmed off at the two slanted ends of the trapezium exactly fills the two corresponding notches needed to complete rectangle EFGH — so Area(ABCD) = Area(EFGH).

Since Area \(=\frac12h(a+b)\), pick any height and pair of parallel sides satisfying \(h(a+b)=288\). For example, with height \(h=12\) cm and parallel sides \(a=10\) cm, \(b=14\) cm: Area \(=\frac12\times12\times(10+14)=\frac12\times12\times24=144\) cm² ✓.

Construct this directly: draw a 14 cm base, mark a point 12 cm perpendicular above one end, draw a 10 cm segment parallel to the base from that point, then join the remaining two vertices to close the trapezium.

General method: a regular hexagon can always be divided from its centre into 6 identical equilateral triangles, each with side equal to the hexagon's own sidelength. Any other dissection of the same hexagon — including one into a trapezium, an equilateral triangle, and a rhombus — must have piece areas that are some whole or half-multiples of this "unit" equilateral triangle's area (since the dividing lines typically run along diagonals connecting vertices, which align with edges of these 6 unit triangles).

To find the exact ratio for your specific figure: count how many of these 6 unit-triangle "slices" each of the three named pieces (trapezium, triangle, rhombus) covers, then express that as a ratio.

Since A is the midpoint of XY, we have \(AY=AX\). At point A, the two lines ZB and XY cross, giving vertically opposite angles \(\angle ZAY=\angle BAX\). Also, since \(ZY\parallel WX\) (and B lies on line WX), the alternate angles \(\angle ZYA=\angle BXA\) are equal.

By ASA, \(\triangle ZAY\cong\triangle BAX\), so these two triangles have equal area.

Now, Area(ZYXW) = Area(△ZWX) + Area(△ZXY) (splitting the trapezium along diagonal ZX), while Area(△ZWB) = Area(△ZWX) + Area(△ZXB) (splitting the larger triangle along the same ZX). Since \(\triangle ZXY\) and \(\triangle ZXB\) both equal \(\triangle ZAY\) and \(\triangle BAX\) respectively in area (each being the corresponding congruent piece plus the shared triangle ZAX), and \(\triangle ZAY\cong\triangle BAX\):

Area \(=21\times29.7=623.7\) cm².

This is an open, hands-on estimation activity — measure your own tabletop's length and width (in cm), multiply to get its area, then divide by 623.7 cm² (the A4 sheet's area) to estimate roughly how many A4 sheets would tile it.

(i) \(5\times2.54=12.7\) cm.

(ii) \(7.4\times2.54=18.796\) cm.

(i) \(5.08\div2.54=2\) in.

(ii) \(11.43\div2.54=4.5\) in.

Since \(1\text{ in}=2.54\text{ cm}\), a 1 in × 1 in square equals a 2.54 cm × 2.54 cm square:

\(10\text{ in}^2=10\times6.4516=64.516\text{ cm}^2\).

Every 6.4516 cm² gives 1 in², so \(161.29\text{ cm}^2=\dfrac{161.29}{6.4516}\text{ in}^2=25\text{ in}^2\).

Since \(1\text{ ft}=12\text{ in}\), a \(1\text{ ft}\times1\text{ ft}\) square equals a \(12\text{ in}\times12\text{ in}\) square: \(1\text{ ft}^2=12^2\text{ in}^2=144\text{ in}^2\).

This is an open research and estimation activity: search for the local area unit used in your specific region (e.g. bigha in parts of North India, cent in Kerala/Tamil Nadu, gunta in parts of the Deccan), then make your own estimate of your school's area and compare it against the school's actual recorded area if available.

These are open research questions with answers that change over time and vary by source — look up your own city/town/village's official area, along with current data on India's and the world's largest and smallest cities by area, from a reliable source such as government census data or an up-to-date reference.