Both a square and an equilateral triangle are regular polygons — all of their sides are equal in length (and all their angles are equal too). Because of this, the perimeter of both shapes can be found the same way:

For the square this gives $4 \times \text{side}$, and for the equilateral triangle it gives $3 \times \text{side}$.

Counting the straight (s) and diagonal (d) segments used to trace the boundary of each shape:

Remember: a diagonal unit (d) is longer than a straight unit (s) — it is actually $s\sqrt2 \approx 1.41s$, since it is the hypotenuse of a right triangle with two sides of length $s$.

Objects like a square tile, a hexagonal nut, a regular-pentagon-shaped clock, or an equilateral-triangle road sign are all regular polygons — measuring one side and counting the number of sides gives the perimeter directly. In general:

This works because every side of a regular polygon has exactly the same length.

Setting up: The original 6 cm × 4 cm chit is cut down the middle of its 4 cm side, giving two identical pieces, each 6 cm × 2 cm. Each piece has perimeter $2\times(6+2)=16$ cm, so the two pieces together have a total perimeter of $16+16=32$ cm — but wherever the pieces touch, that shared edge is hidden inside the new figure and must be subtracted twice (once from each piece):

• Arrangement (a) — the two pieces are joined along their full 2 cm edge (side by side, forming a 12 cm × 2 cm strip). Shared length = 2 cm, so $P = 32 – 2(2) = 28$ cm. ✔ (matches the given example)
• Arrangement (b) — an “L”/”7” shape: one piece lies flat on top, the other stands joined to it along a full 2 cm edge at one end. Shared length = 2 cm, so $P = 32-2(2) = \mathbf{28\text{ cm}}$.
• Arrangement (c) — a “T” shape: the vertical piece sits on the horizontal one, again meeting along its full 2 cm width. Shared length = 2 cm, so $P = 32-2(2) = \mathbf{28\text{ cm}}$.
• Arrangement (d) — the two pieces stand side by side but slide past each other, overlapping (touching) for only 3 cm of their 6 cm length. Shared length = 3 cm, so $P = 32-2(3) = \mathbf{26\text{ cm}}$.

Making a perimeter of 22 cm: Using the same rule, $32-2L=22 \Rightarrow L = 5$ cm. So the two 6 cm × 2 cm pieces must touch along a shared length of exactly 5 cm. This happens if we place the two pieces side-by-side along their long (6 cm) edges — like reforming the original 6 cm × 4 cm rectangle — but slide one piece 1 cm out of line, so only 5 cm of that long edge actually overlaps, leaving a 1 cm step.

Perimeter of b = 28 cm, c = 28 cm, d = 26 cm; a 22 cm figure is made by overlapping the pieces along 5 cm of their long edge.

Measuring 350 m backwards from the finishing line, along each track:

• On the inner track (side 100 m): $100+100+100+50 = 350$ m — so starting point A is 50 m from a corner along the side just before the finish line.
• On the outer track (side 150 m): $125+150+75 = 350$ m — so starting point B is 125 m from the finishing line along the near side (or equivalently, 25 m from the nearer corner).

Both paths total exactly 350 m, so runners starting at A and B finish together at the common finishing line.

Toshi is right — the perimeter is more than 9 units.

The triangle has two straight (grid-aligned) sides, each 3 units long, and one diagonal side. Akshi’s mistake is assuming the diagonal side also measures 3 units just by counting dots — but a diagonal line across a square grid is always longer than a straight side of the same square, since it is the hypotenuse of a right triangle:

So the true perimeter is:

Since $10.24 > 9$, Toshi’s reasoning is correct: the diagonal of a square is always greater than its side, so the perimeter is more than 9 units.

Counting squares using the given convention, the four shapes have areas:

Squares are used as the standard unit of area because:

• No gaps, no overlaps: squares of the same size tile a plane perfectly, covering every bit of the region exactly once — circles always leave curved gaps between them, however tightly packed.
• Consistent count: because circles leave gaps, the number of circles that fit depends on how they are arranged (42 vs 44 in the example) — so “number of circles” isn’t a reliable measure of area. Squares always give the same count for the same region.
• Matches length × breadth: a grid of unit squares lines up naturally with the length and width of a rectangle, which is exactly why Area $=$ length $\times$ breadth works so simply for squares but not for circles or triangles packed the same way.

Area = 24 sq units — the whole-number length–breadth pairs are:

Greatest perimeter: 1 × 24 (P = 50 units) — the thinnest rectangle. Least perimeter: 4 × 6 (P = 20 units) — the rectangle closest to a square.

Area = 32 sq cm — the whole-number pairs are:

Greatest perimeter: 1 × 32 (P = 66 cm). Least perimeter: 4 × 8 (P = 24 cm).

General rule: for a fixed area, the more “stretched out” (thin) a rectangle is, the greater its perimeter; the rectangle whose length and breadth are closest to each other (closest to a square) always has the least perimeter.

Yes — when a rectangle (or a square, which is a special rectangle) is cut along a diagonal, the two triangles formed are congruent (identical in shape and size), so they overlap each other exactly when placed one on top of the other.

Inference: A diagonal divides a rectangle into two triangles of equal area, each exactly half the area of the rectangle:

The triangle’s area is exactly half of the rectangle’s area. Whenever a triangle has the same base and the same height as a rectangle (its apex touching the side parallel to the base), the triangle’s area works out to:

• Area of rectangle ABCD $= 6\times4 = 24$ sq units.
• Triangle BAD is a right triangle with legs $AB=6$ and $AD=4$: $\text{Area}=\frac12\times6\times4=\mathbf{12}$ sq units — exactly half of rectangle ABCD.
• Triangle ABE has base $AB=6$ and height $4$ (the perpendicular distance from E to AB, since E lies on DC which is 4 units above AB): $\text{Area}=\frac12\times6\times4=\mathbf{12}$ sq units — also exactly half of ABCD!

Why are they equal? Splitting triangle ABE at F (the foot of the perpendicular from E) gives two smaller triangles AEF and BEF. Each of these is exactly half of a smaller rectangle (AFED and BFEC respectively):

Any triangle with the same base and height as a rectangle has half its area — no matter where the apex sits along the top side.

Attaching one new unit square always adds all 4 of its own sides to the total boundary, but removes (from the count) every side where it touches the existing figure, since a touching edge becomes hidden inside the shape (subtract 2 per touching side — one from each piece). So:

• (a) Increases: attach the new square so it touches the figure along only 1 side — change $=4-2(1)=+2$, so the perimeter becomes 26 units.
• (b) Decreases: attach it into a notch where it touches 3 sides at once — change $=4-2(3)=-2$, so the perimeter becomes 22 units.
• (c) Stays the same: attach it into a corner/notch where it touches exactly 2 sides — change $=4-2(2)=0$, so the perimeter stays at 24 units.

Since the Master Bedroom + Small Bedroom column is 30 ft tall and the Master Bedroom is 15 ft, the Small Bedroom’s height is $30-15=15$ ft, so its width is $180\div15=12$ ft. The Toilet + Kitchen column is also 30 ft tall; since the Kitchen is 12 ft tall, the Toilet + Utility together take up $30-12=18$ ft… following the plan through:

Total plot size = 35 ft × 30 ft, so the area of Charan’s whole house = $35\times30=$ 1050 sq ft (which also matches the sum of all the individual rooms above).

Total area of Sharan’s house = $42 \times 25 = $ 1050 sq ft — exactly the same as Charan’s house!

However, perimeters differ: Charan’s house perimeter $=2(35+30)=$ 130 ft, while Sharan’s house perimeter $=2(42+25)=$ 134 ft. So even though both houses have the same area, Sharan’s (more rectangular, less square-ish) plot has a slightly greater perimeter — the same “equal area, different perimeter” idea seen throughout this chapter.