Exercises — Perimeter of polygons — regular and irregular

An octagon has $n=8$ sides. "Regular" means every side is the same length, so we may use the shortcut $P=n\cdot s$. $P=8\times 5=40\text{ cm}$ What we did: counted sides ($8$), used the multiply-shortcut because all sides repeat the same value.

Sides differ, so there is no single $s$ to multiply — we simply add each side once. $P=6+9+11=26\text{ cm}$

A pentagon has $n=5$. We know $P=n\cdot s$, but here we know $P$ and want $s$. So we undo the multiplication by dividing. $s=\frac{P}{n}=\frac{45}{5}=9\text{ cm}$ Why divide? Multiplication by $5$ built the perimeter from one side; division by $5$ reverses that to recover the side.

A rectangle (see Quadrilaterals) has two lengths and two widths. It is not regular (length ≠ width), but it has a pattern: two equal long sides, two equal short sides. $P=12+7+12+7=38\text{ m}$ Grouping the pairs: $P=2(12+7)=2\times 19=38\text{ m}$.

First the perimeter: hexagon $n=6$, so $P=6\times 15=90\text{ m}$. Then the cost: $90\text{ m}\times ₹40/\text{m}=₹3600$. Two steps, two units: perimeter is a length (m); cost multiplies that length by a rate (₹/m), and the metres cancel to leave rupees. Compare Units of measurement.

We only have two sides; the third is the slanted side (the hypotenuse). To find it we use the Pythagorean theorem, which links the three sides of a right triangle: ${\text{hypotenuse}}^2=3^2+4^2=9+16=25\Rightarrow \text{hypotenuse}=\sqrt{25}=5\text{ cm}$ Why this tool? Perimeter needs all three sides, but the third one isn't given. Pythagoras is the one tool that hands us a missing side of a right triangle from the other two. Now add all three: $P=3+4+5=12\text{ cm}$

Perimeter is the distance around the boundary you actually walk, so we simply add every edge of the L, cut corners included: $P=10+6+4+2+6+4=32\text{ m}$ Key insight: the notch does not shorten the walk — it adds two extra edges. Never "skip" the indented sides.

From $P=n\cdot s$ we solve for $n$ by dividing: $n=\frac{P}{s}=\frac{56}{8}=7$ Seven equal sides → a regular heptagon. Why divide by $s$ here (not by $n$)? This time the unknown is $n$, and $s$ is known, so we strip away $s$ to reveal how many copies of it fit into $P$.

Let the width be $w$. Then the length is $w+4$. The rectangle's perimeter rule gives $2((w+4)+w)=30.$ Why set up an equation? We have two unknowns tied by one relationship ("$4$ more") and one measured total ($30$); algebra lets both facts act at once. Simplify inside: $2(2w+4)=30\Rightarrow 4w+8=30\Rightarrow 4w=22\Rightarrow w=5.5\text{ cm}$. Then length $=w+4=9.5\text{ cm}$. Check: $2(9.5+5.5)=2\times 15=30\text{ cm}$. ✓

The square alone contributes $3$ outer sides (the fourth is glued and hidden inside): $3\times 6=18$ cm. The triangle (see Triangles) contributes $2$ outer sides (its base is the glued edge): $2\times 6=12$ cm. The shared $6$ cm edge is inside the figure — you cannot walk along it, so it counts for neither. $P=18+12=30\text{ cm}$ Why drop the shared edge? Perimeter is the outer boundary only. Internal seams are not part of the walk around.

Walking one lap: the two long straights ($80$ m each) plus two semicircular ends. Two semicircles of the same diameter join into one full circle, so their combined length is a full circumference (see Circles): $C=\pi d=3.14\times 50=157\text{ m}.$ Why circumference, not $n\cdot s$? A circle has no straight sides, so the polygon shortcut fails; its "perimeter" is the circumference $\pi d$. Add the straights: $P=80+80+157=317\text{ m}.$

Never add mixed units. Convert everything to metres first ($100$ cm $=1$ m): $150$ cm $=1.5$ m, $250$ cm $=2.5$ m (see Units of measurement). $P=2+1.5+3+2.5=9\text{ m}$ Why convert first? Adding "$2+150$" would blend metres with centimetres, giving a meaningless number ($152$ of nothing).

Start from the shortcut $P=n\cdot s$. This is an equation multiplying $s$ by $n$. To isolate $s$, undo the multiplication by dividing both sides by $n$: $\frac{P}{n}=\frac{n\cdot s}{n}=s\Rightarrow s=\frac{P}{n}.$ Test — decagon: $n=10$, $P=60$, so $s=\frac{60}{10}=6\text{ cm}.$ Sanity check: $10\times 6=60$ ✓ — the side and the count rebuild the given perimeter.

Q1.1 ::: $40$ cm Q1.2 ::: $26$ cm Q2.1 ::: $9$ cm Q2.2 ::: $38$ m Q2.3 ::: ₹$3600$ Q3.1 ::: $12$ cm Q3.2 ::: $32$ m Q4.1 ::: $7$ sides (heptagon) Q4.2 ::: width $5.5$ cm, length $9.5$ cm Q4.3 ::: $30$ cm Q5.1 ::: $317$ m Q5.2 ::: $9$ m Q5.3 ::: $s=P/n$; decagon side $=6$ cm