---
title: "Areas and Volumes"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 53
exercises: 12
source: https://one-course.com/books/math/1/en/chapter/53-areas-and-volumes
---

# Chapter 53 — Areas and Volumes

Grade 6 measured rectangles, [right triangles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) and boxes ([Chapter 43](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#ch-g6-measure)). With a pair of scissors — cut a piece here, glue it there — the same ideas now give the [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of [parallelograms](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram), of *all* triangles and of the disk, and the [volumes](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume) of [prisms](#def-g7-areas-prism) and [cylinders](#def-g7-areas-prism).

## 53.1 Area of a parallelogram

**Theorem 53.1 (Parallelogram).**

A [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) with a side of length $b$ (its *base*) and distance $h$ between that side and the opposite one (its *height*) has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area)

$$
A = b \times h .
$$

**Proof by scissors.** Cut off the [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) triangle sticking out on one side and glue it on the other: the [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) becomes a $b \times h$ rectangle with the same [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area). ∎

![The scissors proof: sliding the red triangle from the left end to the right end turns the parallelogram into a rectangle — same area, b × h.](https://one-course.com/images/onecourse/chapters/math-1/g7-areas/fig-ff97504b430f.svg)

*The scissors proof: sliding the red triangle from the left end to the right end turns the [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) into a rectangle — same [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area), $b \times h$.*

**Remark 53.2.**

The height is the *[perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)* distance between the two bases, not the length of the slanted side! A very slanted [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) can have long sides and a small [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area).

## 53.2 Area of a triangle

**Theorem 53.3 (Triangle).**

A triangle with base $b$ and corresponding height $h$ (the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) distance from the opposite [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) to the base line) has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area)

$$
A = \frac{b \times h}{2} .
$$

**Proof.** Two copies of the triangle, one turned by a half-turn around the midpoint of one side, fit together into a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) with base $b$ and height $h$ (that is [central symmetry](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-def) at work, [Proposition 52.4](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#prop-g7-central-preserved)). The triangle is [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of it: $\frac{bh}{2}$. ∎

![A triangle and its half-turned copy (dashed) make a parallelogram: the triangle’s area is b× h/2. Any of the three sides can serve as the base — with its own height.](https://one-course.com/images/onecourse/chapters/math-1/g7-areas/fig-c7863134f268.svg)

*A triangle and its half-turned copy (dashed) make a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram): the triangle’s [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) is $\frac{b\times h}{2}$. Any of the three sides can serve as the base — with *its own* height.*

**Example 53.4.**

A triangle has a base of $9$ cm and the corresponding height measures $4$ cm:

$$
A = \frac{9 \times 4}{2} = \frac{36}{2} = 18 \text{ cm}^2 .
$$

For a *right* triangle, the two legs are a base and its height: we recover the formula of [Proposition 43.6](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#prop-g6-measure-formulas).

## 53.3 Area of a disk

**Theorem 53.5 (Disk).**

A disk of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $r$ has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area)

$$
A = \pi r^2 .
$$

**Idea of proof.** Cut the disk into many thin equal slices and lay them head to tail: they almost form a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) of height $r$ whose base is [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the circumference, $\pi r$ ([Proposition 43.3](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#prop-g6-measure-circle)). Its [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) is close to $\pi r \times r = \pi r^2$, and the approximation becomes perfect as the slices get thinner. A complete proof needs integral calculus, from the High School volume. ∎

![Slicing the disk and unrolling the slices: almost a parallelogram of base π r (half the border) and height r.](https://one-course.com/images/onecourse/chapters/math-1/g7-areas/fig-1cb0774f961c.svg)

*Slicing the disk and unrolling the slices: almost a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) of base $\pi r$ ([half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the border) and height $r$.*

**Example 53.6.**

A round table top has [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $60$ cm. Its [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) is $\pi \times 60^2 = 3600\pi \approx 11\,310$ cm$^2$, a bit more than $1.1$ m$^2$. Watch the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference): the *[perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter)* $2\pi r$ uses $r$, the *[area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area)* $\pi r^2$ uses $r^2$.

## 53.4 Prisms and cylinders

**Definition 53.7 (Prism, cylinder).**

A *prism* is a [solid](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) with two identical [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) polygonal [faces](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) (the *bases*) joined by rectangles; a *cylinder* is the same with disks as bases. The *height* is the distance between the two bases.

**Theorem 53.8 (Volumes).**

For a [prism](#def-g7-areas-prism) or a [cylinder](#def-g7-areas-prism) with base [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $B$ and height $h$:

$$
V = B \times h .
$$

**Proof.** *Admitted at this level.* ∎

**Example 53.9.**

A tent is a [prism](#def-g7-areas-prism) lying on its side: its bases are triangles of base $2$ m and height $1.5$ m, and the tent is $3$ m long.

1. Base [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) : $B = \frac{2 \times 1.5}{2} = 1.5$ m $^2$ .
2. [Volume](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume) : $V = B \times h = 1.5 \times 3 = 4.5$ m $^3$ .

A tin can of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $4$ cm and height $11$ cm: $V = \pi \times 4^2 \times 11 = 176\pi \approx 553$ cm$^3$, roughly [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) a liter.

## 53.5 Exercises

**Exercise 53.1 ★.**

Compute the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) with base $8$ cm and height $5$ cm; of another with base $6.5$ m and height $4$ m.

**Solution of Exercise 53.1.**

$8 \times 5 = 40$ cm$^2$; $6.5 \times 4 = 26$ m$^2$.

**Exercise 53.2 ★.**

A [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) has sides $10$ cm and $6$ cm, and the height relative to the $10$ cm base measures $4$ cm. Compute its [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area). Why is the answer not $60$ cm$^2$?

**Solution of Exercise 53.2.**

$A = 10 \times 4 = 40$ cm$^2$. The answer is not $10 \times 6 = 60$ because the $6$ cm side is slanted: the formula uses the *height* ([perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) distance), which is $4$ cm.

**Exercise 53.3 ★.**

Compute the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of a triangle with base $12$ cm and height $7$ cm; of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with legs $9$ and $6$; of a triangle with base $5.5$ m and height $2$ m.

**Solution of Exercise 53.3.**

$\frac{12 \times 7}{2} = 42$ cm$^2$; $\frac{9 \times 6}{2} = 27$; $\frac{5.5 \times 2}{2} = 5.5$ m$^2$.

**Exercise 53.4 ★.**

Compute the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of a disk of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $5$ cm, then of a half-disk of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $10$ cm (exact values with $\pi$, then rounded to the unit).

**Solution of Exercise 53.4.**

Disk: $\pi \times 5^2 = 25\pi \approx 79$ cm$^2$. Half-disk of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $10$: $\frac{\pi \times 10^2}{2} = 50\pi \approx 157$ cm$^2$.

**Exercise 53.5 ★.**

Compute the [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) *and* the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of a disk of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $3$ cm. Which of the two answers is in cm and which in cm$^2$?

**Solution of Exercise 53.5.**

[Perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter): $2\pi \times 3 = 6\pi \approx 18.8$ cm (a length, in cm). [Area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area): $\pi \times 3^2 = 9\pi \approx 28.3$ cm$^2$ (a surface, in cm$^2$).

**Exercise 53.6 ★.**

Compute the [volume](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume) of a [prism](#def-g7-areas-prism) with base [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $24$ cm$^2$ and height $10$ cm; of a [cylinder](#def-g7-areas-prism) of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $3$ cm and height $8$ cm (exact, then rounded).

**Solution of Exercise 53.6.**

[Prism](#def-g7-areas-prism): $24 \times 10 = 240$ cm$^3$. [Cylinder](#def-g7-areas-prism): $\pi \times 3^2 \times 8 = 72\pi \approx 226$ cm$^3$.

**Exercise 53.7 ★.**

A triangle has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $28$ cm$^2$ and base $8$ cm. Find the corresponding height, writing the equation first.

**Solution of Exercise 53.7.**

$\frac{8 \times h}{2} = 28$, so $4h = 28$ and $h = 7$ cm.

**Exercise 53.8 ★★.**

Draw any triangle and measure carefully the three base–height pairs. Compute $\frac{b \times h}{2}$ for each pair: the three results should agree (up to measuring error). Why?

**Solution of Exercise 53.8.**

The three [products](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) agree because each one computes the *same* quantity — the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of the triangle. This gives a practical check of constructions, and a way of computing a height from another base–height pair (used in [Exercise 53.12](#exo-g7-areas-12)).

**Exercise 53.9 ★★.**

A trapezoid-shaped garden can be split, by one diagonal, into two triangles sharing the same height $h = 20$ m, with bases $30$ m and $18$ m. Compute its total [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area). Can you guess a general formula for the trapezoid?

**Solution of Exercise 53.9.**

The two triangles have [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $\frac{30 \times 20}{2} = 300$ m$^2$ and $\frac{18 \times 20}{2} = 180$ m$^2$: total $480$ m$^2$. General formula suggested: for [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides $a$ and $b$ at distance $h$,

$$
A = \frac{(a + b) \times h}{2}
$$

— the trapezoid formula (here $\frac{(30+18)\times 20}{2} = 480$).

**Exercise 53.10 ★★.**

A cylindrical vase of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $6$ cm contains water to a height of $15$ cm. All the water is poured into an empty [box](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) $18$ cm $\times$ $12$ cm at the base. What water height is reached in the [box](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def)? (Same [volume](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume), different base.)

**Solution of Exercise 53.10.**

Water [volume](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume): $\pi \times 6^2 \times 15 = 540\pi \approx 1\,696$ cm$^3$. [Box](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) base: $18 \times 12 = 216$ cm$^2$. Height reached: $\frac{540\pi}{216} = 2.5\pi \approx 7.9$ cm.

**Exercise 53.11 ★★.**

A pizza of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $30$ cm costs $9$ euros; one of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $40$ cm costs $14$ euros. Compute the two [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) and the price per $100$ cm$^2$. Which pizza is the better deal?

**Solution of Exercise 53.11.**

Radii $15$ and $20$ cm. [Areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area): $225\pi \approx 707$ cm$^2$ and $400\pi \approx 1257$ cm$^2$. Price per $100$ cm$^2$: $\frac{9}{7.07} \approx 1.27$ euros and $\frac{14}{12.57} \approx 1.11$ euros. The large pizza is the better deal — [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) grow with the *square* of the [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle).

**Exercise 53.12 ★★★.**

The two legs of a [right triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) measure $6$ cm and $8$ cm, and its hypotenuse measures $10$ cm. Compute its [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) with the legs, then use that [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) to find the height relative to the *hypotenuse*.

**Solution of Exercise 53.12.**

With the legs: $A = \frac{6 \times 8}{2} = 24$ cm$^2$. With the hypotenuse as base: $24 = \frac{10 \times h}{2} = 5h$, so $h = 4.8$ cm.

## 53.6 Problem: The vanishing square

**Problem 53.1.**

Weekend problem — a famous “proof” that $64 = 65$, unmasked by areas and slopes, with pizzas and rings for dessert

A magician cuts an $8 \times 8$ square into four pieces, slides them around, and reassembles them into a $5 \times 13$ rectangle. The audience counts: $8 \times 8 = 64$, but $5 \times 13 = 65$ — a square unit has appeared out of thin air! This chapter’s [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) formulas are exactly the tools that catch the trick. The problem then puts honest [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) reasoning to work: on trapezoids, on tilted stacks, on pizza economics — and finally returns to the magician for a second, [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) stranger performance.

**Part I — The trick, performed and unmasked.** The four pieces of the $8 \times 8$ square are: two [right triangles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) with legs $8$ and $3$, and two right trapezoids with [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides $5$ and $3$ and height $5$.

1. Compute the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of the original square and the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of the claimed rectangle. State the scandal in one line.
2. Compute the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of each of the four pieces ( [Theorem 53.3](#thm-g7-areas-triangle) ; for the trapezoids, cut or use [Exercise 53.9](#exo-g7-areas-9) ).
3. Add the four [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) . Which of the two figures — square or rectangle — can the pieces really fill?
4. In the rectangle arrangement, a triangle’s hypotenuse (rising $3$ across $8$ ) is supposed to continue a trapezoid’s slanted edge (rising $2$ across $5$ ) in one straight “diagonal”. Test the alignment: are $3$ across $8$ and $2$ across $5$ the same steepness ( [Theorem 49.3](https://one-course.com/books/math/1/en/chapter/49-proportionality#thm-g7-prop-cross) )?
5. Explain where the sixty-fifth square hides: what shape is the gap along the false diagonal, what is its exact [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) , and why does the eye miss it? (How wide, on average, is a sliver of that [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) stretched over the rectangle’s diagonal, roughly $13$ units long?)

**Part II — Honest [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) reasoning.**

6. Draw two [parallel lines](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) $4$ cm apart, a base of $6$ cm on one of them, and three different triangles on that base with apexes at various points of the other line. Compute the three [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) . Why are they all equal, however the apex slides along its line?
7. Prove the trapezoid formula in general: a trapezoid with [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides $B$ and $b$ and height $h$, cut along a diagonal, gives two triangles of the same height $h$. Conclude: $$A = \frac{(B + b) \times h}{2} .$$
8. Apply it to a trapezoid with $B = 9$ cm, $b = 5$ cm, $h = 4$ cm.
9. Carpenters use the same formula in disguise: [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $=$ (average of the two [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides) $\times$ height. Explain why this is the same rule, and recompute question 8 that way.
10. Tilt a neat stack of playing cards into a slanted stack. Explain why the [volume](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume) has not changed, and what this says about the height in the [prism](#def-g7-areas-prism) formula $V = B \times h$ ( [Theorem 53.8](#thm-g7-areas-volume) ): which height must one take for a slanted stack — the slanted length or the vertical one? (Compare [Exercise 53.2](#exo-g7-areas-2) , one dimension up.)

**Part III — Rings, pizzas, and the magician’s return.**

11. A circular pond of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $3$ m sits at the center of a circular lawn of [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $5$ m. Compute the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of the grass ring (exact with $\pi$ , then rounded to the m $^2$ ).
12. Find the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) of the single disk whose [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) equals that ring’s [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) exactly. (The three radii you now have form a famous triple, whose story is told in [Chapter 58](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#ch-g8-pythagoras) .)
13. Which is more pizza: one pizza of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $40$ cm, or two pizzas of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $28$ cm? Compute and compare.
14. To *double* a pizza’s [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) , by roughly what factor must its [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) grow? Check the candidate $1.4$ (recall that [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) follow the *square* of the scaling factor, [Problem 44.1](https://one-course.com/books/math/1/en/chapter/44-proportionality-and-data#pb-g6-propdata-1) ), and say where the exact factor — the number whose square is $2$ — makes its official entrance ( [Example 58.9](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#ex-g8-pythagoras-diagonal) ).
15. The magician returns: he cuts a $13 \times 13$ square with the numbers $5$ , $8$ , $13$ playing the roles that $3$ , $5$ , $8$ played before, and reassembles an $8 \times 21$ rectangle. Compute both [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) : what vanishes this time? The numbers $3, 5, 8, 13, 21$ are consecutive terms of the sequence of [Problem 47.1](https://one-course.com/books/math/1/en/chapter/47-fractions-comparing-and-adding#pb-g7-fractions-1) — state the pattern of the magician’s gains and losses, and the moral: why do [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) never lie?

**Solution of Problem 53.1.**

**1.** Square: $8 \times 8 = 64$. Rectangle: $5 \times 13 = 65$. The same four pieces appear to cover $64$ square units in one figure and $65$ in the other: one unit of [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) has been created from nothing — supposedly.

**2.** Each triangle: $\frac{8 \times 3}{2} = 12$. Each trapezoid: $\frac{(5 + 3) \times 5}{2} = 20$.

**3.** $12 + 12 + 20 + 20 = 64$: the pieces total $64$, so they can fill the square exactly — and can *not* fill the $65$-unit rectangle. The rectangle picture must contain a hole.

**4.** Same steepness would mean $3$ across $8$ [proportional](https://one-course.com/books/math/1/en/chapter/44-proportionality-and-data#def-g6-propdata-def) to $2$ across $5$; cross [products](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def): $3 \times 5 = 15$ and $2 \times 8 = 16$. Not equal: the hypotenuse and the slanted edge do *not* line up. The “diagonal” of the rectangle is a lie — two slightly different slopes meeting at a shallow angle.

**5.** Along the false diagonal the four pieces leave a long, extremely thin gap — a sliver in the shape of a very flat [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) — whose [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) is exactly the missing $65 - 64 = 1$ square unit. Stretched along a diagonal about $13$ units long, its average width is about $1 \div 13 \approx 0.08$ units: on a real drawing, thinner than the pencil line that hides it.

**6.** Each triangle has base $6$ cm and height $4$ cm — the distance between the [parallels](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp), wherever the apex sits — so each [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) is $\frac{6 \times 4}{2} = 12$ cm$^2$ ([Theorem 53.3](#thm-g7-areas-triangle)). Sliding the apex along the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) changes the shape, never the base or the height: equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) forever.

**7.** The diagonal cuts the trapezoid into a triangle with base $B$ and one with base $b$, both of height $h$ (the distance between the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides):

$$
A = \frac{B \times h}{2} + \frac{b \times h}{2}
= \frac{(B + b) \times h}{2} .
$$

**8.** $A = \frac{(9 + 5) \times 4}{2} = \frac{56}{2} =
28$ cm$^2$.

**9.** The average of the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides is $\frac{B + b}{2}$, so (average) $\times h =
\frac{(B+b) \times h}{2}$: the same formula, factored differently. Check: average $= \frac{9+5}{2} = 7$, and $7 \times 4 = 28$ cm$^2$.

**10.** The slanted stack contains exactly the same cards — the same layers, each of the same [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) — so the same [volume](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume). In $V = B \times h$, the height must be measured *[perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)* to the base (the vertical height of the stack), not along the slant: exactly as the [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram)’s [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) wants the [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) height, not the slanted side ([Exercise 53.2](#exo-g7-areas-2)), one dimension up.

**11.** Ring $= \pi \times 5^2 - \pi \times 3^2 = 25\pi -
9\pi = 16\pi \approx 50$ m$^2$ ([Theorem 53.5](#thm-g7-areas-disk)).

**12.** A disk of [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $16\pi$ has [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $4$, since $\pi \times 4^2 = 16\pi$. The radii $3$, $4$, $5$: the most famous triple in mathematics, starring in [Chapter 58](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#ch-g8-pythagoras).

**13.** One $40$ cm pizza: [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $20$, [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $\pi \times 400 \approx 1\,257$ cm$^2$. Two $28$ cm pizzas: [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) $14$, together $2 \times \pi \times 196 = 392\pi \approx 1\,232$ cm$^2$. The single large pizza is *more* pizza than the two mediums.

**14.** Scaling the [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) by $1.4$ [scales](https://one-course.com/books/math/1/en/chapter/49-proportionality#def-g7-prop-scale) the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) by $1.4^2 = 1.96$ — almost the double. The exact factor is the number whose square is $2$, about $1.414$: the diagonal number of [Example 58.9](https://one-course.com/books/math/1/en/chapter/58-the-pythagorean-theorem#ex-g8-pythagoras-diagonal). (Rule of thumb: [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) again as wide is twice the pizza, nearly.)

**15.** $13 \times 13 = 169$, but $8 \times 21 = 168$: this time one square unit *disappears* — the pieces overlap in a thin sliver instead of leaving a gap. With consecutive terms $3, 5, 8, 13, 21$ of the Hemachandra–Fibonacci sequence, the [products](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) flip between one more and one less than the square ($5 \times 13 = 65 =
8^2 + 1$, then $8 \times 21 = 168 = 13^2 - 1$): the magician alternately “gains” and “loses” a unit. [Areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) never lie: pieces totalling $64$ cover $64$, wherever they slide — every missing or extra unit is hiding in a sliver, waiting for a slope check to expose it.