Pythagoras' Theorem in Real-World Problems

Solution

The wall meets the ground at a right angle at $C$, so the ladder $AB$ is the hypotenuse.

$x^2 = 2.4^2 + 0.5^2 = 5.76 + 0.25 = 6.01$

$x = \sqrt{6.01} = 2.451\ldots = 2.45$ (3 s.f.)

Solution

The ladder is the hypotenuse, and the height $h$ m is a shorter side.

$h^2 = 10^2 - 2^2 = 100 - 4 = 96$

$h = \sqrt{96} = 9.797\ldots = 9.80$ (3 s.f.)

Solution

(a) At first the ladder $FT$ is the hypotenuse of $\triangle OFT$: $OT^2 = 5^2 - 1.4^2 = 25 - 1.96 = 23.04\text{,}$ so $OT = \sqrt{23.04} = 4.8$ m.

(b) Afterwards the ladder $GU$ is the hypotenuse of $\triangle OGU$, with $OU = 4$ m: $OG^2 = 5^2 - 4^2 = 25 - 16 = 9\text{,}$ so $OG = 3$ m.

The foot has moved from $1.4$ m to $3$ m from the wall: $FG = 3 - 1.4 = 1.6$ m.

The top slid down $4.8 - 4 = 0.8$ m, but the foot moved out $1.6$ m: the two distances are not the same.

Solution

North and east are at right angles, so $\angle CNE = 90^{\circ}$ and $CE$ is the hypotenuse.

$CE^2 = 8^2 + 11^2 = 64 + 121 = 185$

$CE = \sqrt{185} = 13.60\ldots = 13.6$ (3 s.f.)

Solution

Distance $=$ speed $\times$ time: $PQ = 12 \times 3 = 36$ km, $QR = 12 \times 2 = 24$ km, $PS = 18 \times 2 = 36$ km and $ST = 18 \times 3 = 54$ km.

(a) East and north are at right angles, so $\angle QPS = 90^{\circ}$. $QS^2 = 36^2 + 36^2 = 1296 + 1296 = 2592\text{,}$ so $QS = \sqrt{2592} = 50.91\ldots = 50.9$ km (3 s.f.).

(b) $R$ and $T$ are not on a right triangle yet: make one. From $R$, go due east to $U$, directly below $T$.

$T$ is $54$ km east of $P$ and $R$ is $36$ km east of $P$, so $RU = 54 - 36 = 18$ km. $T$ is $36$ km north of $P$ and $R$ is $24$ km south of it, so $UT = 36 + 24 = 60$ km.

$RT^2 = 18^2 + 60^2 = 324 + 3600 = 3924\text{,}$ so $RT = \sqrt{3924} = 62.64\ldots = 62.6$ km (3 s.f.).

Solution

Combine the north and south stages first: $3$ km north and $1$ km south leave him $3 - 1 = 2$ km north of the start. He is also $5$ km east of it.

So the start, the point $2$ km due north of it, and the end make a right-angled triangle with sides $2$ km and $5$ km.

$d^2 = 2^2 + 5^2 = 4 + 25 = 29$

$d = \sqrt{29} = 5.385\ldots = 5.39$ (3 s.f.)

Solution

On the map, $AB^2 = 6^2 + 8^2 = 36 + 64 = 100\text{,}$ so $AB = 10$ cm.

Each centimetre on the map stands for $0.5$ km, so $10$ cm stands for $10 \times 0.5 = 5$ km.

(Or change both lengths first: $3$ km east and $4$ km north, so $\sqrt{9 + 16} = 5$ km.)

Solution

Along the edges: $AB + BC = 80 + 60 = 140$ m.

The corner $B$ of the rectangle is a right angle, so $AC$ is the hypotenuse of $\triangle ABC$: $AC^2 = 80^2 + 60^2 = 6400 + 3600 = 10\,000\text{,}$ so $AC = 100$ m.

The straight path is $140 - 100 = 40$ m shorter.