Solve Exercise 18, Question 1a).

The triangle is right-angled at $B$. We are given the hypotenuse $AC = 40$ cm and a leg $BC = 32$ cm. We need to find the other leg $AB$.
Using the Pythagorean theorem $AB^2 + BC^2 = AC^2$:

$$AB^2 + 32^2 = 40^2$$ $$AB^2 + 1024 = 1600$$ $$AB^2 = 1600 - 1024$$ $$AB^2 = 576$$ $$AB = \sqrt{576}$$ $$AB = 24 \text{ cm}$$

• The unknown side is $\boxed{24 \text{ cm}}$.

The triangle is right-angled at $D$. We are given the hypotenuse $EF = 24$ cm and a leg $DE = 10$ cm. We need to find the other leg $DF$.
Using the Pythagorean theorem $DE^2 + DF^2 = EF^2$:

$$10^2 + DF^2 = 24^2$$ $$100 + DF^2 = 576$$ $$DF^2 = 576 - 100$$ $$DF^2 = 476$$ $$DF = \sqrt{476}$$

To simplify $\sqrt{476}$, we find its prime factors: $476 = 4 \times 119$.

$$DF = \sqrt{4 \times 119} = \sqrt{4} \times \sqrt{119}$$ $$DF = 2\sqrt{119} \text{ cm}$$

• The unknown side is $\boxed{2\sqrt{119} \text{ cm}}$.

The triangle is right-angled at $P$. We are given the hypotenuse $QR = 25$ cm and a leg $PQ = 15$ cm. We need to find the other leg $PR$.
Using the Pythagorean theorem $PQ^2 + PR^2 = QR^2$:

$$15^2 + PR^2 = 25^2$$ $$225 + PR^2 = 625$$ $$PR^2 = 625 - 225$$ $$PR^2 = 400$$ $$PR = \sqrt{400}$$ $$PR = 20 \text{ cm}$$

• The unknown side is $\boxed{20 \text{ cm}}$.

The triangle is right-angled at $A$. We are given the hypotenuse $BC = 18$ cm and a leg $AC = 12$ cm. We need to find the other leg $AB$.
Using the Pythagorean theorem $AB^2 + AC^2 = BC^2$:

$$AB^2 + 12^2 = 18^2$$ $$AB^2 + 144 = 324$$ $$AB^2 = 324 - 144$$ $$AB^2 = 180$$ $$AB = \sqrt{180}$$

To simplify $\sqrt{180}$, we find its prime factors: $180 = 36 \times 5$.

$$AB = \sqrt{36 \times 5} = \sqrt{36} \times \sqrt{5}$$ $$AB = 6\sqrt{5} \text{ cm}$$

• The unknown side is $\boxed{6\sqrt{5} \text{ cm}}$.

The triangle is right-angled at $N$. We are given the hypotenuse $MO = 21$ cm and a leg $NO = 15$ cm. We need to find the other leg $MN$.
Using the Pythagorean theorem $MN^2 + NO^2 = MO^2$:

$$MN^2 + 15^2 = 21^2$$ $$MN^2 + 225 = 441$$ $$MN^2 = 441 - 225$$ $$MN^2 = 216$$ $$MN = \sqrt{216}$$

To simplify $\sqrt{216}$, we find its prime factors: $216 = 36 \times 6$.

$$MN = \sqrt{36 \times 6} = \sqrt{36} \times \sqrt{6}$$ $$MN = 6\sqrt{6} \text{ cm}$$

• The unknown side is $\boxed{6\sqrt{6} \text{ cm}}$.

The triangle is right-angled at $F$. We are given the hypotenuse $GH = 12$ cm and a leg $FG = 7$ cm. We need to find the other leg $FH$.
Using the Pythagorean theorem $FG^2 + FH^2 = GH^2$:

$$7^2 + FH^2 = 12^2$$ $$49 + FH^2 = 144$$ $$FH^2 = 144 - 49$$ $$FH^2 = 95$$ $$FH = \sqrt{95} \text{ cm}$$

The number 95 has no perfect square factors ($95 = 5 \times 19$), so it cannot be simplified further.

• The unknown side is $\boxed{\sqrt{95} \text{ cm}}$.

The ladder, wall, and ground form a right-angled triangle.
The length of the ladder is the hypotenuse ($c = 4.25$ m).
The distance from the wall to the foot of the ladder is one leg ($a = 2$ m).
The height above the ground is the other leg ($b$).
Using the Pythagorean theorem $a^2 + b^2 = c^2$:

$$2^2 + b^2 = 4.25^2$$ $$4 + b^2 = 18.0625$$ $$b^2 = 18.0625 - 4$$ $$b^2 = 14.0625$$ $$b = \sqrt{14.0625}$$ $$b = 3.75 \text{ m}$$

• The height above the ground is $\boxed{3.75 \text{ m}}$.

To determine if a triangle is right-angled, we check if the square of the longest side is equal to the sum of the squares of the other two sides ($c^2 = a^2 + b^2$). The right angle is opposite the longest side.

• a) $\triangle ABC$ with $a=9$ cm, $b=7$ cm and $c=4$ cm. The longest side is $a=9$ cm. $a^2 = 9^2 = 81$. $b^2 + c^2 = 7^2 + 4^2 = 49 + 16 = 65$. Since $81 \neq 65$, the triangle is not right-angled.

• b) $\triangle DEF$ with $d=5$ cm, $e=13$ cm and $f=12$ cm. The longest side is $e=13$ cm. $e^2 = 13^2 = 169$. $d^2 + f^2 = 5^2 + 12^2 = 25 + 144 = 169$. Since $169 = 169$, the triangle is right-angled. The right angle is opposite side $e$ (which is $DF$), so the right angle is at $\boxed{\angle E}$.

• c) $\triangle STR$ with $s=20$ cm, $t=16$ cm and $r=10$ cm. The longest side is $s=20$ cm. $s^2 = 20^2 = 400$. $t^2 + r^2 = 16^2 + 10^2 = 256 + 100 = 356$. Since $400 \neq 356$, the triangle is not right-angled.

• d) $\triangle PQR$ with $p=24$ cm, $q=25$ cm and $r=7$ cm. The longest side is $q=25$ cm. $q^2 = 25^2 = 625$. $p^2 + r^2 = 24^2 + 7^2 = 576 + 49 = 625$. Since $625 = 625$, the triangle is right-angled. The right angle is opposite side $q$ (which is $PR$), so the right angle is at $\boxed{\angle Q}$.

The ladder, wall, and ground form a right-angled triangle.
The height against the wall is one leg ($a = 6$ m).
The distance from the wall to the foot of the ladder is the other leg ($b = 2.5$ m).
The length of the ladder is the hypotenuse ($c$).
Using the Pythagorean theorem $c^2 = a^2 + b^2$:

$$c^2 = 6^2 + 2.5^2$$ $$c^2 = 36 + 6.25$$ $$c^2 = 42.25$$ $$c = \sqrt{42.25}$$ $$c = 6.5 \text{ m}$$

• The ladder must be $\boxed{6.5 \text{ m}}$ long.