Pythagoras in two dimensions

(a) 13.0 cm

1. Pythagoras: the hypotenuse is opposite the right angle.
   $$c^2 = 7^2 + 11^2 = 49 + 121 = 170$$
2. $$c = \sqrt{170} = 13.038\ldots$$
3. Correct to 1 decimal place: 13.0 cm.

• M1 Writing $7^2 + 11^2$.
• M1 Square-rooting their sum of squares.
• A1 Correct answer: 13.0 cm.

(a) Yes: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$.

1. Square the two shorter sides and add.
   $$5^2 + 12^2 = 25 + 144 = 169$$
2. Square the longest side.
   $$13^2 = 169$$
3. They are equal, so the triangle is right-angled.

• M1 $5^2 + 12^2$ and $13^2$ both worked out.
• C1 "Yes" because $5^2 + 12^2 = 13^2$ (169).

(a) $24$ cm

1. $$25^{2}-7^{2}$$
2. $$\sqrt{576}$$
3. By Pythagoras, the missing side squared is $25^{2}$ $-$ $7^{2}.$
4. 625 $-$ 49 = 576, so the length is $\sqrt{576}$ = 24 cm.

• P1 Establishing $25^{2}-7^{2}$ or an equivalent valid method.
• P1 Establishing $\sqrt{576}$ or an equivalent valid method.
• A1 Correct answer: $24$ cm

(a) $8$ m

1. The ladder is the hypotenuse, so subtract the ground-distance square.
   $$10^{2}-6^{2}$$
2. Take the positive square root.
   $$\sqrt{64}$$
3. Therefore $8$ m.

• P1 The ladder is the hypotenuse, so subtract the ground-distance square.
• P1 Take the positive square root.
• A1 Correct answer: $8$ m

(b) $2$ m

1. Subtract the original height from the ladder’s full length.
   $$10-8$$
2. Therefore $2$ m.

• P1 Subtract the original height from the ladder’s full length.
• A1 Correct answer: $2$ m

(a) $15$ cm

1. The diagonal forms a right triangle with the two sides.
   $$\sqrt{9^{2}+12^{2}}$$
2. Therefore $15$ cm.

• M1 The diagonal forms a right triangle with the two sides.
• A1 Correct answer: $15$ cm

(b) $6$ cm

1. Subtract the diagonal from the two-side route.
   $$9+12-15$$
2. Therefore $6$ cm.

• P1 Subtract the diagonal from the two-side route.
• A1 Correct answer: $6$ cm

(a) $12^2+16^2=20^2$, so the converse of Pythagoras establishes a right angle opposite the longest side.

1. Add the squares of the two shorter sides.
   $$12^{2}+16^{2}$$
2. Compare with the square of the longest side.
   $$20^{2}$$
3. Therefore $12^2+16^2=20^2$, so the converse of Pythagoras establishes a right angle opposite the longest side.

• P1 Add the squares of the two shorter sides.
• P1 Compare with the square of the longest side.
• C1 Correct conclusion with supporting reasoning: $12^2+16^2=20^2$, so the converse of Pythagoras establishes a right angle opposite the longest side.

(b) $96$ cm²

1. The verified perpendicular sides can serve as base and height.
   $$12\times 16/2$$
2. Therefore $96$ cm².

• P1 The verified perpendicular sides can serve as base and height.
• A1 Correct answer: $96$ cm²

(a) 30 cm

1. The longest side, $x + 8$, is the hypotenuse.
   $$x^2 + (x + 7)^2 = (x + 8)^2$$
2. Expand.
   $$x^2 + x^2 + 14x + 49 = x^2 + 16x + 64$$
3. Simplify.
   $$x^2 - 2x - 15 = 0$$
4. Factorise.
   $$(x - 5)(x + 3) = 0$$
5. $x > 0$, so $x = 5$: the sides are 5, 12 and 13.
6. $$5 + 12 + 13 = 30$$

• P1 Pythagoras with $x + 8$ as the hypotenuse.
• P1 Simplifying to $x^2 - 2x - 15 = 0$.
• P1 Solving to $x = 5$ and rejecting $-3$.
• A1 Correct answer: 30 cm.

(a) $300$ cm²

1. $$(70/2)^{2}-25^{2}$$
2. $$600/2$$
3. Let the side lengths be x and y. The perimeter gives x + y = 35.
4. Pythagoras gives $x^{2}$ + $y^{2}$ = $25^{2}$ = 625.
5. Expand (x + y)$^{2}$: $35^{2}$ = $x^{2}$ + $y^{2}$ + 2xy = 625 + 2xy.
6. Thus 2xy = 1225 $-$ 625 = 600, so the area xy = 300 cm².

• P1 Establishing $(70/2)^{2}-25^{2}$ or an equivalent valid method.
• P1 Establishing $600/2$ or an equivalent valid method.
• A1 Correct answer: $300$ cm²