Foundation geometry and equipment clinic

(a) $1.65\text{ m}^2$

1. Rectangle.
   $$1.2 \times 0.9 = 1.08$$
2. Semicircle, radius 0.6 m.
   $$\tfrac{1}{2} \times \pi \times 0.6^2 = 0.5654\ldots$$
3. $$1.08 + 0.5654\ldots = 1.6454\ldots$$

• P1 Rectangle area 1.08.
• P1 Semicircle area with radius 0.6.
• A1 $1.65\text{ m}^2$.

(a) 20 m

1. Jess: $40 + 30 = 70$ m.
2. Ben: Pythagoras.
   $$\sqrt{40^2 + 30^2} = \sqrt{2500} = 50$$
3. $$70 - 50 = 20$$

• P1 Jess's distance, 70 m.
• P1 Ben's distance by Pythagoras, 50 m.
• A1 Correct answer: 20 m.

(a) $108^\circ$

1. Exterior angle $360 \div 5 = 72$, so interior $180 - 72 = 108$.

• M1 $360 \div 5$ or $(5 - 2) \times 180 \div 5$.
• A1 Correct answer: $108^\circ$.

(b) $360 \div 108$ is not a whole number, so the angles cannot make $360^\circ$.

1. Three angles make $324^\circ$ and four make $432^\circ$: neither is $360^\circ$.

• C1 Linking the angle $108^\circ$ to 360: 360 is not a multiple of 108.

(a) $105$°

1. Adjacent angles on a straight line sum to 180 degrees.
   $$180-75$$
2. Therefore $105$°.

• M1 Adjacent angles on a straight line sum to 180 degrees.
• A1 Correct answer: $105$°

(b) It is $75^\circ$ because corresponding angles between parallel lines are equal.

1. It is $75^\circ$ because corresponding angles between parallel lines are equal.

• C1 Correct conclusion with supporting reasoning: It is $75^\circ$ because corresponding angles between parallel lines are equal.

(a) Construct the internal angle bisector with equal arcs. Draw an arc centred at O of radius 20 cm. Its intersection with the bisector inside the angle is P.

1. Equal distances from the two rays locate P on the internal angle bisector.
2. The distance OP places P on a circle centred at O.
3. Construct the internal angle bisector with equal arcs. Draw an arc centred at O of radius 20 cm. Its intersection with the bisector inside the angle is P.

• M1 Equal distances from the two rays locate P on the internal angle bisector.
• M1 The distance OP places P on a circle centred at O.
• C1 Correct conclusion with supporting reasoning: Construct the internal angle bisector with equal arcs. Draw an arc centred at O of radius 20 cm. Its intersection with the bisector inside the angle is P.

(b) $45$°

1. Halve the right angle.
   $$90/2$$
2. Therefore $45$°.

• M1 Halve the right angle.
• A1 Correct answer: $45$°

(a) $221$ cm²

1. Average the parallel sides and multiply by the perpendicular height.
   $$(15+19)\times 13/2$$
2. Therefore $221$ cm².

• M1 Average the parallel sides and multiply by the perpendicular height.
• A1 Correct answer: $221$ cm²

(b) $13$ cm

1. Rearrange the triangle area formula.
   $$2\times (221)/34$$
2. Therefore $13$ cm.

• P1 Rearrange the triangle area formula.
• A1 Correct answer: $13$ cm

(a) $60$ cm

1. The diagonal forms a right triangle with the two sides.
   $$\sqrt{36^{2}+48^{2}}$$
2. Therefore $60$ cm.

• M1 The diagonal forms a right triangle with the two sides.
• A1 Correct answer: $60$ cm

(b) $24$ cm

1. Subtract the diagonal from the two-side route.
   $$36+48-60$$
2. Therefore $24$ cm.

• P1 Subtract the diagonal from the two-side route.
• A1 Correct answer: $24$ cm

(a) $70$°

1. Subtract the two clockwise bearings from the same north line.
   $$170-100$$
2. Therefore $70$°.

• M1 Subtract the two clockwise bearings from the same north line.
• A1 Correct answer: $70$°

(b) $55$°

1. The equal sides make the base angles equal.
   $$\frac{180-70}{2}$$
2. Therefore $55$°.

• M1 The equal sides make the base angles equal.
• A1 Correct answer: $55$°