Areas of triangles, quadrilaterals and composites

(a) $63$ cm²

1. $$14\times 9/2$$
2. The area of a triangle is half $\times$ base $\times$ perpendicular height.
3. Area = $\frac{1}{2}$ $\times$ 14 $\times$ 9 = 63 cm².

• P1 Establishing $14\times 9/2$ or an equivalent valid method.
• A1 Correct answer: $63$ cm²

(a) £432

1. Area of a trapezium (on the formula sheet).
   $$\tfrac{1}{2}(9 + 15) \times 8 = 12 \times 8 = 96$$
2. Cost.
   $$96 \times 4.50 = 432$$

• P1 Using $\frac{1}{2}(a + b)h$ with 9, 15 and 8.
• P1 Area $= 96\text{ m}^2$.
• P1 Multiplying their area by 4.50.
• A1 Correct answer: £432.

(a) $23$ cm

1. $$93\times 2/6$$
2. $$31-8$$
3. For parallel sides a and b and height h, area = $\frac{1}{2}$(a + b)h.
4. 93 = $\frac{1}{2}$(8 + b) $\times$ 6, so 8 + b = 31.
5. b = 31 $-$ 8 = 23 cm.

• P1 Establishing $93\times 2/6$ or an equivalent valid method.
• P1 Establishing $31-8$ or an equivalent valid method.
• A1 Correct answer: $23$ cm

(a) $32$ cm²

1. Average the parallel sides and multiply by the perpendicular height.
   $$(6+10)\times 4/2$$
2. Therefore $32$ cm².

• M1 Average the parallel sides and multiply by the perpendicular height.
• A1 Correct answer: $32$ cm²

(b) $4$ cm

1. Rearrange the triangle area formula.
   $$2\times (32)/16$$
2. Therefore $4$ cm.

• P1 Rearrange the triangle area formula.
• A1 Correct answer: $4$ cm

(a) $54$ m²

1. Find the full area and subtract the removed area.
   $$9\times 7-3\times 3$$
2. Therefore $54$ m².

• M1 Find the full area and subtract the removed area.
• A1 Correct answer: $54$ m²

(b) $32$ m

1. The two new inner edges replace equal lengths removed from the outer boundary.
   $$2\times (9+7)$$
2. Therefore $32$ m.

• P1 The two new inner edges replace equal lengths removed from the outer boundary.
• A1 Correct answer: $32$ m

(a) $6$ cm

1. Divide area by the corresponding base.
   $$72/12$$
2. Therefore $6$ cm.

• M1 Divide area by the corresponding base.
• A1 Correct answer: $6$ cm

(b) $42$ cm

1. The boundary uses the sloping sides, not the perpendicular height.
   $$2\times (12+9)$$
2. Therefore $42$ cm.

• M1 The boundary uses the sloping sides, not the perpendicular height.
• A1 Correct answer: $42$ cm

(a) $80$ cm²

1. Split along the 16 cm diagonal into two triangles whose heights add to 10 cm.
   $$16\times 10/2$$
2. Therefore $80$ cm².

• M1 Split along the 16 cm diagonal into two triangles whose heights add to 10 cm.
• A1 Correct answer: $80$ cm²

(b) $42$ cm

1. Divide the area by the width to find the length.
   $$80/5$$
2. Add all four sides.
   $$2\times (16+5)$$
3. Therefore $42$ cm.

• P1 Divide the area by the width to find the length.
• P1 Add all four sides.
• A1 Correct answer: $42$ cm

(a) $x = 4$

1. Area left = square minus triangle.
   $$(2x)^2 - \tfrac{1}{2}x^2 = 4x^2 - \tfrac{1}{2}x^2 = \tfrac{7}{2}x^2$$
2. Set equal to 56.
   $$\tfrac{7}{2}x^2 = 56 \;\Rightarrow\; x^2 = 16$$
3. $x$ is a length, so $x = 4$ (not $-4$).

• P1 An expression for the area left, $(2x)^2 - \frac{1}{2}x^2$.
• P1 Setting their expression equal to 56 and reaching $x^2 = 16$.
• A1 Correct answer: $x = 4$ only.

(b) $8\text{ cm}^2$

1. $\frac{1}{2} \times 4 \times 4 = 8$. Check: $64 - 8 = 56$.

• B1 $8\text{ cm}^2$.