2D Area & Perimeter: WJEC GCSE Numeracy Higher June 2025 Unit 1 Q13

Numeracy Higher June 2025 Unit 1 Q13

WJECHigherCurrent spec15 marks2D Area & Perimeter Pythagoras Volume & Surface Area

13. Pizza-Rea is a company that makes outdoor pizza ovens.

(a) The diagram below shows the design for the metal front plate of its latest pizza oven.

The front of the oven is symmetrical.
It has a rectangular hole, through which pizzas are placed in the oven.

Pizza-Rea wants to calculate an estimate of the area of the metal used for this front plate.

Shaded symmetrical front plate with a curved top rising to a point, vertical sides and a flat base, with a white rectangular hole in the middle

Diagram not drawn to scale

The diagram below is a sketch of the front plate of the pizza oven drawn on coordinate axes.

The front plate turned on its side and drawn on x and y axes, with the point of the curve at the origin and the rectangular hole crossing the x-axis

Diagram not drawn to scale

Pizza-Rea have accurately drawn one of the symmetrical halves of the front plate on graph paper.

Graph paper, x (cm) 0 to 50, y (cm) 0 to 40: shaded metal region under a curve from (0, 0) through about (10, 22), (20, 33), (30, 39) to (35, 40), then y = 40 to x = 50 and down to the x-axis; a white hole from x = 25 to x = 45, height 20

Using 5 strips of equal width, calculate an estimate of the area of metal shown in the graph.
Hence calculate the total area of metal in the complete front plate. [5]

(b) Pizzas are cooked in the inner part of the pizza oven.
The inner part is in the shape of a half-cylinder.

A hole is needed in the top of the oven to let hot air out.
The straight-line distance from the centre of this hole to either of the front bottom corners of the oven should be less than 50 cm.

The dotted line, \(AB\), in the diagram below represents one of these straight-line distances.

Half-cylinder oven with front diameter 60 cm; hole A on the top, 20 cm back from the top of the front semicircle; B is a front bottom corner; dotted line AB

Diagram not drawn to scale

Show that the distance from \(A\) to \(B\) is less than 50 cm. [4]

(c) Pizza-Rea supplies spiked legs for the pizza oven.
Each leg is made by connecting a cuboid to a pyramid.
The perpendicular height of the pyramid is 42 cm.

The diagrams below show the design of the spiked legs.

Spiked leg: cuboid with top x by x/2 and height 16 cm on top of a pyramid of perpendicular height 42 cm; side view shows width x/2, 16 cm and 42 cm

Diagrams not drawn to scale

The volume of each leg is 750 cm³.
Calculate the length \(x\).
Give your answer in the form \(a\sqrt{b}\), where \(a\) and \(b\) are integers and \(b\) is as small as possible. [6]

Mark scheme (a)

Scheme	Marks	Comments
(Area =) \(\dfrac{1}{2}\)×10×(0+40 + 2(22+33+39+40)) OR \(\dfrac{1}{2}\) × 10 × (40+44+66+78+80)	M2	If 5 strips of unequal width used, all marks available, but penalise -1 from accuracy marks gained M1 for at most 1 slip in substitution of values from the vertical axis OR M1 for 1 of the vertical readings omitted with all others correct
= 1540 (cm²)	A1	FT from M1. May be implied in further working
(Area of the front plate = ) 1540 − 20×20 (×2)	m1	FT ‘their 1540’ provided M1 or M2 previously awarded
= 2280 (cm²)	A1	FT ‘their 1540’ If final m0A0 awarded, and again FT ‘their 1540’ award SC1 for any one of the following: (2×1540 − 20×20 =) 2680 (cm²) (1540 − 2×20×20 =) 740 (cm²)

Alternative method 1

Scheme	Marks	Comments
\(\dfrac{(0 + 22)}{2}\)×10 + \(\dfrac{(22 + 33)}{2}\)×10 + \(\dfrac{(33 + 39)}{2}\)×10 + \(\dfrac{(39 + 40)}{2}\)×10 + \(\dfrac{(40 + 40)}{2}\)×10 [110 + 275 + 360 + 395 + 400]	M2	If 5 strips of unequal width used, all marks available, but penalise -1 from accuracy marks gained Each area may be seen as the sum of the area of a rectangle and triangle M1 for the sum of these 5 areas with at most one error (may be repeated) in the substitution of values from the vertical axis OR M1 for the sight of 5 correct areas with the intention to add them (possibly omitting one)
= 1540 (cm²)	A1	FT from M1
(Area of the front plate = ) 1540 − 20×20 (×2)	m1	FT ‘their 1540’ provided M1 or M2 previously awarded
= 2280 (cm²)	A1	FT ‘their 1540’ If final m0A0 awarded, and again FT ‘their 1540’ award SC1 for any one of the following: (2×1540 − 20×20 =) 2680 (cm²) (1540 − 2×20×20 =) 740 (cm²)

Alternative method 2

Scheme	Marks	Comments
\(\dfrac{(0 + 22)}{2}\)×10 + \(\dfrac{(22 + 33)}{2}\)×10 + (\(\dfrac{(33 + 39)}{2}\)×10 − 5×20 ) + (\(\dfrac{(39 + 40)}{2}\)×10 − 10×20 ) or + \(\dfrac{(19 + 20)}{2}\)×10 + (\(\dfrac{(40 + 40)}{2}\)×10 − 5×20 ) [110 + 275 + 260 + 195 + 300]	M3	If 5 strips of unequal width used, all marks available, but penalise -1 from accuracy marks gained Each area may be seen as the sum of the area of a rectangle and triangle M2 for the sum of these 5 areas with 1 one error (possibly repeated) in the substitution of values from the vertical axis or 1 error in the values used for the rectangle being subtracted OR M2 for the sight of 5 correct areas with the intention to add them (possibly omitting one) M1 for the sum of these 5 areas with 1 one error (possibly repeated) in the substitution of values from the vertical axis and 1 error in the values used for the rectangle being subtracted
= 1140 (cm²)	A1	FT from M2
(Area of the front plate = ) 2280 (cm²)	A1	FT provided M2 or M3 previously awarded FT 2 × ‘their 1140’

Mark scheme (b)

Scheme	Marks	Comments
Correct use of Pythagoras in 1 of the planes 30² + 30² or \(\sqrt{30^2 + 30^2}\) OR 30² + 20² or \(\sqrt{30^2 + 20^2}\)	M1	May be embedded in further working
(Distance² =) 30² + 30² + 20²	M1	May be seen as 2 applications of Pythagoras
Distance² = 2200 or (Distance =) √2200 (= 46.9…)	A1	CAO
Correct comparison statement e.g. ‘2200 is less than 2500’, or ‘2200 is less than 50²’, or ‘√2200 is less than √2500’, or ‘√22 is less than 5’, or ‘√22 is less than √25’	A1	FT ‘their 2200’ provided > 900 and M1M1 previously awarded

Alternative method 1

Scheme	Marks	Comments
50² − 30² or \(\sqrt{50^2 - 30^2}\)	M1
(Distance to hole² =) 50² − 30² − 30²	M1	May be seen as 2 applications of Pythagoras
Distance to hole² = 700 or (Distance to hole =) √700	A1	CAO
Correct comparison statement e.g. ‘700 is greater than 400’, or ‘700 is greater than 20²’, or ‘√700 is greater than √400’, or ‘√7 is greater than 2’, or ‘√7 is greater than √4’	A1	FT ‘their 700’ provided < 2500 and M1M1 previously awarded

Alternative method 2

Scheme	Marks	Comments
50² − 20² or \(\sqrt{50^2 - 20^2}\) OR 50² − 30² or \(\sqrt{50^2 - 30^2}\)	M1	May be embedded in further working
(Distance to hole² =) 50² − 20² − 30²	M1	May be seen as 2 applications of Pythagoras
Dist to hole² = 1200 or (Distance to hole =) √1200	A1	CAO
Correct comparison statement e.g. ‘1200 is greater than 900’, or ‘1200 is greater than 30²’, or ‘√1200 is greater than √900’, or ‘√12 is greater than 3’, or ‘√12 is greater than √9’	A1	FT ‘their 1200’ provided < 2500 and M1M1 previously awarded

Alternative method 3

Scheme	Marks	Comments
50² − 30² or \(\sqrt{50^2 - 30^2}\) OR 20² + 30² or \(\sqrt{20^2 + 30^2}\)	M1
Comparison of 50² − 30² with 20² + 30² OR \(\sqrt{50^2 - 30^2}\) with \(\sqrt{20^2 + 30^2}\)	M1
= 1600 AND 1300 or √1600 AND √1300 AND a correct statement e.g. 1600 > 1300	A2	A1 for any one of the following: 1600 AND 1300 or √1600 AND √1300 without a correct statement 1600 or 1300 correctly evaluated with a correct statement

Alternative method 4

Scheme	Marks	Comments
50² − 20² or \(\sqrt{50^2 - 20^2}\) OR 30² + 30² or \(\sqrt{30^2 + 30^2}\)	M1
Comparison of 50² − 20² with 30² + 30² OR \(\sqrt{50^2 - 20^2}\) with \(\sqrt{30^2 + 30^2}\)	M1
= 2100 AND 1800 or √2100 AND √1800 AND a correct statement e.g. 2100 > 1800	A2	A1 for any one of the following: 2100 AND 1800 or √2100 AND √1800 without a correct statement 2100 or 1800 correctly evaluated with a correct statement

Mark scheme (c)

Scheme	Marks	Comments
\(\dfrac{1}{3} \times x \times \dfrac{x}{2} \times 42 \;+\; x \times \dfrac{x}{2} \times 16 = 750\) or \(7x^2 + 8x^2 = 750\) or equivalent	M2	M1 for any one of the following: \(\dfrac{1}{3} \times x \times \dfrac{x}{2} \times 42\) + …….. = 750 or …………. + \(x \times \dfrac{x}{2} \times 16\) = 750 or \(\dfrac{1}{3} \times x \times \dfrac{x}{2} \times 42 + x \times \dfrac{x}{2} \times 16\)
\(15x^2 = 750\)	m1	FT from M1 if possible Needs to be of the form \(ax^2 = 750\)
\(x^2 = 50\) or \(x = \sqrt{50}\)	A1	CAO
\(x = 5\sqrt{2}\) (cm)	B2	For B2 or B1, FT ‘their derived 50’ for equivalent level of difficulty provided it comes from \(ax^2 = 750\) B1 for writing 50 as a product of 2 or more factors where one of the factors OR the product of 2 of their factors is a square number e.g. 25 × 2, 5 × 5 × 2, OR B1 for writing \(\sqrt{50}\) as a product of 2 or more surds where one of the surds OR the product of 2 of their surds gives a whole number e.g. \(\sqrt{25} \times \sqrt{2}\), \(\sqrt{5} \times \sqrt{5} \times \sqrt{2}\)