-
Either paper 3 marks
Shape \(ABCDE\) is drawn on a centimetre grid.
(a)Complete this statement by writing the name of one side of the shape.
\(AB : \_\_\_\_\_\_ = 3 : 1\)
(b)On the centimetre grid below, draw a rectangle with the same area as shape \(ABCDE\).
You cannot draw on this page. Draw (b) on paper, then show the answer and check your drawing against the model.
Show answer
(a) CD (or DC)
(b) Any rectangle of area \(14\) cm\(^2\), for example \(2\) cm by \(7\) cm(b)Counting squares on the grid, \(AB = 6\) cm.
For \(AB : \_\_\_\_\_\_ = 3 : 1\), the missing length is \(6 \div 3 = 2\) cm.
The side that is \(2\) cm long is \(CD\).
Shape \(ABCDE\) is a rectangle \(6\) cm by \(3\) cm with a right-angled triangle cut from one corner. The triangle has legs \(4\) cm and \(2\) cm.
Area of the rectangle \(= 6 \times 3 = 18\) cm\(^2\)
Area of the triangle \(= \frac{1}{2} \times 4 \times 2 = 4\) cm\(^2\)
Area of shape \(ABCDE = 18 - 4 = 14\) cm\(^2\)
Any rectangle whose length times width is \(14\) will do, for example \(2\) cm by \(7\) cm.
- (b) M1area of shape \(ABCDE\) \(14\) cm\(^2\), shown by a shape of that area
- (b) A1rectangle of area \(14\) cm\(^2\) drawn, for example \(2\) by \(7\)
-
Either paper 5 marks
Here is an L-shape. Every corner of the shape is a right angle.
The dashed line splits it into two rectangles.
(a)Calculate the area of the shape.
(b)Calculate the perimeter of the shape.
Show answer
(a) \(80\ \text{cm}^2\)
(b) \(48\ \text{cm}\)The dashed line cuts the shape into a rectangle along the bottom and a rectangle standing on top of it.
Bottom rectangle: \(10 \times 4 = 40\)
Top rectangle: \(4 \times 10 = 40\)
Add them:
\(40 + 40 = 80\)
The area is \(80\) cm².
The perimeter is the outside only, so the dashed line is not part of it.
\(10 + 4 + 6 + 10 + 4 + 14 = 48\)
The perimeter is \(48\) cm.
- (a) M1bottom rectangle \(10 \times 4 = 40\)
- (a) M1top rectangle \(4 \times 10 = 40\)
- (a) A1total area \(80\) cm²
- (b) M1adding all six outside sides, with the missing sides \(6\) and \(10\) found
- (b) A1perimeter \(48\) cm
-
Either paper 5 marks
Here is a shape made from two rectangles. Every corner is a right angle.
(a)Calculate the length marked \(x\).
(b)Calculate the length marked \(y\).
(c)Calculate the area of the shape.
Show answer
(a) \(7\ \text{cm}\)
(b) \(8\ \text{cm}\)
(c) \(182\ \text{cm}^2\)The right-hand side of the shape is \(14\) cm tall, and the bottom-left part of it takes up \(7\) cm of that height.
\(x = 14 - 7 = 7\)
\(x = 7\) cm
The bottom of the shape is \(17\) cm across, and the top part takes up \(9\) cm of it.
\(y = 17 - 9 = 8\)
\(y = 8\) cm
The dashed line splits the shape into two rectangles.
Left rectangle: \(8 \times 7 = 56\)
Right rectangle: \(9 \times 14 = 126\)
\(56 + 126 = 182\)
The area is \(182\) cm².
- (c) M1one rectangle area, \(8 \times 7 = 56\) or \(9 \times 14 = 126\)
- (c) M1both rectangle areas found and added
- (c) A1total area \(182\) cm²
-
Either paper 5 marks
Ben is covering the floor of his hallway with carpet. The floor is the L-shape below, and every corner is a right angle.
(a)Calculate the area of the floor.
(b)The carpet cost £10 for each square metre.
Work out how much he pays.
Show answer
(a) \(74\ \text{m}^2\)
(b) \(\text{£}740\)Split the floor along the dashed line into two rectangles.
\(10 \times 5 = 50\) m²
\(4 \times 6 = 24\) m²
\(50 + 24 = 74\)
The area is \(74\) m².
Each square metre costs \(£10\), and there are \(74\) of them.
\(74 \times £10 = £740\)
Ben pays \(£740\).
- (a) M1bottom rectangle \(10 \times 5 = 50\)
- (a) M1top rectangle \(4 \times 6 = 24\)
- (a) A1total area \(74\) m²
- (b) M1area times price \(74 \times 10\)
- (b) A1cost \(£740\)
-
Either paper 6 marks
The diagram shows a shape with two vertical sides. The dashed line splits it into a rectangle and a right-angled triangle.
(a)Work out the perimeter of the shape.
(b)Work out the area of the shape.
Show answer
(a) \(72\ \text{cm}\)
(b) \(240\ \text{cm}^2\)The perimeter goes round the outside only, so the dashed line does not count.
\(12 + 28 + 20 + 12 = 72\)
The perimeter is \(72\) cm.
The dashed line is \(12\) cm long, the same as the bottom, so the rectangle below it is \(12\) cm by \(12\) cm.
\(\text{Rectangle} = 12 \times 12 = 144\)
The triangle above it has a base of \(12\) cm. Its height is what is left of the taller side:
\(28 - 12 = 16\) cm
\(\text{Triangle} = \frac{1}{2} \times 12 \times 16 = 96\)
Add the two parts:
\(144 + 96 = 240\)
The area is \(240\) cm².
- (a) M1sum of the four outside sides \(12 + 28 + 20 + 12\)
- (a) A1perimeter \(72\) cm
- (b) M1rectangle area \(12 \times 12 = 144\)
- (b) M1height of triangle \(16\) cm
- (b) M1triangle area \(96\)
- (b) A1total area \(240\) cm²
-
Either paper 5 marks
The diagram shows a rectangle with a rectangular piece cut out of it.
Every corner of the shape is a right angle.
(a)Work out the area of the shape.
(b)Work out the perimeter of the shape.
Show answer
(a) \(138\ \text{cm}^2\)
(b) \(62\ \text{cm}\)This shape is easier to take apart than to build up: start from the whole rectangle and subtract the piece that has been cut out.
The whole rectangle is \(13\) cm by \(12\) cm.
\(13 \times 12 = 156\ \text{cm}^2\)
The piece cut out is \(3\) cm by \(6\) cm.
\(3 \times 6 = 18\ \text{cm}^2\)
\(156 - 18 = 138\ \text{cm}^2\)
The perimeter is the whole way round the outside, so add all eight sides.
\(13 + 12 + 6 + 6 + 3 + 6 + 4 + 12 = 62\) cm
Notice what the notch did. The \(3\) cm it took out of the top is put straight back by the \(3\) cm along the bottom of the notch, so the only change is the two sides of the notch: \(2 \times 6 = 12\) cm. Cutting the notch wider would not change the perimeter at all.
- (a) M1whole rectangle \(13 \times 12 = 156\)
- (a) M1notch \(3 \times 6 = 18\)
- (a) A1area \(138\) cm²
- (b) M1adding all eight outside sides
- (b) A1perimeter \(62\) cm
-
Either paper 3 marks
The diagram shows a rectangle with a rectangular piece cut from one corner.
All the corners are right angles. Measurements are in centimetres.
The area of the shape is \(332\ \text{cm}^2\).
Work out the value of \(x\).
Show answer
\(7\ \text{cm}\)The shape is the whole rectangle with the notch removed.
Whole rectangle: \(20 \times 18 = 360\ \text{cm}^2\)
The notch is \(x\) cm by \(4\) cm, so its area is \(4x\ \text{cm}^2\).
Whole rectangle minus notch equals the shape:
\(360 - 4x = 332\)
\(4x = 360 - 332 = 28\)
\(x = 28 \div 4 = 7\) cm
- M1area of the whole rectangle \(20 \times 18 = 360\)
- M1equation: whole rectangle minus the notch, \(360 - 4x = 332\)
- A1\(x = 7\)
-
Either paper 3 marks
Here is a compound shape made from two rectangles. All measurements are in centimetres.
Show that the area of the shape can be written as \(8x^2 + 43x + 56\) cm².
Show answer
\((3x + 9)(2x + 4) + (x + 4)(2x + 5) = 8x^2 + 43x + 56\)Split the shape into two rectangles, as shown by the dashed line.
The left-hand side is \((4x + 9)\) cm tall, and the lower rectangle already takes up \((2x + 4)\) cm of that height, so the top rectangle's height is:
\((4x + 9) - (2x + 4) = 2x + 5\)
Bottom rectangle: \((3x + 9)(2x + 4) = 6x^2 + 30x + 36\)
Top rectangle: \((x + 4)(2x + 5) = 2x^2 + 13x + 20\)
Add the two areas and collect like terms:
\(6x^2 + 30x + 36 + 2x^2 + 13x + 20 = 8x^2 + 43x + 56\)
So the area of the shape is \(8x^2 + 43x + 56\) cm², as required.
- M1area of one rectangle, \((3x + 9)(2x + 4)\)
- M1area of the other, \((x + 4)(2x + 5)\)
- A1expands and adds to \(8x^2 + 43x + 56\)
-
Either paper 3 marks
Shape \(ABCDE\) is drawn on a centimetre grid.
(a)Complete this statement by writing the name of one side of the shape.
\(AB : \_\_\_\_\_\_ = 5 : 1\)
(b)On the centimetre grid below, draw a rectangle with the same area as shape \(ABCDE\).
You cannot draw on this page. Draw (b) on paper, then show the answer and check your drawing against the model.
Show answer
(a) EA (or AE)
(b) Any rectangle of area \(42\) cm\(^2\), for example \(6\) cm by \(7\) cm(b)Counting squares on the grid, \(AB = 10\) cm.
For \(AB : \_\_\_\_\_\_ = 5 : 1\), the missing length is \(10 \div 5 = 2\) cm.
The side that is \(2\) cm long is \(EA\).
Shape \(ABCDE\) is a rectangle \(10\) cm by \(6\) cm with a right-angled triangle cut from one corner. The triangle has legs \(9\) cm and \(4\) cm.
Area of the rectangle \(= 10 \times 6 = 60\) cm\(^2\)
Area of the triangle \(= \frac{1}{2} \times 9 \times 4 = 18\) cm\(^2\)
Area of shape \(ABCDE = 60 - 18 = 42\) cm\(^2\)
Any rectangle whose length times width is \(42\) will do, for example \(6\) cm by \(7\) cm.
- (b) M1area of shape \(ABCDE\) \(42\) cm\(^2\), shown by a shape of that area
- (b) A1rectangle of area \(42\) cm\(^2\) drawn, for example \(6\) by \(7\)
-
Either paper 5 marks
The diagram shows an L-shape. Every corner of the shape is a right angle.
The dashed line splits it into two rectangles.
(a)Calculate the area of the shape.
(b)Calculate the perimeter of the shape.
Show answer
(a) \(85\ \text{cm}^2\)
(b) \(42\ \text{cm}\)The dashed line cuts the shape into a rectangle along the bottom and a rectangle standing on top of it.
Bottom rectangle: \(10 \times 6 = 60\)
Top rectangle: \(5 \times 5 = 25\)
Add them:
\(60 + 25 = 85\)
The area is \(85\) cm².
The perimeter is the outside only, so the dashed line is not part of it.
\(10 + 6 + 5 + 5 + 5 + 11 = 42\)
The perimeter is \(42\) cm.
- (a) M1bottom rectangle \(10 \times 6 = 60\)
- (a) M1top rectangle \(5 \times 5 = 25\)
- (a) A1total area \(85\) cm²
- (b) M1adding all six outside sides, with the missing sides \(5\) and \(5\) found
- (b) A1perimeter \(42\) cm