Question

end - of - unit review
1 work out the area of each of these shapes.
a
work out: i the circumference ii the area of these circles.
a radius = 4 cm b diameter = 12 cm
use π = 3.14.
round your answers correct to 1 d.p.
3 show how you would check your answers to question 1, using a single number instead of π.
4 a circular tile has a circumference of 48.2 cm.
work out the diameter of the tile.
give your answer correct to the nearest millimetre.
5 work out the area of each of these compound shapes.
a
6 work out the area of the pink - shaded region.
7 work out:
a the volume
b the surface area of this cuboid.
8 the diagram shows a cuboid.
the height of the cuboid is 1.2 cm.
the end face of the cuboid is a square.
the volume of the cuboid is 5760 mm³.
work out the surface area of the cuboid.
9 a sketch the net of this shape.
b use your net to work out the surface area.

Explanation:

- ## Step1: Recall triangle - area formula
The formula for the area of a triangle is \(A=\frac{1}{2}\times base\times height\).
Here, the base \(b = 15.2\) cm and the height \(h = 8.8\) cm.
\(A=\frac{1}{2}\times15.2\times8.8\).
- ## Step2: Perform the calculation
First, \(\frac{1}{2}\times15.2 = 7.6\).
Then, \(7.6\times8.8=66.88\) \(cm^{2}\).
- # Answer: \(66.9\) \(cm^{2}\) (rounded to 1 d.p.)
2. **b. Area of parallelogram**:
- # Explanation:
- ## Step1: Recall parallelogram - area formula
The formula for the area of a parallelogram is \(A = base\times height\).
Here, the base \(b = 7\) cm and the height \(h = 4\) cm.
\(A=7\times4\).
- ## Step2: Perform the calculation
\(A = 28\) \(cm^{2}\).
- # Answer: \(28\) \(cm^{2}\)
3. **c. Area of trapezium**:
- # Explanation:
- ## Step1: Recall trapezium - area formula
The formula for the area of a trapezium is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the lengths of the parallel - sides and \(h\) is the height.
Here, \(a = 16\) m, \(b = 24\) m and \(h = 8\) m.
\(A=\frac{(16 + 24)\times8}{2}\).
- ## Step2: Simplify the expression
First, \(16+24 = 40\).
Then, \(\frac{40\times8}{2}=160\) \(m^{2}\).
- # Answer: \(160\) \(m^{2}\)
4. **2a. Circumference of circle with radius \(r = 4\) cm**:
- # Explanation:
- ## Step1: Recall circumference formula
The formula for the circumference of a circle is \(C = 2\pi r\).
Given \(r = 4\) cm and \(\pi=3.14\).
\(C = 2\times3.14\times4\).
- ## Step2: Perform the calculation
\(2\times3.14\times4=25.12\approx25.1\) cm (rounded to 1 d.p.).
- # Answer: \(25.1\) cm
- **2a. Area of circle with radius \(r = 4\) cm**:
- # Explanation:
- ## Step1: Recall area formula
The formula for the area of a circle is \(A=\pi r^{2}\).
Given \(r = 4\) cm and \(\pi = 3.14\).
\(A=3.14\times4^{2}\).
- ## Step2: Perform the calculation
First, \(4^{2}=16\).
Then, \(3.14\times16 = 50.24\approx50.2\) \(cm^{2}\) (rounded to 1 d.p.).
- # Answer: \(50.2\) \(cm^{2}\)
5. **2b. Circumference of circle with diameter \(d = 12\) cm**:
- # Explanation:
- ## Step1: Recall circumference formula in terms of diameter
The formula for the circumference of a circle is \(C=\pi d\).
Given \(d = 12\) cm and \(\pi=3.14\).
\(C=3.14\times12\).
- ## Step2: Perform the calculation
\(C = 37.68\approx37.7\) cm (rounded to 1 d.p.).
- # Answer: \(37.7\) cm
- **2b. Area of circle with diameter \(d = 12\) cm**:
- # Explanation:
- ## Step1: Find the radius
Since \(r=\frac{d}{2}\), if \(d = 12\) cm, then \(r = 6\) cm.
- ## Step2: Recall area formula
The formula for the area of a circle is \(A=\pi r^{2}\).
Given \(r = 6\) cm and \(\pi=3.14\).
\(A=3.14\times6^{2}\).
- ## Step3: Perform the calculation
First, \(6^{2}=36\).
Then, \(3.14\times36 = 113.04\approx113.0\) \(cm^{2}\) (rounded to 1 d.p.).
- # Answer: \(113.0\) \(cm^{2}\)
6. **4. Diameter of circular tile with circumference \(C = 48.2\) cm**:
- # Explanation:
- ## Step1: Recall circumference formula
The formula for the circumference of a circle is \(C=\pi d\).
We need to solve for \(d\), so \(d=\frac{C}{\pi}\).
Given \(C = 48.2\) cm and \(\pi = 3.14\).
\(d=\frac{48.2}{3.14}\).
- ## Step2: Perform the calculation
\(d=\frac{48.2}{3.14}\approx15.35\) cm.
Converting to millimeters, \(d\approx154\) mm (to the nearest millimeter).
- # Answer: \(154\) mm
7. **5a. Area of compound - shape**:
- # Explanation:
- ## Step1: Split the shape
We can split the shape into two rectangles. One rectangle with dimensions \(5\) cm by \(4\) cm and another with dimensions \((8 - 5)=3\) cm by \(3\) cm.
- ## Step2: Calculate the areas of rectangles
The area of the first rectangle \(A_1=5\times4 = 20\) \(cm^{2}\).
The area of the second rectangle \(A_2=3\times3 = 9\) \(cm^{2}\).
- ## Step3: Find the total area
\(A=A_1 + A_2=20 + 9=29\) \(cm^{2}\).
- # Answer: \(29\) \(cm^{2}\)
8. **5b. Area of compound - shape**:
- # Explanation:
- ## Step1: Split the shape
The shape is composed of a rectangle and a semi - circle.
The rectangle has dimensions \(4\) cm by \(8\) cm and the semi - circle has a diameter of \(8\) cm (so radius \(r = 4\) cm).
- ## Step2: Calculate the area of the rectangle
The area of the rectangle \(A_1=4\times8 = 32\) \(cm^{2}\).
- ## Step3: Calculate the area of the semi - circle
The area of a full - circle is \(A_{circle}=\pi r^{2}\), so the area of a semi - circle \(A_2=\frac{1}{2}\pi r^{2}\).
Given \(r = 4\) cm and \(\pi = 3.14\), \(A_2=\frac{1}{2}\times3.14\times4^{2}=25.12\) \(cm^{2}\).
- ## Step4: Find the total area
\(A=A_1+A_2=32 + 25.12=57.12\approx57.1\) \(cm^{2}\) (rounded to 1 d.p.).
- # Answer: \(57.1\) \(cm^{2}\)
9. **6. Area of the pink - shaded region**:
- # Explanation:
- ## Step1: Calculate the area of the large rectangle
The area of the large rectangle \(A_1=15\times10 = 150\) \(cm^{2}\).
- ## Step2: Calculate the area of the small rectangle
The area of the small rectangle \(A_2=6\times5 = 30\) \(cm^{2}\).
- ## Step3: Find the shaded area
\(A=A_1 - A_2=150 - 30 = 120\) \(cm^{2}\).
- # Answer: \(120\) \(cm^{2}\)
10. **7a. Volume of cuboid with length \(l = 10\) cm, width \(w = 4\) cm and height \(h = 5\) cm**:
- # Explanation:
- ## Step1: Recall volume formula
The formula for the volume of a cuboid is \(V=l\times w\times h\).
\(V=10\times4\times5\).
- ## Step2: Perform the calculation
\(10\times4\times5 = 200\) \(cm^{3}\).
- # Answer: \(200\) \(cm^{3}\)
- **7b. Surface area of cuboid**:
- # Explanation:
- ## Step1: Recall surface - area formula
The formula for the surface area of a cuboid is \(SA = 2(lw+lh+wh)\).
Here, \(l = 10\) cm, \(w = 4\) cm and \(h = 5\) cm.
\(SA=2(10\times4 + 10\times5+4\times5)\).
- ## Step2: Calculate the values inside the parentheses
\(10\times4 = 40\), \(10\times5 = 50\) and \(4\times5 = 20\).
\(10\times4+10\times5 + 4\times5=40 + 50+20 = 110\).
- ## Step3: Find the surface area
\(SA=2\times110 = 220\) \(cm^{2}\).
- # Answer: \(220\) \(cm^{2}\)
11. **8. Surface area of cuboid**:
- First, convert the volume to \(cm^{3}\). Since \(1\) \(cm^{3}=1000\) \(mm^{3}\), \(V = 5760\) \(mm^{3}=5.76\) \(cm^{3}\).
- Let the side of the square end - face be \(x\) cm. The volume of the cuboid \(V=x^{2}\times1.2\).
- # Explanation:
- ## Step1: Find the side of the square end - face
We know that \(x^{2}\times1.2=5.76\).
Then \(x^{2}=\frac{5.76}{1.2}=4.8\), so \(x=\sqrt{4.8}\approx2.19\) cm.
- ## Step2: Recall surface - area formula
The surface area of the cuboid \(SA = 2x^{2}+4xh\).
Substitute \(x\approx2.19\) cm and \(h = 1.2\) cm.
\(SA=2\times4.8+4\times2.19\times1.2\).
\(2\times4.8 = 9.6\), \(4\times2.19\times1.2=10.512\).
\(SA=9.6 + 10.512=20.112\approx20.1\) \(cm^{2}\).
- # Answer: \(20.1\) \(cm^{2}\)

(Note: The net - drawing part (question 9) is a visual task and not included in the step - by - step calculation part. To work out the surface area from the net, you would find the area of each face in the net and sum them up.)

Answer:

1. a. Area of triangle:

• # Explanation:
• ## Step1: Recall triangle - area formula

The formula for the area of a triangle is \(A=\frac{1}{2}\times base\times height\).
Here, the base \(b = 15.2\) cm and the height \(h = 8.8\) cm.
\(A=\frac{1}{2}\times15.2\times8.8\).

• ## Step2: Perform the calculation

First, \(\frac{1}{2}\times15.2 = 7.6\).
Then, \(7.6\times8.8=66.88\) \(cm^{2}\).

• # Answer: \(66.9\) \(cm^{2}\) (rounded to 1 d.p.)

1. b. Area of parallelogram:

• # Explanation:
• ## Step1: Recall parallelogram - area formula

The formula for the area of a parallelogram is \(A = base\times height\).
Here, the base \(b = 7\) cm and the height \(h = 4\) cm.
\(A=7\times4\).

• ## Step2: Perform the calculation

\(A = 28\) \(cm^{2}\).

• # Answer: \(28\) \(cm^{2}\)

1. c. Area of trapezium:

• # Explanation:
• ## Step1: Recall trapezium - area formula

The formula for the area of a trapezium is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the lengths of the parallel - sides and \(h\) is the height.
Here, \(a = 16\) m, \(b = 24\) m and \(h = 8\) m.
\(A=\frac{(16 + 24)\times8}{2}\).

• ## Step2: Simplify the expression

First, \(16+24 = 40\).
Then, \(\frac{40\times8}{2}=160\) \(m^{2}\).

• # Answer: \(160\) \(m^{2}\)

1. 2a. Circumference of circle with radius \(r = 4\) cm:

• # Explanation:
• ## Step1: Recall circumference formula

The formula for the circumference of a circle is \(C = 2\pi r\).
Given \(r = 4\) cm and \(\pi=3.14\).
\(C = 2\times3.14\times4\).

• ## Step2: Perform the calculation

\(2\times3.14\times4=25.12\approx25.1\) cm (rounded to 1 d.p.).

• # Answer: \(25.1\) cm
• 2a. Area of circle with radius \(r = 4\) cm:
• # Explanation:
• ## Step1: Recall area formula

The formula for the area of a circle is \(A=\pi r^{2}\).
Given \(r = 4\) cm and \(\pi = 3.14\).
\(A=3.14\times4^{2}\).

• ## Step2: Perform the calculation

First, \(4^{2}=16\).
Then, \(3.14\times16 = 50.24\approx50.2\) \(cm^{2}\) (rounded to 1 d.p.).

• # Answer: \(50.2\) \(cm^{2}\)

1. 2b. Circumference of circle with diameter \(d = 12\) cm:

• # Explanation:
• ## Step1: Recall circumference formula in terms of diameter

The formula for the circumference of a circle is \(C=\pi d\).
Given \(d = 12\) cm and \(\pi=3.14\).
\(C=3.14\times12\).

• ## Step2: Perform the calculation

\(C = 37.68\approx37.7\) cm (rounded to 1 d.p.).

• # Answer: \(37.7\) cm
• 2b. Area of circle with diameter \(d = 12\) cm:
• # Explanation:
• ## Step1: Find the radius

Since \(r=\frac{d}{2}\), if \(d = 12\) cm, then \(r = 6\) cm.

• ## Step2: Recall area formula

The formula for the area of a circle is \(A=\pi r^{2}\).
Given \(r = 6\) cm and \(\pi=3.14\).
\(A=3.14\times6^{2}\).

• ## Step3: Perform the calculation

First, \(6^{2}=36\).
Then, \(3.14\times36 = 113.04\approx113.0\) \(cm^{2}\) (rounded to 1 d.p.).

• # Answer: \(113.0\) \(cm^{2}\)

1. 4. Diameter of circular tile with circumference \(C = 48.2\) cm:

• # Explanation:
• ## Step1: Recall circumference formula

The formula for the circumference of a circle is \(C=\pi d\).
We need to solve for \(d\), so \(d=\frac{C}{\pi}\).
Given \(C = 48.2\) cm and \(\pi = 3.14\).
\(d=\frac{48.2}{3.14}\).

• ## Step2: Perform the calculation

\(d=\frac{48.2}{3.14}\approx15.35\) cm.
Converting to millimeters, \(d\approx154\) mm (to the nearest millimeter).

• # Answer: \(154\) mm

1. 5a. Area of compound - shape:

• # Explanation:
• ## Step1: Split the shape

We can split the shape into two rectangles. One rectangle with dimensions \(5\) cm by \(4\) cm and another with dimensions \((8 - 5)=3\) cm by \(3\) cm.

• ## Step2: Calculate the areas of rectangles

The area of the first rectangle \(A_1=5\times4 = 20\) \(cm^{2}\).
The area of the second rectangle \(A_2=3\times3 = 9\) \(cm^{2}\).

• ## Step3: Find the total area

\(A=A_1 + A_2=20 + 9=29\) \(cm^{2}\).

• # Answer: \(29\) \(cm^{2}\)

1. 5b. Area of compound - shape:

• # Explanation:
• ## Step1: Split the shape

The shape is composed of a rectangle and a semi - circle.
The rectangle has dimensions \(4\) cm by \(8\) cm and the semi - circle has a diameter of \(8\) cm (so radius \(r = 4\) cm).

• ## Step2: Calculate the area of the rectangle

The area of the rectangle \(A_1=4\times8 = 32\) \(cm^{2}\).

• ## Step3: Calculate the area of the semi - circle

The area of a full - circle is \(A_{circle}=\pi r^{2}\), so the area of a semi - circle \(A_2=\frac{1}{2}\pi r^{2}\).
Given \(r = 4\) cm and \(\pi = 3.14\), \(A_2=\frac{1}{2}\times3.14\times4^{2}=25.12\) \(cm^{2}\).

• ## Step4: Find the total area

\(A=A_1+A_2=32 + 25.12=57.12\approx57.1\) \(cm^{2}\) (rounded to 1 d.p.).

• # Answer: \(57.1\) \(cm^{2}\)

1. 6. Area of the pink - shaded region:

• # Explanation:
• ## Step1: Calculate the area of the large rectangle

The area of the large rectangle \(A_1=15\times10 = 150\) \(cm^{2}\).

• ## Step2: Calculate the area of the small rectangle

The area of the small rectangle \(A_2=6\times5 = 30\) \(cm^{2}\).

• ## Step3: Find the shaded area

\(A=A_1 - A_2=150 - 30 = 120\) \(cm^{2}\).

• # Answer: \(120\) \(cm^{2}\)

1. 7a. Volume of cuboid with length \(l = 10\) cm, width \(w = 4\) cm and height \(h = 5\) cm:

• # Explanation:
• ## Step1: Recall volume formula

The formula for the volume of a cuboid is \(V=l\times w\times h\).
\(V=10\times4\times5\).

• ## Step2: Perform the calculation

\(10\times4\times5 = 200\) \(cm^{3}\).

• # Answer: \(200\) \(cm^{3}\)
• 7b. Surface area of cuboid:
• # Explanation:
• ## Step1: Recall surface - area formula

The formula for the surface area of a cuboid is \(SA = 2(lw+lh+wh)\).
Here, \(l = 10\) cm, \(w = 4\) cm and \(h = 5\) cm.
\(SA=2(10\times4 + 10\times5+4\times5)\).

• ## Step2: Calculate the values inside the parentheses

\(10\times4 = 40\), \(10\times5 = 50\) and \(4\times5 = 20\).
\(10\times4+10\times5 + 4\times5=40 + 50+20 = 110\).

• ## Step3: Find the surface area

\(SA=2\times110 = 220\) \(cm^{2}\).

• # Answer: \(220\) \(cm^{2}\)

1. 8. Surface area of cuboid:

• First, convert the volume to \(cm^{3}\). Since \(1\) \(cm^{3}=1000\) \(mm^{3}\), \(V = 5760\) \(mm^{3}=5.76\) \(cm^{3}\).
• Let the side of the square end - face be \(x\) cm. The volume of the cuboid \(V=x^{2}\times1.2\).
• # Explanation:
• ## Step1: Find the side of the square end - face

We know that \(x^{2}\times1.2=5.76\).
Then \(x^{2}=\frac{5.76}{1.2}=4.8\), so \(x=\sqrt{4.8}\approx2.19\) cm.

• ## Step2: Recall surface - area formula

The surface area of the cuboid \(SA = 2x^{2}+4xh\).
Substitute \(x\approx2.19\) cm and \(h = 1.2\) cm.
\(SA=2\times4.8+4\times2.19\times1.2\).
\(2\times4.8 = 9.6\), \(4\times2.19\times1.2=10.512\).
\(SA=9.6 + 10.512=20.112\approx20.1\) \(cm^{2}\).

• # Answer: \(20.1\) \(cm^{2}\)

(Note: The net - drawing part (question 9) is a visual task and not included in the step - by - step calculation part. To work out the surface area from the net, you would find the area of each face in the net and sum them up.)