Preliminary Preparation (HSC SSCE Mathematics Standard): Revision Notes

Worked Example: Finding the length of the hypotenuse

Find the length of the hypotenuse, correct to two decimal places.

Solution:

Write Pythagoras' theorem:

$h^2 = a^2 + b^2$

Substitute the length of the sides:

$h^2 = 5^2 + 9^2$

Calculate the value of $h^2$:

$h^2 = 25 + 81$

$h^2 = 106$

Take the square root to find $h$:

$h = \sqrt{106}$

$h \approx 10.30 \text{ cm}$

Express answer correct to two decimal places:

The length of the hypotenuse is 10.30 cm.

Worked Example: Finding the length of a shorter side

What is the value of $x$ correct to one decimal place?

Solution:

Write Pythagoras' theorem:

$h^2 = a^2 + b^2$

Substitute the length of the sides:

$12^2 = x^2 + 5^2$

Make $x^2$ the subject:

$x^2 = 12^2 - 5^2$

$x^2 = 144 - 25$

$x^2 = 119$

Take the square root to find $x$:

$x = \sqrt{119}$

$x \approx 10.9 \text{ mm}$

Express answer correct to one decimal place:

The value of $x$ is 10.9 mm.

Worked Example: Finding the perimeter of a rectangle

Find the perimeter of the following rectangle.

Solution:

The shape is a rectangle, so use the formula:

$P = 2(l + b)$

Substitute the values for $l$ and $b$ where $l = 8$ and $b = 3$:

$P = 2(8 + 3)$

Evaluate:

$P = 2 \times 11$

$P = 22 \text{ m}$

Write the answer in words:

The perimeter of the rectangle is 22 m.

Worked Example: Finding the perimeter of a triangle

Find the perimeter of the triangle. Answer correct to one decimal place.

Solution:

Find the length of the hypotenuse or $h$:

Write Pythagoras' theorem:

$h^2 = a^2 + b^2$

Substitute the length of the sides:

$h^2 = 4.2^2 + 3.7^2$

Evaluate the value of $h$:

$h = \sqrt{4.2^2 + 3.7^2} \approx 5.6 \text{ cm}$

Add the lengths of sides to find the perimeter:

$P = 3.7 + 4.2 + 5.6$

$P = 13.5 \text{ cm}$

Express answer correct to one decimal place:

The perimeter of the triangle is 13.5 cm.

Worked Example: Finding the circumference of a circle

Find the perimeter of a circle with a radius of 9 mm. Answer correct to two decimal places.

Solution:

The shape is a circle, so use the formula:

$C = 2\pi r$

Substitute the value for $r$ where $r = 9$:

$C = 2 \times \pi \times 9$

Evaluate:

$C \approx 56.55 \text{ mm}$

Write the answer in words:

The perimeter of the circle is 56.55 mm.

Worked Example: Finding the perimeter of a semicircle

Find the perimeter of a semicircle with a diameter of 4 m. Answer correct to two decimal places.

Solution:

The shape is a semicircle, so use the formula:

$C = \frac{\pi d}{2}$

Substitute the value for $d$ where $d = 4$ to find the curved distance:

$C = \frac{\pi \times 4}{2}$

Evaluate:

$C \approx 6.28 \text{ m}$

Add the curved distance to the diameter:

$P = 6.28 + 4$

Evaluate:

$P = 10.28 \text{ m}$

Write the answer in words:

The perimeter is 10.28 m.

Worked Example: Finding the perimeter of composite shapes

Find the perimeter of each of these shapes.

Solution:

Part a (L-shaped figure):

Find the unknown side lengths using the measurements given:

The missing horizontal side: $8 - 2 = 6$ cm

The missing vertical side: $6 + 5 = 11$ cm

Add the lengths of all the edges to find the perimeter:

$P = 2 + 6 + 6 + 5 + 8 + 11$

Evaluate:

$P = 38 \text{ cm}$

Write the answer in words:

The perimeter is 38 cm.

Part b (Rectangle with semicircle):

Use the formula for the curved length:

$C = \frac{2\pi r}{4}$

Substitute the value for $r$ where $r = 5$:

$C = \frac{2 \times \pi \times 5}{4}$

Evaluate:

$C \approx 7.85 \text{ m}$

Add the curved length to other edges:

$P = 7.85 + 5 + 5 + 5 + 5$

Evaluate:

$P = 27.85 \text{ m}$

Write the answer in words:

The perimeter is 27.85 m.

Worked Example: Finding the area of a triangle

Find the area of the triangle.

Solution:

The shape is a triangle, so use the formula:

$A = \frac{1}{2}bh$

Substitute the values for $b$ and $h$ where $b = 8.1$ and $h = 5.5$:

$A = \frac{1}{2} \times 8.1 \times 5.5$

Evaluate:

$A = 22.275 \text{ m}^2$

Write the answer using the correct units:

The area of the triangle is 22.275 m².

Worked Example: Finding the area of a trapezium

Find the area of the quadrilateral.

Solution:

The shape is a trapezium, so use the formula:

$A = \frac{1}{2}(a + b)h$

Substitute the values for $a$, $b$, and $h$:

$A = \frac{1}{2}(2.9 + 5.1) \times 3$

Evaluate:

$A = \frac{1}{2} \times 8 \times 3$

$A = 12 \text{ cm}^2$

Write the answer using the correct units:

The area of the shape is 12 cm².

Worked Example: Finding the area of a parallelogram

Find the area of the following quadrilateral.

Solution:

The shape is a parallelogram, so use the formula:

$A = bh$

Substitute the values for $b$ and $h$ where $b = 6.5$ and $h = 4$:

$A = 6.5 \times 4$

Evaluate:

$A = 26 \text{ mm}^2$

Write the answer using the correct units:

The area of the shape is 26 mm².

Worked Example: Finding the area of a circle

Find the area of a circle with a radius of 5 metres. Give your answer correct to one decimal place.

Solution:

The shape is a circle, so use the formula:

$A = \pi r^2$

Substitute the value for $r$ where $r = 5$:

$A = \pi \times 5^2$

Evaluate correct to one decimal place:

$A = \pi \times 25$

$A \approx 78.5 \text{ m}^2$

Write the answer using the correct units:

The area of the circle is 78.5 m².

Worked Example: Finding the area of a composite shape

Find the area of the composite shape. Answer correct to one decimal place.

Solution:

Divide the shape into a rectangle and a semicircle.

Use the formula for the rectangle:

$A = lb$

Substitute where $l = 12$ and $b = 10$:

$A = 12 \times 10$

Evaluate:

$A = 120 \text{ cm}^2$

Use the formula for the semicircle:

$A = \frac{1}{2}\pi r^2$

Substitute where $r = 5$:

$A = \frac{1}{2} \times \pi \times 5^2$

Evaluate:

$A = \frac{1}{2} \times \pi \times 25$

$A \approx 39.3 \text{ cm}^2$

Add the area of the rectangle to the semicircle:

$A = 120 + 39.3$

Evaluate:

$A = 159.3 \text{ cm}^2$

Write the answer using the correct units:

The area of the shape is 159.3 cm².