Multi-step right-triangle problems | Year 10

01 / LEARN

Understand the idea.

Choose one triangle at a time

Pythagoras applies to one right triangle at a time. Identify the right angle first; the side opposite it is the hypotenuse. A rectangular box needs a triangle across its base and another through its height.

Name the length shared by the two triangles. Calculate it from the triangle with enough information, then use that same length in the second triangle. Its role can change from hypotenuse to leg.

c^{2} = a^{2} + b^{2}

Finish the whole problem

A calculated side may be only an intermediate result. Re-read the request: a diagonal is a straight line, whereas a perimeter follows edges and an attachment allowance is extra material.

Keep exact roots or stored calculator values until the last calculation. Every hypotenuse should exceed either leg, and a length answer needs an ordinary length unit rather than a squared unit.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two linked triangles

A box is 12 cm long, 9 cm wide and 20 cm high. Find its space diagonal.

Worked solution

1
Find the diagonal across the base.
$d^{2} = 12^{2} + 9^{2} = 225$
The length and width are perpendicular legs of the base triangle.
2
Take its positive square root.
$d = \sqrt{225} = 15 cm$
A physical length is positive.
3
Combine the base diagonal with the height.
$D^{2} = 15^{2} + 20^{2} = 625$
The vertical height is perpendicular to the horizontal base diagonal.
4
Find the space diagonal.
$D = \sqrt{625} = 25 cm$
This hypotenuse runs between opposite corners of the box.

Answer

25 cm

EXAMPLE 02

Add the allowance

A cable joins the top of an 8 m pole to a ground point 6 m from its base. Allow 1 m extra for attaching it. How much cable is needed?

Worked solution

1
Model the taut cable as a hypotenuse.
$c^{2} = 8^{2} + 6^{2} = 100$
The upright pole and level ground form perpendicular legs.
2
Find the taut length.
$c = \sqrt{100} = 10 m$
The positive root measures the straight span.
3
Add the separate attachment allowance.
$L = 10 + 1 = 11 m$
The extra metre is cable beyond the triangular span.

Answer

11 m

03 / YOUR TURN

Try it yourself.

Check your understanding

A rectangle is 5 m by 12 m. Find its diagonal.

Show answer & reasoning

13 m

The diagonal is the hypotenuse: $d = \sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13 m$ . It is longer than either side.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Find the diagonal of a 9 cm by 12 cm rectangle.
2 marks
A cuboid has perpendicular edges 3 m, 4 m and 12 m. Find its space diagonal.
3 marks
A 13 m ladder has its foot 5 m from a wall. How high does it reach?
2 marks
A rectangular garden has diagonal 26 m and width 10 m. Find its perimeter.
3 marks
A symmetrical roof has span 16 m and rise 6 m. Find the total length of its two sloping edges.
3 marks
A student claims a right triangle with hypotenuse 17 cm and leg 8 cm has other leg 19 cm. Correct the result and explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the diagonal of a 9 cm by 12 cm rectangle.
2 marks
$d^{2} = 9^{2} + 12^{2} = 225$ .
$d = 15$ cm.
15 cm
A cuboid has perpendicular edges 3 m, 4 m and 12 m. Find its space diagonal.
3 marks
The base diagonal is $\sqrt{3^{2} + 4^{2}} = 5$ m.
The space diagonal is $\sqrt{5^{2} + 12^{2}} = 13$ m.
13 m
A 13 m ladder has its foot 5 m from a wall. How high does it reach?
2 marks
$h^{2} = 13^{2} - 5^{2} = 144$ .
$h = 12$ m.
12 m
A rectangular garden has diagonal 26 m and width 10 m. Find its perimeter.
3 marks
The other side is $\sqrt{26^{2} - 10^{2}} = 24$ m.
The perimeter is $2 (24 + 10) = 68$ m.
68 m
A symmetrical roof has span 16 m and rise 6 m. Find the total length of its two sloping edges.
3 marks
Each half has run 8 m.
Each slope is $\sqrt{8^{2} + 6^{2}} = 10$ m.
There are two slopes.
20 m
A student claims a right triangle with hypotenuse 17 cm and leg 8 cm has other leg 19 cm. Correct the result and explain.
3 marks
$x^{2} = 17^{2} - 8^{2} = 225$ .
$x = 15$ cm.
Subtract the known leg square from the hypotenuse square.
15 cm; a leg cannot exceed the hypotenuse.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

In this lesson

Understand the idea.

Choose one triangle at a time

Finish the whole problem

See the method in action.

Two linked triangles

Find the diagonal across the base.

Take its positive square root.

Combine the base diagonal with the height.

Find the space diagonal.

Add the allowance

Model the taut cable as a hypotenuse.

Find the taut length.

Add the separate attachment allowance.

Try it yourself.

Check your understanding

Keep practising