Linked triangles | Year 10

01 / LEARN

Understand the idea.

Use the shared side as a bridge

A shared side connects the information in linked triangles. Solve the triangle containing enough known measurements first. Label the shared length once so that the second triangle receives exactly the same value.

Each triangle still has its own hypotenuse, opposite its own right angle. A length that is a leg in one triangle need not have the same role in another; identify the geometry before writing each equation.

A new height is not a drop

Distinguish a position from a change in position. A ladder's new height measures where its top finishes; its drop is the difference between the initial and final heights. The ladder length remains the hypotenuse in both positions.

For adjacent triangles, a full base may require adding the two horizontal segments after finding the shared height. Write what each intermediate result measures and delay rounding until the final sum or difference.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Shared height

Two right triangles share a vertical height. The left has base 5 cm and hypotenuse 13 cm. The right base is 9 cm. Find the right hypotenuse and full base.

Worked solution

1
Find the shared height from the left triangle.
$h^{2} = 13^{2} - 5^{2} = 144$
The known hypotenuse square equals the sum of the leg squares.
2
Use the height in the right triangle.
$h = 12 cm, c^{2} = 12^{2} + 9^{2} = 225$
Both triangles have the same vertical height.
3
State the right hypotenuse.
$c = \sqrt{225} = 15 cm$
The positive root is the required side length.
4
Combine the base segments.
$B = 5 + 9 = 14 cm$
The left and right base segments lie end to end.

Answer

15 cm; base 14 cm

EXAMPLE 02

A change in height

A 13 m ladder moves from base distance 5 m to base distance 12 m. How far does its top drop?

Worked solution

1
Find the initial height.
$h_{1} = \sqrt{13^{2} - 5^{2}} = 12 m$
The fixed ladder is the hypotenuse at the first position.
2
Find the final height.
$h_{2} = \sqrt{13^{2} - 12^{2}} = 5 m$
The ladder length is unchanged while the horizontal distance increases.
3
Calculate the drop.
$h_{1} - h_{2} = 12 - 5 = 7 m$
A drop is the initial height minus the final height.

Answer

7 m

03 / YOUR TURN

Try it yourself.

Check your understanding

A right triangle has height 8 m and base 15 m. Find its hypotenuse.

Show answer & reasoning

17 m

The height and base are perpendicular legs, so $c = \sqrt{8^{2} + 15^{2}} = \sqrt{289} = 17 m$ .

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6 worksheet questions · 17 marks

Open worksheet Worked solutions

Read the questions online

Two right triangles share a height. One has base 8 cm and hypotenuse 17 cm; the other base is 20 cm. Find its hypotenuse.
3 marks
A 10 m ladder moves from base distance 6 m to 8 m. Find its top drop.
3 marks
A 12 m mast has ground anchors 5 m and 9 m from its base on opposite sides. Find both cable lengths and anchor separation.
3 marks
A symmetrical triangular roof has span 20 m and rise 9 m. Find either slope to 0.1 m.
3 marks
Points on level ground are 20 m and 40 m from a 12 m tower. Find both elevation angles to ${0.1}^{\circ}$ .
3 marks
Two adjacent right triangles have bases 3 m and 4 m and share a height. Why cannot 7 m be used as the base of each triangle?
2 marks

Read the worked solutions

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

Two right triangles share a height. One has base 8 cm and hypotenuse 17 cm; the other base is 20 cm. Find its hypotenuse.
3 marks
Height $= \sqrt{17^{2} - 8^{2}} = 15$ cm.
Other hypotenuse $= \sqrt{15^{2} + 20^{2}} = 25$ cm.
25 cm
A 10 m ladder moves from base distance 6 m to 8 m. Find its top drop.
3 marks
Heights are $\sqrt{100 - 36} = 8$ m and $\sqrt{100 - 64} = 6$ m.
Drop $= 8 - 6 = 2$ m.
2 m
A 12 m mast has ground anchors 5 m and 9 m from its base on opposite sides. Find both cable lengths and anchor separation.
3 marks
Lengths are $\sqrt{12^{2} + 5^{2}} = 13$ m and $\sqrt{12^{2} + 9^{2}} = 15$ m.
Anchor separation $= 5 + 9 = 14$ m.
13 m and 15 m; 14 m apart
A symmetrical triangular roof has span 20 m and rise 9 m. Find either slope to 0.1 m.
3 marks
Half-span is 10 m.
Slope $= \sqrt{10^{2} + 9^{2}} = \sqrt{181}$ m.
13.5 m
Points on level ground are 20 m and 40 m from a 12 m tower. Find both elevation angles to ${0.1}^{\circ}$ .
3 marks
Angles are ${tan}^{- 1} (\frac{12}{20})$ and ${tan}^{- 1} (\frac{12}{40})$ .
The nearer point has the larger angle.
${31.0}^{\circ}$ and ${16.7}^{\circ}$
Two adjacent right triangles have bases 3 m and 4 m and share a height. Why cannot 7 m be used as the base of each triangle?
2 marks
Each calculation uses sides of one right triangle.
The bases remain 3 m and 4 m respectively.
7 m is the full span, not the base of either individual triangle.

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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.

Use the shared side as a bridge

A new height is not a drop

See the method in action.

Shared height

Find the shared height from the left triangle.

Use the height in the right triangle.

State the right hypotenuse.

Combine the base segments.

A change in height

Find the initial height.

Find the final height.

Calculate the drop.

Try it yourself.

Check your understanding

Keep practising