Linked Pythagoras problems with worked solutions | Year 10

Name the quantity you need at each step. A shared side links two calculations, but each use of Pythagoras still needs its own right triangle.

Use an intermediate side in a second right triangle.
Calculate a change in height from two positions.
Combine a horizontal diagonal and vertical height to find a space diagonal.

The idea

Pythagoras’ theorem applies separately to each right triangle: the square of the hypotenuse equals the sum of the squares of its perpendicular legs. In a linked problem, a length found in one triangle becomes an input to the next. Decide whether the question asks for a final length, a sum or a difference before writing the answer.

Before you begin: Identify a hypotenuse, calculate squares and square roots, and solve a one-triangle Pythagoras problem.

Worked examples

01 Start with the method

A 39 m ladder has its foot 15 m from a vertical wall. It slides until its foot is 36 m from the wall. How far does its top move down?

h_{initial}^{2} + 15^{2} = 39^{2},

h_{final}^{2} + 36^{2} = 39^{2}

Step 1

The ladder is the hypotenuse in both positions. First calculate the original vertical height.

h_{initial} = \sqrt{39^{2} - 15^{2}} = \sqrt{1296} = 36 m

Step 2

Use the new horizontal distance to calculate the final vertical height.

h_{final} = \sqrt{39^{2} - 36^{2}} = \sqrt{225} = 15 m

Step 3

The question asks for the drop, so subtract the final height from the initial height.

drop = 36 - 15 = 21 m

Step 4

The ladder length stays fixed while its foot moves farther out. A lower final height is consistent with that movement.

15^{2} + 36^{2} = 39^{2}

A mistake to avoid

The final height is 15 m, but the requested drop is 21 m. Label each intermediate quantity so a correct calculation does not answer the wrong question.

02Carry a shared height into another triangle

Two right triangles stand on opposite sides of the same vertical height h. The left base is 5 cm and its hypotenuse is 13 cm. The right base is 9 cm. Find h, the right hypotenuse s and the combined base.

h^{2} + 5^{2} = 13^{2},

h^{2} + 9^{2} = s^{2}

Step 1

Use the left triangle to find the shared height.

h = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12 cm

Step 2

The right triangle has perpendicular legs 12 cm and 9 cm. Its hypotenuse is a new length.

s = \sqrt{12^{2} + 9^{2}} = \sqrt{225} = 15 cm

Step 3

Because the bases lie on opposite sides of the shared height, add their lengths.

combined base = 5 + 9 = 14 cm

Result

h = 12 cm

s = 15 cm

base = 14 cm

03Find a rectangular room’s space diagonal

A rectangular room has a 3 m by 4 m floor and a vertical height of 12 m. Find the straight-line distance from one floor corner to the opposite ceiling corner.

d_{floor}^{2} = 3^{2} + 4^{2}

Step 1

First find the diagonal across the horizontal floor.

d_{floor} = \sqrt{3^{2} + 4^{2}} = 5 m

Step 2

That horizontal diagonal is perpendicular to the 12 m vertical edge, forming a second right triangle.

D^{2} = 5^{2} + 12^{2} = 169

Step 3

Take the positive square root because D is a distance. It exceeds each individual edge length.

D = \sqrt{169} = 13 m

Result

13 m

Practice questions with answers

Work these out first, then open each solution to compare your method. These questions are untimed.

Question 1

A 25 m ladder moves from a foot distance of 7 m to 15 m. Find the top’s drop.

h_{1} = \sqrt{25^{2} - 7^{2}},

h_{2} = \sqrt{25^{2} - 15^{2}}

Show solution 1

h_{1} = 24

h_{2} = 20

drop = 4 m

The two right triangles give heights 24 m and 20 m. The drop is their difference, not either height.

Question 2

Two right triangles share a height of 8 cm and stand on opposite sides of it. Their bases are 6 cm and 15 cm. Find both hypotenuses and the combined base.

a^{2} = 8^{2} + 6^{2},

b^{2} = 8^{2} + 15^{2}

Show solution 2

a = 10 cm

b = 17 cm

base = 21 cm

Apply Pythagoras separately to the 6–8 and 8–15 triangles. Add 6 + 15 for the combined base.

Question 3

A rectangular room has edges 6 m, 8 m and 24 m. Find its space diagonal.

D^{2} = 6^{2} + 8^{2} + 24^{2}

Show solution 3

D = \sqrt{676} = 26 m

The floor diagonal is 10 m. The second right triangle has legs 10 m and 24 m, giving hypotenuse 26 m.

Question 4

A symmetric roof cross-section has horizontal half-span 5 m and vertical rise 12 m. Find the combined length of its two equal sloping sides.

s^{2} = 5^{2} + 12^{2}

Show solution 4

s = 13 m,

2 s = 26 m

Each half is a 5–12–13 right triangle. The question asks for both slopes together, so double 13.

Question 5

Two right triangles stand on opposite sides of a shared 15 cm height. Their hypotenuses are 17 cm and 25 cm. Find the combined base.

a^{2} = 17^{2} - 15^{2},

b^{2} = 25^{2} - 15^{2}

Show solution 5

a = 8 cm

b = 20 cm

a + b = 28 cm

The base lengths are √64 and √400. Add them because they extend in opposite directions from the shared height.

Common questions

When should I add or subtract squares?

Add the two leg squares when finding the hypotenuse. When a leg is unknown, subtract the known leg’s square from the hypotenuse’s square, then take the positive root for a length.

Can I apply Pythagoras to any triangle?

No. It requires a right angle. Identify the two perpendicular legs and the opposite hypotenuse each time, including in a second triangle formed inside a larger shape.

Why keep intermediate answers exact?

Rounding an intermediate side can alter the final answer. Keep a square root or full calculator value until the final step, then round only if requested.