Pythagoras' Theorem practice questions

Secondary 2 topics, working towards the O-Level Mathematics 4052 syllabus. Every question below has a full worked solution.

When a student gets a question wrong in Math Amigo, it does not show the answer straight away. It reads the working, identifies the step that went wrong, and asks a question that points at the correction. The worked solution is there once a genuine attempt has been made.

Practise pythagoras' theorem free

What this topic covers

Worked examples

Example 1

Two vertical poles are 55 m and 1414 m tall and stand 1212 m apart on level ground. A cable connects the tops of the two poles. Find the length of the cable.
Show the worked solution
  • $\text{Height difference} = 14 - 5 = 9$ m
  • $\text{The cable, the height difference, and the ground distance form a right triangle.}$
  • $\text{Cable}^2 = \text{height difference}^2 + \text{ground distance}^2 \quad \scriptsize\textit{Pythagoras' theorem}$
  • $\text{Cable}^2 = 9^2 + 12^2$
  • $\text{Cable}^2 = 81 + 144$
  • $\text{Cable}^2 = 225$
  • $\text{Cable} = \sqrt{225}$
  • $\text{Cable} = 15$ m

Answer: 1515

Example 2

A rectangular monitor is 3030 cm wide and 1616 cm tall. An ant walks from one corner along the edges to the diagonally opposite corner. A second ant walks directly along the diagonal. How much further does the first ant walk?
Show the worked solution
  • $\text{Edge path} = 30 + 16 = 46$ cm
  • $\text{Diagonal}^2 = 30^2 + 16^2 \quad \scriptsize\textit{Pythagoras' theorem}$
  • $= 900 + 256$
  • $= 1156$
  • $\text{Diagonal} = 34$ cm
  • $\text{Difference} = 46 - 34$
  • $= 12$ cm

Answer: 1212

Example 3

A triangular field PQRPQR has PQ=40PQ = 40 m, QR=42QR = 42 m and PR=58PR = 58 m. Show that ∠PQR=90°PQR = 90° and find the area of the field.
Show the worked solution
  • $\text{The longest side is } PR = 58$ m
  • $PR^2 = 58^2 = 3364$
  • $PQ^2 + QR^2 = 40^2 + 42^2$
  • $= 1600 + 1764$
  • $= 3364$
  • $\text{Since } PR^2 = PQ^2 + QR^2, \text{ the right angle is at Q.} \quad \scriptsize\textit{converse of Pythagoras' theorem}$
  • $\text{Area} = \frac{1}{2} \times PQ \times QR$
  • $= \frac{1}{2} \times 40 \times 42$

Answer: 840840

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Which syllabus does this follow?

O-Level Mathematics (4052), for Singapore secondary students. Math Amigo covers Secondary 1 to 4, both O-Level Mathematics and Additional Mathematics.

Will it just show my child the answer?

No. On a wrong answer it looks at the working, points to the step that went wrong, and asks a question that helps the student find the correction. The full worked solution is available after a genuine attempt.

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