What this quiz covers
Grade 8 — Triangle Geometry & Pythagoras Quiz 1 draws its questions at random from a bank of 7 questions, with full worked solutions for every one.
Worked examples
A few of the question types, with full solutions, so you know what to expect.
In the diagram, \(CA\) is produced (extended) beyond \(A\). The angle of \(130^\circ\) and angle \(x\) lie on this straight line at \(A\). Calculate the value of \(x\).
At \(A\), the \(130°\) and \(x\) lie next to each other on the straight line \(CA\) produced.
Angles on a straight line add up to \(180°\):
\(x = 180° - 130°\)
Which reason correctly justifies the size of \(x\) in the diagram?
- Angles on a straight line (sum to \(180°\))
- Angles in a triangle (sum to \(180°\))
- Vertically opposite angles are equal
- Corresponding angles on parallel lines
Answer: Angles on a straight line (sum to \(180°\))
The \(130°\) and \(x\) sit side by side on one straight line (\(CA\) produced).
Two angles on a straight line are supplementary — they add up to \(180°\).
*That is exactly why we wrote \(x = 180° - 130°\).
In \(\triangle ABC\): the angle at \(A\) is \(x = 50°\), the angle at \(C\) is \(y°\), and the angle at \(B\) is \((y - 20)°\). Calculate the value of \(y\).
The three interior angles of \(\triangle ABC\) are \(50°\), \(y°\) and \((y-20)°\).
Angles in a triangle add up to \(180°\):
\(50° + y + (y - 20°) = 180°\)
\(2y + 30° = 180°\)
\(2y = 150°\)
In \(\triangle ABC\), \(y = 75°\) and the angle at \(B\) is \((y - 20)°\). Calculate the size of \(\hat{B}\).
\(\hat{B} = y - 20°\)
\(\hat{B} = 75° - 20°\)
*Check: \(50° + 75° + 55° = 180°\) ✓
In the diagram, \(PQ = QR\) and \(PS \perp QR\) (so \(\hat{S} = 90°\)). \(PQ = 15\) and \(PS = 9\). Calculate the length of \(QS\).
\(\triangle PQS\) is right-angled at \(S\), and \(PQ\) is the hypotenuse.
Theorem of Pythagoras: \(PQ^2 = QS^2 + PS^2\)
\(15^2 = QS^2 + 9^2\)
\(225 = QS^2 + 81\)
\(QS^2 = 225 - 81 = 144\)