iGCSE Mathematics (0580) - C6.1 Pythagoras’ theorem- Exam Style Questions Paper 1- New Syllabus
▶️ Answer/Explanation
(a) A triangular prism consists of 2 triangular bases and 3 rectangular sides, giving a total of 5 faces. Counting the boundary lines, it has 3 edges on each triangle plus 3 connecting lengths along its body, totaling 9 edges.
(b) The perpendicular height splits the isosceles base $AB$ in half, creating a right-angled triangle with a base of $3\text{ cm}$ ($\frac{6}{2}$) and a height of $4\text{ cm}$. Applying Pythagoras’ theorem: $BC = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}$.
(c) To complete the net from the main $6 \times 6$ base rectangle already given, construct the remaining adjacent shapes accurately: add two triangles with a base of 6 and height of 4, and two side rectangles of dimensions $5 \times 6$.
✅ Answer:
(a) 5 faces and 9 edges
(b) $\sqrt{3^2 + 4^2} = 5$
(c) Correct net drawn completely
(b) The perpendicular height splits the isosceles base $AB$ in half, creating a right-angled triangle with a base of $3\text{ cm}$ ($\frac{6}{2}$) and a height of $4\text{ cm}$. Applying Pythagoras’ theorem: $BC = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}$.
(c) To complete the net from the main $6 \times 6$ base rectangle already given, construct the remaining adjacent shapes accurately: add two triangles with a base of 6 and height of 4, and two side rectangles of dimensions $5 \times 6$.
✅ Answer:
(a) 5 faces and 9 edges
(b) $\sqrt{3^2 + 4^2} = 5$
(c) Correct net drawn completely
▶️ Answer/Explanation
Answers:
(a) \( x = 29.1 \) (or 29.05…)
(b) \( k = 17 \)
Explanation:
(a) Finding \( x \):
- Given a right-angled triangle with:
- Adjacent side to angle \( 47^\circ = 20 \) cm
- Hypotenuse \( = x \) cm
- Use the cosine ratio:
\( \cos(47^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{20}{x} \) - Solve for \( x \):
\( x = \frac{20}{\cos(47^\circ)} \approx \frac{20}{0.682} \approx 29.3 \, \text{cm} \)
(Note: More precise calculation gives \( x \approx 29.05 \))
(b) Finding \( k \):
- Given a right-angled triangle with:
- Opposite side to angle \( 30^\circ = k \) cm
- Adjacent side \( = 30 \) cm
- Use the tangent ratio:
\( \tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{k}{30} \) - Solve for \( k \):
\( k = 30 \times \tan(30^\circ) = 30 \times 0.577 \approx 17.32 \)
(Exact value: \( k = 30 \times \frac{1}{\sqrt{3}} \approx 17.32 \), but likely rounded to \( 17 \))
