Question 13
Which of the following is a planar graph with 5 vertices and 4 faces?
That’s it. Step through the hints below to see why.
Not quite. (B) has four enclosed triangles, but the region outside the graph is a face too. That makes 5 faces. Euler’s formula agrees: $f=2-5+8=5$.
Not quite. Count the dots: (C) has 4 vertices, not 5. It has 3 faces, the two triangles and the outside.
Not quite. (D) has 5 vertices, but it never closes a region, so its only face is the outside. The 4 in (D) is its number of edges.
Pick a graph and count it piece by piece. Then switch the outside face off to see the count that is easy to get wrong.
The question asks for two things at once, 5 vertices and 4 faces. Do the easy count first.
(A), (B) and (D) each have 5 dots. (C) has 4, so (C) is out before you count a single face.
A face is a region bounded by edges, and the unbounded region outside the graph always counts as one face.
- (A): three triangles plus the outside, 4 faces. The top of (A) is open, so the wedge above the centre is part of the outside.
- (B): four triangles plus the outside, 5 faces.
- (D): no closed regions, so the outside is its only face.
Check with Euler’s formula from the formula book, $v+f-e=2$. Rearranged, it gives the faces from counts you can’t get wrong:
$$f=2-v+e$$
$$\begin{aligned}\text{(A): }f&=2-5+7=4\\\text{(B): }f&=2-5+8=5\\\text{(D): }f&=2-5+4=1\end{aligned}$$
Only (A) has 5 vertices and 4 faces.
(C) has only 4 vertices. The other three each have 5 vertices, and (A), (B) and (D) have 7, 8 and 4 edges. Euler’s formula $v+f-e=2$ gives $f=2-v+e$:
$$\begin{aligned}\text{(A): }f&=2-5+7=4\\\text{(B): }f&=2-5+8=5\\\text{(D): }f&=2-5+4=1\end{aligned}$$
(A) has three triangular faces and the outside face, 4 in all. Each graph is drawn with no edges crossing, so all of them are planar.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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