What is the perimeter of the figure shown below ? AJ = 10 cm, JI = 12 cm, AB = x, CD = x + 1, EF = x + 2, GH = x + 3, BC = DE = FG = HI = y

Correct — A, (a) 44 cm.

The figure printed with this question is a closed outline shaped like a descending staircase, with ten labelled corners and every angle a right angle. Reading round it: A is the top-left corner; AB runs right along the top to B; BC drops to C; CD runs right to D; DE drops to E; EF runs right to F; FG drops to G; GH runs right to H; HI drops to I, the bottom-right corner; IJ runs left along the bottom to J, the bottom-left corner; and JA closes the figure up the left side. The two overall dimensions are given in the stem: AJ, the left side, is 10 cm, and JI, the bottom, is 12 cm.

Now group the sides by direction, which is the whole trick.

The four steps that run rightward — AB, CD, EF and GH — between them carry the outline from the left edge to the right edge. Their lengths therefore add up to the width of the figure, which is the bottom side: AB + CD + EF + GH = JI = 12 cm.

The four steps that drop downward — BC, DE, FG and HI — between them carry the outline from the top edge to the bottom edge. Their lengths add up to the height, which is the left side: BC + DE + FG + HI = AJ = 10 cm.

The perimeter is those eight sides plus the bottom and the left side:

P = 12 + 10 + 12 + 10 = 44 cm.

In other words a rectilinear staircase has exactly the same perimeter as the rectangle that encloses it, 2 × (10 + 12) = 44 cm, and that is the point of the question. The individual step sizes never matter.

The algebra confirms it if you would rather see the numbers. The horizontal runs are x, x + 1, x + 2 and x + 3, so 4x + 6 = 12 and x = 1.5; the vertical drops are all y, so 4y = 10 and y = 2.5. Adding all ten sides: (1.5 + 2.5 + 3.5 + 4.5) + 4(2.5) + 12 + 10 = 12 + 10 + 12 + 10 = 44 cm. The keyed option is 44 cm.

Why the others are wrong

• (b)48 cm — 48 cm is the perimeter of a square of side 12, and it is the figure a candidate reaches by using the dimension 12 for both the width and the height — reading the number written under the bottom of the drawing and losing the 10 written beside the left side. It is worth being explicit about why no such answer can be right: the projections argument fixes the perimeter at twice the sum of the two given dimensions, so with 10 and 12 the answer must be 44 and nothing else. There is also a rough check available in a few seconds. The staircase is contained inside a rectangle 12 wide and 10 high, and its perimeter cannot exceed that rectangle’s; any option larger than 44 is therefore impossible before a single step length has been considered.
• (c)54 cm — 54 cm is 44 plus another 10, which is the total of the four vertical drops or the length of the left side — so it is the number produced by counting the height twice. That kind of slip is easy to make when a candidate tries to add the ten sides one at a time instead of grouping them: with four drops of y, a left side of 10 and the equation 4y = 10 in front of him, it is natural to write the 10 down once for the sum of the drops and again for the side itself. Grouping by direction avoids it, because each direction is accounted for exactly twice — once for the steps and once for the opposite side. The answer to a rectilinear perimeter problem of this kind is always twice the sum of the two overall dimensions.
• (d)58 cm — 58 cm is 14 more than the correct perimeter and corresponds to no consistent reading of the figure. It is the largest option on the list and belongs to the same family as the other two wrong answers, all of which are bigger than 44. That is not an accident: the item is built on the fact that a staircase looks as though it should have a longer boundary than the plain rectangle around it, since it has ten sides instead of four. The intuition is wrong. Every step outward is matched by a step inward, and the total distance travelled left-to-right and top-to-bottom is unchanged, so the two figures have identical perimeters. Once that is seen, all three larger options can be rejected together without any arithmetic at all.

The idea being tested is the projection of a rectilinear outline onto its two axes. In any closed figure made only of horizontal and vertical segments, the horizontal segments running one way must, in total, cover the same distance as the horizontal segments running the other way, and the same is true of the vertical segments. So the perimeter of a rectilinear polygon that has no re-entrant overhangs is exactly the perimeter of its bounding rectangle: twice the width plus twice the height.

That single fact answers a whole class of school and recruitment problems — the staircase, the L-shape, the T-shape, the plus-shape, the room with a rectangular alcove — provided the outline never doubles back on itself. It is worth testing the limit: the rule holds when every horizontal side faces up or down consistently with the enclosing rectangle, and it fails for a figure with a notch cut into it in a way that adds sides running against the direction of travel, where the extra sides must be added separately.

The second lesson is about what a question gives you and what it needs. This stem supplies x, y and four expressions in x, none of which is required; a candidate who begins by solving for x and y will get the right answer, but slowly, and with several chances to slip. The habit worth building on a quantitative paper is to ask what the question is really about before starting to compute, because the arithmetic is usually the least interesting part of it.

The area, incidentally, does depend on the step sizes — which is why the same figure with an area question would be a much harder problem.

Key facts

• In the printed figure A is the top-left corner and the outline runs A-B-C-D-E-F-G-H-I-J and back to A, all corners being right angles; I is the bottom-right corner and J the bottom-left.
• The stem gives AJ = 10 cm for the left side and JI = 12 cm for the bottom, with AB = x, CD = x + 1, EF = x + 2, GH = x + 3 and BC = DE = FG = HI = y.
• The four rightward steps together span the width, so AB + CD + EF + GH = 12 cm; the four downward steps together span the height, so BC + DE + FG + HI = 10 cm.
• The perimeter is therefore 12 + 10 + 12 + 10 = 44 cm, which is exactly the perimeter of the enclosing rectangle, 2 × (10 + 12).
• Solving for the variables gives x = 1.5 cm and y = 2.5 cm, and substituting them returns the same 44 cm — the individual step sizes cancel out.
• The rule holds for any rectilinear outline without re-entrant sides: its perimeter equals that of its bounding rectangle, although its area does not.

Common traps

• Assuming a staircase must have a longer boundary than the rectangle around it. All three wrong options are larger than 44 cm.
• Solving for x and y first. The values are never needed, and each extra step is a chance to slip.
• Counting one of the two given dimensions twice while adding the sides one by one, which turns 44 into 54.
• Using the single dimension written under the bottom of the drawing for both directions, which gives the perimeter of a square instead.

• practice — not a real PYQ

  A room is shaped like the letter L. Its overall width is 9 m and its overall depth is 7 m, and all its corners are right angles. What is the perimeter of the room ?

  • (a)23 m
  • (b)28 m
  • (c)32 m
  • (d)It cannot be found without the dimensions of the notch

  Answer(c) 32 m — an L-shape is a rectilinear outline with no re-entrant sides, so its perimeter equals that of its bounding rectangle, 2 × (9 + 7) = 32 m. The dimensions of the notch affect the area but not the perimeter, which is why option (d) is wrong; options (a) and (b) are smaller than the enclosing rectangle’s perimeter, which is impossible.

• practice — not a real PYQ

  Four horizontal steps of a staircase-shaped figure measure x, x + 2, x + 4 and x + 6, and together they span a width of 20 cm. What is the value of x ?

  • (a)2 cm
  • (b)2.5 cm
  • (c)3 cm
  • (d)5 cm

  Answer(a) 2 cm — the four lengths sum to 4x + 12, and the steps together span the width, so 4x + 12 = 20, giving 4x = 8 and x = 2 cm. Option (b) comes from dividing 20 by 8, option (c) from ignoring one of the increments, and option (d) from dividing the width by four and forgetting the increments altogether.