Math Detours · Grade 7 · Around the corner · 9

Slide the top, keep the area

Push a stack of cards sideways and its area does not change. That is why every triangle is half a rectangle.

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Tonight's idea

Look at a stack of playing cards side on: a rectangle. Push the top sideways so it leans. No card got longer or thicker. Each one only slid. So the side of the stack covers the same area as before.

a stack of cardspushed sideways → slide the top corner →baseheight
Each card slides; none changes size. The two triangles share a base and a height, so they have the same area.

This move is a shear. With thinner and thinner cards the stepped edge becomes a straight slope. The shape is a parallelogram, with the same base and height as the rectangle.

Area of a parallelogram = base × height. The height is measured straight up, not along the slanted side.

Triangles shear too. Keep the base still and slide the top corner along a line parallel to the base. Each thin strip of the triangle slides and keeps its length, so the area stays the same. Stop when the corner is straight above one end of the base. Now it is a right-angled triangle: exactly half of a rectangle.

Area of any triangle = \(\tfrac12\) × base × height.

Watch one: a triangle that leans right over

A triangle has corners \((0, 0)\), \((6, 0)\) and \((9, 4)\). The top corner is not above the base. Base 6, height 4. Slide the top corner along its row to \((0, 4)\). The triangle is now half of a 6 by 4 rectangle, so its area is 12.

Check: the right-angled triangle \((0,0)\), \((9,0)\), \((9,4)\) has area 18. Cut off the triangle \((6,0)\), \((9,0)\), \((9,4)\), area 6. That leaves 12.

Make a triangle on the dots, then remake it with the top corner moved along its row: mathdetours.com/demos/pick.html.

Your turn

  1. A parallelogram has base 7 cm, height 4 cm and slanted sides of 5 cm. (a) Find its area. (b) Sam says \(7 \times 5 = 35\) cm². What shape has that area, and why is it bigger?
  2. A triangle has its base from \((0, 0)\) to \((5, 0)\). Find its area when the top corner is at (a) \((0, 6)\); (b) \((3, 6)\); (c) \((40, 6)\); (d) \((3, 3)\).
  3. A parallelogram has corners \((0, 0)\), \((5, 0)\), \((8, 4)\) and \((3, 4)\). (a) Find its area. (b) Shear it: the bottom stays still and \((3, 4)\) slides to \((0, 4)\). Where does \((8, 4)\) go, and what shape is it now? (c) Cut the first parallelogram from \((5, 0)\) to \((3, 4)\). What is the area of each piece?
  4. P (4, 6)Q (6, 3)R (4, 0) (0, 0)(10, 0)
    The field in question 4.
    Two farmers share a field 10 wide and 6 tall, split by a bent fence from \(P\) to \(Q\) to \(R\). They want a straight fence from \(P\), with nobody gaining or losing land. (a) Triangle \(PQR\) has base \(PR\). Slide \(Q\) parallel to \(PR\) until it reaches the bottom edge. Where does it land? (b) Where should the new fence go, and why is that fair? (c) Check: find the area of the left-hand field before and after.

Stretch: same area, different edge

Shearing keeps the area but changes the perimeter (the distance round the edge). Every triangle with base 8 and height 3 has area 12. Which one has the shortest perimeter, and how long is it? (Think of a 3-4-5 right-angled triangle.) Is there a longest?

Answers

  1. (a) \(7 \times 4 = 28\) cm². (b) 35 cm² is the area of a 7 by 5 rectangle. Stand the slanted sides upright and the shape gets taller: 5 cm high, not 4. Leaning it back to 4 cm high loses area. Shearing keeps the height. Rotating the sides does not.
  2. (a), (b) and (c) all have base 5 and height 6: \(\tfrac12 \times 5 \times 6 = 15\). The top corner only slid along the line 6 units up. (d) The height is now 3: \(\tfrac12 \times 5 \times 3 = 7.5\).
  3. (a) Base 5, height 4: area 20. (b) The top slides 3 to the left, so \((8, 4)\) goes to \((5, 4)\). The shape is a 5 by 4 rectangle, area 20 again. (c) Two triangles, each with base 5 and height 4: area 10 each. One has its base along the bottom, the other along the top.
  4. (a) \(PR\) runs straight up the line 4 across. \(Q\) is 6 across, so it slides straight down to \(S = (6, 0)\). (b) Straight from \(P\) to \(S\). Triangles \(PQR\) and \(PSR\) have the same base \(PR\) and the same height 2, so the same area: \(\tfrac12 \times 6 \times 2 = 6\). The left field swaps one for the other. (c) Before: a 4 by 6 rectangle plus triangle \(PQR\), \(24 + 6 = 30\). After: the same rectangle plus triangle \(PSR\), \(24 + 6 = 30\). Each farmer has half of the 60.

Stretch. The one with its top corner above the middle of the base. It splits into two 3-4-5 triangles, so its sides are 5, 5 and 8: perimeter 18. Slide the top corner either way and one side grows by more than the other shrinks. There is no longest: slide the corner far away and both slanted sides grow without limit, while the area stays 12.

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