What happens if the parallelogram is ‘thin’ (see the figure) and the foot of the perpendicular from C to AD does not lie on side AD? The construction then does not seem to work. How do we fix this ‘gap’?
Answer: Slide the top side along its own line in small steps: replace ABCD by A′BCD′ with A′A = D′D. A congruent triangle is removed on one side and added on the other, so the area stays the same while the base and height stay the same. Repeat until the parallelogram is ‘fat’, then use the usual cut-and-move. So the area is still base × height.
Step-by-step solution
Idea: Moving the top side along its line never changes the base or the height, and each move swaps a triangle for a congruent one, so the area never changes.
- The gap: the usual proof cuts off a right triangle at one end and moves it to the other end. In a thin parallelogram, the perpendicular from C meets the line AD outside the side AD, so that cut does not lie inside the shape.1 mark
- The fix: pick D′ on DA close to D, and A′ on DA extended with A′A = D′D. Then A′BCD′ is a parallelogram with the same base BC and the same height. Triangle CDD′ is cut off and triangle BAA′ is added; these are congruent (BA = CD, AA′ = DD′, and ∠BAA′ = ∠CDD′ as corresponding angles). So area(A′BCD′) = area(ABCD).1 mark
- Repeat this step as often as needed. Each time the top side slides further back over the base, until the foot of the perpendicular from C lands on the top side. Now the usual construction works and gives area = base × height. Since the area never changed, the thin parallelogram also has area = base × height.1 mark
- Another fix in one step: put the thin parallelogram (base b, height h, top side shifted sideways by s) inside a rectangle of width b + s and height h. The rectangle minus two right triangles (each 12sh) is the parallelogram: (b + s)h − sh = bh.
Slide the top side step by step (A′A = D′D each time); each step swaps congruent triangles, so the area, base and height are unchanged. Once the parallelogram is fat enough the usual construction applies, so the area is still base × height.
Answer to write in the exam
Take D′ on DA and A′ on DA extended with A′A = D′D.
△CDD′ ≅ △BAA′ (SAS: CD = BA, DD′ = AA′, ∠CDD′ = ∠BAA′)
⇒ ar(A′BCD′) = ar(ABCD); base BC and height unchanged.
Repeat until the foot of the perpendicular from C lies on the top side.
∴ Area of thin parallelogram = base × height
Common mistakes that cost marks
- Thinking a thin parallelogram has a smaller area formula. Area = base × height for every parallelogram.
- Measuring the slanting side instead of the perpendicular height.
- Sliding the top side by different amounts at A and D, which no longer gives a parallelogram.
How this can come in the exam
MCQ (1 mark)
A parallelogram has base 12 cm and height 3 cm, and its slant sides are 20 cm long. Its area is
- 36 cm2
- 60 cm2
- 240 cm2
- 18 cm2
Show answer
(A) 36 cm2
Area = base × height = 12 × 3 = 36 cm2; the slant side does not enter.
Try one yourself
A thin parallelogram has base 5 cm, height 2 cm and its top side shifted 9 cm to the right of its base. Use an enclosing rectangle to find its area.
Show answer
Rectangle (5 + 9) × 2 = 28; minus two triangles 12 × 9 × 2 = 9 each: 28 − 18 = 10 cm2 = 5 × 2 ✓.
More questions like this
- The area of a rectangle can be found when we know the lengths of its sides. Is the same true for a parallelogram? That is, can we find the area of a parallelogram when we know the lengths of its sides? Why or why not? (Hint: What happens to the area of a parallelogram if we decrease or increase the angle between the adjacent sides while keeping the lengths fixed?)
- You may wonder, like earlier, is there a gap in our argument? What would we do if angle EFG is obtuse and the triangle were shaped like triangle EFG in the figure? Please work out the answer to this question.
- Do you see why the two triangles fit together to make a parallelogram? (If you study the angles in the figure (e.g., ∠B’C’A’ and ∠BCA), you will see why this is so. Keep in mind the criterion by which we check whether two lines are parallel.)
- Since ΔABD and ΔACD have equal area, you may wonder — Can we divide ΔABD using straight cuts into two or more pieces that we can then rearrange to exactly cover ΔACD? What do you think? Is it possible?
- Suppose we are given two polygons P and Q with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g.,