NCERT Solutions for Class 9th Maths Chapter 6 Think and Reflect — Area of a Parallelogram
Book page 131 Updated on2026-09-08
Q1.
What happens if the parallelogram is ‘thin’ (Fig. 6.18) 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
The formula is still base × height; only the proof needs repair. The fix is to slide the parallelogram back into a "fat" one by repeated shearing, one step at a time.
The standard proof cuts off the right triangle at one end and slides it to the other end to make a rectangle. That works only if the foot of the perpendicular from C lands on the segment AD. In a very slanted parallelogram it lands outside AD, so there is no triangle to cut.
Choose D′ on DA close to D, and A′ on DA produced, with A′A = D′D. Then A′BCD′ is a parallelogram (A′D′ ∥ BC and A′D′ = AD = BC). ΔCDD′ ≅ ΔBAA′ (CD = BA, DD′ = AA′, ∠CDD′ = ∠BAA′) ∴ ar(A′BCD′) = ar(ABCD) − ar(ΔCDD′) + ar(ΔBAA′) = ar(ABCD)
One shearing step: A′BCD′ (dashed) has the same base BC, the same height and the same area as ABCD, but leans less.
Each step keeps the area unchanged and makes the parallelogram less slanted. Repeat as often as needed until the foot of the perpendicular does land inside the base — then the original cut-and-slide proof applies, and the area is base × height.
Why it happens: sliding the top side along its own line is a shear. A shear moves every point parallel to the base, so it changes neither the base nor the height, and by the congruence above it changes no area either. The formula bh cannot notice a shear — which is exactly why base × height, and not the side lengths, is the right thing to multiply.
Tip: this is the same "the argument breaks in an awkward case, so fix the case" discipline you will use in geometry all through Class 9 and 10. A formula is not proved until every configuration is covered.