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NCERT Solutions for Class 8th Maths Chapter 7 Parallelogram — In-text Questions

Book page 161 Updated on2026-09-19

Q1.
Give a method to convert a parallelogram into a rectangle of equal area. You can try this using a cut-out of a parallelogram.
Answer

One cut and one slide.

  1. In parallelogram ABCD, drop a perpendicular from A to the side DC. Call the foot X, so AX ⊥ DC. AX is a height of the parallelogram.
  2. Cut along AX. This separates ∆AXD from the trapezium ABCX.
  3. Slide ∆AXD across to the right-hand end and fit it against BC. The result is a rectangle.
Area of the rectangle = Area of the parallelogram, because nothing was added or removed — only moved
Why it happens: the piece that is cut off on the left is exactly the piece that is missing on the right. Extending XC and dropping BY ⊥ XC produces ∆BYC, and ∆AXD ≅ ∆BYC by RHS, so ∆AXD fits over ∆BYC perfectly. This process of cutting a figure and rearranging the pieces into a different figure of the same area is called dissection.
Q2.
Can ∆AXD and ABCX fit together, as shown in the figure, to get a rectangle?
A DX AB XCY
The two pieces of the parallelogram: ∆AXD on the left, and the trapezium ABCX on the right with ∆BYC (dashed, shaded) shown in the place where ∆AXD would fit.
Answer

Yes. The test is to find the triangle that would complete ABCX into a rectangle, and then check whether ∆AXD is congruent to it.

Extend XC to the right, and drop BY ⊥ XC.
ABYX now has four right angles → a rectangle,
and ∆BYC is exactly the missing corner piece.

BY = AX  (opposite sides of rectangle ABYX)
∠BYC = ∠AXD = 90°
BC = AD  (opposite sides of parallelogram ABCD)
So by RHS, ∆BYC ≅ ∆AXD

Since the two triangles are congruent, ∆AXD can be laid exactly over the gap ∆BYC. So yes — the two pieces fit together into rectangle ABYX.

Q3.
We need another right angle to get a rectangle (what about the fourth angle?)
Answer

The fourth angle looks after itself.

The four angles of a quadrilateral add up to 360°
If three of them are 90°, then the fourth = 360° − (90° + 90° + 90°) = 90°

So only three right angles need to be arranged; the fourth is forced. That is why constructing BY ⊥ XC is enough to finish the rectangle.

Q4.
Is ∆AXD congruent to it?
Answer

Yes — ∆AXD ≅ ∆BYC by the RHS congruency criterion.

R — right angle: ∠AXD = ∠BYC = 90°
H — hypotenuse: AD = BC (opposite sides of the parallelogram)
S — side: AX = BY (opposite sides of the rectangle ABYX)

So ∆AXD ≅ ∆BYC, and hence Area (∆AXD) = Area (∆BYC)
Why it happens: congruent figures can be laid one on top of the other, so they must have the same area. That is what makes the dissection honest — the piece that leaves the left-hand end is the exact size of the hole it fills at the right-hand end.
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