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NCERT Solutions Ganita Prakash (Part 2) Chapter 7 – 164Parallelogram — Figure it Out

Book page 162 Updated on2026-09-05

Q1.
Observe the parallelograms in the figure below. (i) What can we say about the areas of all these parallelograms? (ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?
Answer

(i) All seven have the same area. Every one of them is drawn on a base of the same length, between the same pair of parallel lines — so every one has the same base and the same height.

Area = base × height, and both are the same for (a) to (g)
So Area (a) = Area (b) = … = Area (g)

(ii) Their perimeters are all different. The base stays fixed, but the two slanting sides get longer as the parallelogram leans over further.

  • Minimum perimeter: (a) — the one that leans least, so its slant sides are closest to being upright.
  • Maximum perimeter: (g) — the one that leans the most, so its slant sides are the longest.
Why it happens: the slant side is the hypotenuse of a right triangle whose vertical leg is the fixed height and whose horizontal leg is the lean. The more the parallelogram leans, the longer that hypotenuse — while the height, and therefore the area, does not budge. Equal areas with unequal perimeters, once again.
Q2.
Find the areas of the following parallelograms: [Figure (i): base 7 cm, height 4 cm. Figure (ii): base 5 cm, height 3 cm. Figure (iii): base 5 cm, height 4.8 cm. Figure (iv): base 2 cm, height 4.4 cm.]
Answer

Each time, multiply a side by its own height.

ParallelogramBaseHeight to that baseArea
(i)7 cm4 cm7 × 4 = 28 cm²
(ii)5 cm3 cm5 × 3 = 15 cm²
(iii)5 cm4.8 cm5 × 4.8 = 24 cm²
(iv)2 cm4.4 cm2 × 4.4 = 8.8 cm²
Tip: in (ii) the diagonal drawn across the figure is not needed at all. In (iii) and (iv) the height is drawn to the slanting side, so that slanting side is the base — do not reach for the horizontal side instead.
Why it happens: the formula base × height is only valid when the height is measured perpendicular to the chosen base. Pairing a base with the wrong height gives a number that means nothing.
Q3.
Find QN.
Answer

PQRS is a parallelogram with SR = 12 cm, QM ⊥ SR with QM = 6 cm, PS = 7.6 cm, and QN ⊥ PS.

Using SR as base: Area (PQRS) = SR × QM = 12 × 6 = 72 cm²

Using PS as base: Area (PQRS) = PS × QN = 7.6 × QN

So   7.6 × QN = 72
QN = 72 ÷ 7.6 = 720/76 = 180/19
QN ≈ 9.47 cm
Why it happens: the parallelogram has one area, so both base–height pairs must give 72 cm². PS (7.6 cm) is shorter than SR (12 cm), so its height QN must be correspondingly longer than 6 cm — and indeed 9.47 > 6.
Check it yourself: 7.6 × 9.47 = 71.97 ≈ 72 ✓
Q4.
Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.]
Answer

The rectangle has the greater area.

Rectangle: area = 5 × 4 = 20 cm²

Parallelogram on the same base of 5 cm:
its height is the perpendicular distance from the opposite side,
which is less than the slanting side of 4 cm
So area = 5 × height < 5 × 4 = 20 cm²
Why it happens: the 4 cm side of the parallelogram is a slant, not a height. A perpendicular is always the shortest segment from a point to a line, so the height is strictly shorter than the 4 cm side — unless the parallelogram is standing upright, in which case it is the rectangle. The more it leans, the smaller its area, even though its four sides never change length.
Did you know? Push the top of a rectangular gate sideways and it becomes a parallelogram with the same four sides but less area. That is why a diagonal brace is nailed onto a gate — to stop it collapsing.
Q5.
Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?
Answer

Given ∆ABC with base BC and height h, its area is ½ × BC × h. We want a rectangle of area BC × h.

Method 1 — the enclosing rectangle. Draw the line l through A parallel to BC, then drop BE ⊥ BC and CD ⊥ BC meeting l at E and D. The rectangle BCDE has base BC and height h, so its area is BC × h = twice the triangle.

Method 2 — two copies. Make a second copy of ∆ABC and rotate it through a half-turn about the midpoint of AC. The two copies join into a parallelogram of base BC and height h. Now dissect that parallelogram into a rectangle (cut along a height and slide).

Method 3 — choose your own shape. Any rectangle whose two sides multiply to BC × h will do, for instance one of base 2 × BC and height h/2.

Area (∆ABC) = ½ × BC × h
Area (rectangle) = BC × h = 2 × Area (∆ABC) ✓
Why it happens: the triangle formula already contains the factor ½. Removing that factor — by keeping the base and height but building a rectangle instead — is precisely doubling the area.
Q6.
[Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle.
Answer

Cut the triangle at half its height and fold the top down.

  1. In ∆ABC with base BC, mark M and N, the midpoints of AB and AC. Join MN — it is parallel to BC and lies at height h/2.
  2. Cut along MN.
  3. Drop the small triangle ∆AMN down onto the base strip: turn it half a turn about M to fill the left gap, and cut-and-turn the other part about N to fill the right gap.
The pieces close up into a rectangle with base BC and height h/2
Area = BC × h/2 = ½ × BC × h = Area (∆ABC) ✓
Why it happens: a half-turn about the midpoint of a side is a rigid motion, so no area is gained or lost. The trapezium that is left after the cut is short of exactly the two corner triangles that the top piece supplies. This is a dissection: the triangle and the rectangle are made of the very same pieces.
Q7.
[Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it? [Hint: Show that triangles ∆ADB and ∆ADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.]
Answer

Cut along the axis of symmetry, then fit the two right triangles together along their slant sides.

  1. In the isosceles ∆ABC (AB = AC), let AD be the perpendicular from A to BC. Then D is the midpoint of BC, and ∆ADB ≅ ∆ADC by RHS.
  2. Cut along AD. You now hold two right triangles, each with legs AD (the height) and DB = DC (half the base).
  3. Keep ∆ADC where it is. Turn ∆ADB over and lay its slant side AB along the slant side AC, so that the two right angles land at opposite corners.
The result is a rectangle with sides AD and DC
Area = AD × DC = AD × (½ BC) = ½ × BC × AD = Area (∆ABC) ✓
Why it happens: a rectangle is cut by a diagonal into two right triangles with the same legs. Here the two legs are AD and DC, and both of our pieces have exactly those legs — so each is a "half rectangle". Placing them slant-against-slant, with the right angles diagonally opposite, restores the rectangle they came from. Since only cutting and moving are involved, the area is unchanged.
Try This: cut a paper isosceles triangle along its height and try to make the rectangle without turning a piece over. You will find you must flip one piece — the two halves are mirror images of each other.
Q8.
[Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection.
Answer

Run Q7 backwards.

  1. Take rectangle PQRS with base PQ = p and height QR = q.
  2. Cut along the diagonal PR into two right triangles.
  3. Keep one of them. Turn the other over and join it to the first along the side of length q, so that the two right angles sit side by side and their bases form one straight line.
The result is an isosceles triangle with base 2p and height q
Area = ½ × 2p × q = pq = Area of the rectangle ✓
Why it happens: the two right angles at the join add to 180°, so the two bases really do line up into a single straight base. The two slanting sides are the two halves of the diagonal PR, which are equal — which is exactly what makes the triangle isosceles.
Q9.
Which has greater area — an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area — two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.
Answer

The square is greater in both cases.

Let the common sidelength be s.

Square: area = s²

Equilateral triangle: its height h is the perpendicular from a vertex to the opposite side.
A perpendicular is the shortest distance from a point to a line, so h < s.
Area = ½ × s × h < ½ × s × s = s²/2

One triangle < s²/2 < s²  →  the square is larger
Two triangles < 2 × s²/2 = s²  →  the square is still larger

The exact values confirm it. By the Baudhāyana–Pythagoras theorem,

h = √(s² − (s/2)²) = (√3/2)s ≈ 0.866 s
Area of one equilateral triangle = (√3/4)s² ≈ 0.433 s²
Two of them ≈ 0.866 s², and the square is 1.000 s²
Why it happens: the square has two of its sides meeting at a right angle, so one side is the full height for the other. In the equilateral triangle the sides lean at 60°, so the height falls short of the side — and that shortfall is what costs the triangle its area. Notice that two triangles come surprisingly close to the square (about 87%) without ever reaching it.
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