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NCERT Solutions Ganita Prakash (Part 2) Chapter 7 – 170Trapezium — Figure it Out

Book page 169 Updated on2026-09-05

Q1.
Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
Answer
Area of a rhombus = ½ × product of the diagonals
= ½ × 20 × 15
= 150 cm²
Check it yourself: the diagonals cut the rhombus into 4 congruent right triangles with legs 10 cm and 7.5 cm. Each has area ½ × 10 × 7.5 = 37.5 cm², and 4 × 37.5 = 150 cm². ✓
Q2.
Give a method to convert a rectangle into a rhombus of equal area using dissection.
Answer

Run the rhombus dissection of pages 164–165 backwards.

Let the rectangle have sides m and n, so its area is mn. Build the rhombus whose diagonals are m and 2n:

  1. Draw a segment AC of length m, and mark its midpoint O.
  2. Through O draw a line perpendicular to AC, and mark B and D on it with OB = OD = n, so BD = 2n.
  3. Join AB, BC, CD, DA. Since the diagonals bisect each other at right angles, ABCD is a rhombus.
Area of the rhombus = ½ × AC × BD = ½ × m × 2n = mn = area of the rectangle ✓

The dissection itself: cut the rhombus along BD into the two isosceles triangles ∆ABD and ∆CBD, cut each along its axis of symmetry, and reassemble the four right triangles into a rectangle of sides m (= AC) and n (= ½BD).

Why it happens: the same four pieces make both figures. The dissection is reversible, so a construction that turns a rhombus into a rectangle can be read the other way to turn a rectangle into a rhombus.
Q3.
Find the areas of the following figures: [Figure (i): a trapezium with parallel sides 10 ft and 7 ft, height 16 ft. Figure (ii): parallel sides 24 m and 36 m, height 14 m. Figure (iii): parallel sides 14 in and 6 in, height 10 in. Figure (iv): parallel sides 12 ft and 18 ft, height 8 ft.]
Answer

Each is a trapezium, so use Area = ½ × height × (sum of the parallel sides).

FigureParallel sidesHeightArea
(i)10 ft and 7 ft16 ft½ × 16 × 17 = 136 ft²
(ii)24 m and 36 m14 m½ × 14 × 60 = 420 m²
(iii)14 in and 6 in10 in½ × 10 × 20 = 100 in²
(iv)12 ft and 18 ft8 ft½ × 8 × 30 = 120 ft²
(i) 136 ft²   (ii) 420 m²   (iii) 100 in²   (iv) 120 ft²
Tip: in (i) the trapezium is drawn tilted and in (iii) the parallel sides are the two vertical ones. The height is always the perpendicular distance between the parallel sides — the dashed segment, never a slanting side.
Q4.
[Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.
Answer

Let ABCD be an isosceles trapezium with AB = a ‖ DC = b, AD = BC, and height h.

  1. Drop AF ⊥ DC and BE ⊥ DC. Now ABEF is a rectangle of sides a and h, and the two end triangles ∆AFD and ∆BEC are congruent (RHS: right angles at F and E, AD = BC, AF = BE).
  2. Cut off both end triangles. Each has legs h and (b − a)/2.
  3. Turn one of them over and fit it against the other along the slant side. They form a small rectangle of sides h and (b − a)/2.
  4. Push that small rectangle against ABEF. Both have height h, so they line up into a single rectangle.
Width of the new rectangle = a + (b − a)/2 = (a + b)/2
Area = h × (a + b)/2 = ½ h(a + b) ✓
Why it happens: it is the trapezium's own symmetry that makes this work — the two end triangles are congruent, so together they are exactly a rectangle. That is not true of a general trapezium, where the two end triangles have different bases.
Q5.
Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area — Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH? [Hint: If ∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ, then the trapezium and rectangle have equal areas.] [Math Talk]
Answer

Find the midpoints of the two slanting sides and draw the perpendiculars through them.

  1. Mark I, the midpoint of AD, and J, the midpoint of BC.
  2. Through I draw a line perpendicular to DC. It meets the line AB at H and the line DC at G.
  3. Through J draw a line perpendicular to DC. It meets AB at E and DC at F.
  4. EFGH is the required rectangle.
∠AHI = ∠DGI = 90°  (both lines are ⊥ to DC, and AB ‖ DC)
∠AIH = ∠DIG  (vertically opposite angles at I)
AI = DI  (I is the midpoint of AD)
So ∆AHI ≅ ∆DGI  (ASA), and in the same way ∆BEJ ≅ ∆CFJ

Going from the trapezium to the rectangle, we add ∆AHI and ∆BEJ and remove ∆DGI and ∆CFJ. The added pieces are congruent to the removed ones, so the area is unchanged.

Area (EFGH) = Area (ABCD) = ½ h(a + b)
and indeed HE = (a + b)/2, GF = (a + b)/2, height = h
Why it happens: the perpendicular through the midpoint of a slanting side cuts off a triangle above the trapezium exactly as big as the one it leaves out below. The midpoint is what makes the two triangles congruent — try any other point on AD and the two triangles no longer match.
Q6.
Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm². [Math Talk]
Answer

Start from a rectangle of area 144 cm² and open it out into a trapezium.

  1. Choose a rectangle of area 144 cm², say 16 cm × 9 cm.
  2. Keep the height at h = 9 cm. The trapezium must satisfy ½ × 9 × (a + b) = 144, so a + b = 32 cm.
  3. Split 32 cm any way you like — say a = 12 cm and b = 20 cm.
  4. Draw DC = 20 cm; draw a parallel line 9 cm above it; mark AB = 12 cm anywhere on that line; join AD and BC.
Area = ½ × 9 × (12 + 20) = ½ × 9 × 32 = 144 cm² ✓
Try This: other choices work equally well — h = 8 cm with a + b = 36 (say 15 cm and 21 cm), or h = 12 cm with a + b = 24 (say 10 cm and 14 cm). Every one of them has area 144 cm², and they look completely different.
Why it happens: the area fixes only the product h × (a + b). The height and the sum of the parallel sides can be traded off against each other, and the sum can be split in infinitely many ways — so there are infinitely many trapeziums of area 144 cm².
Q7.
A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.
Answer

Trapezium : equilateral triangle : rhombus = 3 : 1 : 2.

Join the centre of the regular hexagon to its six vertices. This gives six congruent equilateral triangles, and every one of the three regions is made up of a whole number of them.

Trapezium = 3 parts Triangle = 1 part Rhombus = 2 parts
The dashed lines split the hexagon into six equal equilateral triangles. Each region is a whole number of them.

Let each small equilateral triangle have area t, so the hexagon has area 6t. Take the hexagon's side as s.

RegionMade ofAreaShare of the hexagon
Trapezium3 small triangles3t½
Equilateral triangle1 small trianglet1/6
Rhombus2 small triangles2t1/3
Trapezium: parallel sides s and 2s, height (√3/2)s → area = ½ × (√3/2)s × 3s = (3√3/4)s²
Equilateral triangle of side s → area = (√3/4)s²
Rhombus: two such triangles → area = (√3/2)s²

Ratio = (3√3/4) : (√3/4) : (√3/2) = 3 : 1 : 2
Why it happens: the long diagonal of a regular hexagon splits it into two halves, which is why the trapezium is 3 parts out of 6. The remaining half is then cut by a line from a vertex to the centre of that diagonal, and the centre of the hexagon lies on it — so the cut separates one small triangle from two. Notice 3 + 1 + 2 = 6, the whole hexagon. ✓
Q8.
ZYXW is a trapezium with ZY‖WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of ∆ZWB.
Answer

B is the point where the line ZA, extended, meets the line WX extended. The whole proof is one congruence.

ZY ‖ WX, so ZY ‖ XB, and YB (the line ZAB) is a transversal:
∠YZA = ∠XBA  (alternate angles)
∠ZYA = ∠BXA  (alternate angles)
YA = XA  (A is the midpoint of XY)

So   ∆ZYA ≅ ∆BXA  (AAS)  →  Area (∆ZYA) = Area (∆BXA)

Now build both figures out of the same pieces:

Trapezium ZYXW = region ZWXA + ∆ZYA
∆ZWB             = region ZWXA + ∆AXB

The two added triangles have equal areas, so
Area (ZYXW) = Area (∆ZWB)  ∎
Why it happens: the congruence also gives XB = ZY. So WB = WX + XB = b + a, and ∆ZWB has base a + b and the same height h as the trapezium. Its area is ½h(a + b) — so this construction is one more proof of the trapezium formula, this time by turning the trapezium into a single triangle.
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