NCERT Solutions Ganita Prakash (Part 2) Chapter 7 Trapezium — In-text Questions

Book page 167 Updated on2026-09-05

Q1.

Can the area of the trapezium be expressed in terms of a, b and h?

Answer

Yes. Replace x + y using the longer parallel side.

The lower parallel side is made of three parts: b = x + a + y
Subtract a from both sides: x + y = b − a

Area WXYZ = ½ h(x + y + 2a)
= ½ h(b − a + 2a)
= ½ h(a + b)

So Area of a trapezium = ½ × height × (sum of the parallel sides).

Why it happens: x and y depend on how the trapezium leans, but a and b do not. Eliminating x + y is what turns a formula about one particular picture into a formula about every trapezium.

Q2.

Will this formula hold for a trapezium that looks like this? There are different ways of approaching this, which are sketched below. Complete the arguments. Approach 1: Rectangle and Triangles — Area ABCD = Area ABED + Area ∆BEC, Area ABED = Area ABEF – Area ∆AFD. Approach 2: Parallelogram and Triangle — Draw BG ‖ AD.

Answer

Yes, the formula still holds, even when the trapezium leans so far that the foot of one perpendicular falls outside the base. Here are both arguments completed, with AB = a, DC = b and height h.

Approach 1 — Rectangle and Triangles. F and E are the feet of the perpendiculars from A and B, and ABEF is a rectangle with FE = AB = a.

Area ABEF = a h, Area (∆AFD) = ½ × FD × h, Area (∆BEC) = ½ × EC × h

Area ABED = Area ABEF − Area (∆AFD) = a h − ½ × FD × h
Area ABCD = Area ABED + Area (∆BEC) = a h − ½ × FD × h + ½ × EC × h
= ½ h (2a + EC − FD)

Now DE = FE − FD = a − FD, and b = DC = DE + EC = a − FD + EC
so EC − FD = b − a

Area ABCD = ½ h(2a + b − a) = ½ h(a + b) ✓

Approach 2 — Parallelogram and Triangle. Draw BG ‖ AD, with G on DC.

ABGD has AB ‖ DG and BG ‖ AD, so it is a parallelogram
DG = AB = a, so GC = b − a

Area ABGD = a h
Area (∆BGC) = ½ × (b − a) × h

Area ABCD = a h + ½(b − a)h = ½ h(2a + b − a) = ½ h(a + b) ✓

Why it happens: in Approach 1 the triangle at one end is subtracted rather than added, because F falls outside DC. The algebra takes care of it: only the difference EC − FD survives, and that difference is always b − a. Approach 2 avoids the difficulty entirely by never leaving the trapezium.

Q3.

Will Approach 2 work for any type of trapezium? [Math Talk]

Answer

Yes — provided you draw the parallel line from an endpoint of the shorter parallel side.

Name the parallel sides so that a ≤ b
Then DG = a ≤ b = DC, so G really does land on the segment DC
Parallelogram of area a h + triangle of area ½(b − a)h
= ½ h(a + b) for every trapezium

If a = b, then G coincides with C, the triangle vanishes, and the trapezium is a parallelogram — area ah, which is what ½h(a+a) gives. ✓
If you carelessly draw the line from the longer side, G falls beyond C and you would have to subtract a triangle instead of adding one. The final formula is the same, but the picture is harder to read.

Why it happens: Approach 2 never uses a perpendicular foot, so it does not care how far the trapezium leans. It only needs the parallel line BG to stay inside the figure, which it does as long as the piece cut off equals the shorter parallel side.

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