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

Book page 166 Updated on2026-09-05

Q1.

Simplify the expression to show that we get the same formula for the area of a rhombus in terms of its diagonals.

Answer

Area = ½ × AO × BD + ½ × CO × BD
= ½ × BD × (AO + CO) [taking ½ × BD common]
= ½ × BD × AC [since AO + CO = AC]

Area of a rhombus = ½ × product of its diagonals

Tip: the same simplification works even when O is not the midpoint — all that is needed is that AC ⊥ BD and that O lies between A and C. So the formula ½d₁d₂ holds for any quadrilateral whose diagonals are perpendicular, such as a kite.

Q2.

Find the areas of the following trapeziums by breaking them into figures whose areas can be computed. [Figures: trapezium ABCD with M on DC; trapezium PQRS with T and U on SR; trapezium WXYZ with M and N on ZY.]

Answer

In each figure the dividing lines have already been drawn. Read off the pieces and add.

Trapezium	Pieces	Area
ABCD, with BM ⊥ DC	rectangle ABMD + ∆BMC	AB × AD + ½ × MC × BM
PQRS, with PT ⊥ SR and QU ⊥ SR	∆PST + rectangle PQUT + ∆QUR	½ × ST × PT + TU × PT + ½ × UR × QU
WXYZ, with WM ⊥ ZY and XN ⊥ ZY	∆WZM + rectangle WXNM + ∆XNY	½ × ZM × WM + MN × WM + ½ × NY × XN

Measure the lengths marked in your book and substitute them.

Why it happens: the two perpendiculars from the ends of the shorter parallel side cut the trapezium into a rectangle in the middle with a right triangle at each end. Since the two perpendiculars are equal (both equal the height h), all three pieces share the same height — which is what makes the pieces add up so tidily into ½h(a + b).

Tip: in the first trapezium AD is itself perpendicular to DC, so only one end triangle appears. The middle piece is then the rectangle ABMD.

Q3.

Consider a trapezium WXYZ with WX ‖ ZY. Find its area.

Answer

Construct WM ⊥ ZY and XN ⊥ ZY, and let MZ = x, WM = XN = h, WX = a, NY = y.

Area WXYZ = Area (∆WMZ) + Area WXNM + Area (∆XNY)
= ½ × MZ × WM + WX × WM + ½ × NY × XN
= ½ hx + ha + ½ hy
= h(½x + a + ½y)
= h((x + y + 2a)/2)
= ½ h(x + y + 2a)

Now bring in the second parallel side. Let ZY = b. Since MN = WX = a,

b = x + a + y → x + y = b − a

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

Why it happens: the two end triangles have different bases x and y, but only their sum ever appears — and that sum is fixed at b − a however the trapezium leans. So the individual pieces do not matter; only the two parallel sides and the height do.

Q4.

Is WXNM a rectangle?

Answer

Yes.

WM ⊥ ZY and XN ⊥ ZY by construction, so ∠WMN = ∠XNM = 90°

WX ‖ ZY, and WM is a transversal, so the interior angles on the same side add to 180°:
∠MWX + ∠WMN = 180° → ∠MWX = 90°
In the same way, ∠NXW = 90°

All four angles are 90°, so WXNM is a rectangle

Consequently WM = XN, so both perpendiculars equal the height h, and MN = WX = a.

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