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

Book page 166 Updated on2026-09-19

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.
AB DMC PQ STUR WX ZMNY
Three trapeziums, each already cut by the vertical lines dropped from the ends of the shorter parallel side: ABCD splits into a rectangle and a triangle; PQRS and WXYZ each split into two triangles and a rectangle.
Answer

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

TrapeziumPiecesArea
ABCD, with BM ⊥ DCrectangle ABMD + ∆BMCAB × AD + ½ × MC × BM
PQRS, with PT ⊥ SR and QU ⊥ SR∆PST + rectangle PQUT + ∆QUR½ × ST × PT + TU × PT + ½ × UR × QU
WXYZ, with WM ⊥ ZY and XN ⊥ ZY∆WZM + rectangle WXNM + ∆XNY½ × 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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