"Mathematics — Mensuration" — Practice

Question 1

The perimeter of a rectangle with length 15 cm and breadth 9 cm is:
(a) 48 cm  (b) 24 cm  (c) 135 sq cm  (d) 39 cm

Accepted Answer

Perimeter = 2(l+b) = 2(15+9) = 2×24 = 48 cm.

Answer: (a) 48 cm. Trap (b) 24 cm comes from forgetting to double the sum, just computing l+b. Trap (c) 135 sq cm comes from computing the AREA (15×9=135) instead of the perimeter — the central perimeter/area confusion this chapter warns about. Trap (d) 39 cm comes from doubling only the length and forgetting to double the breadth: 2(15)+9=39.

Question 2

The area of a triangle with base 12 cm and height 7 cm is:
(a) 42 cm²  (b) 84 cm²  (c) 19 cm²  (d) 9.5 cm²

Accepted Answer

Area = ½ × base × height = ½ × 12 × 7 = 42 cm².

Answer: (a) 42 cm². Trap (b) 84 cm² comes from forgetting the ½ factor entirely: 12×7=84. Trap (c) 19 cm² comes from wrongly adding the base and height instead of multiplying them: 12+7=19. Trap (d) 9.5 cm² comes from wrongly averaging the base and height (½×(12+7)=9.5), confusing the triangle formula with a trapezium-style averaging step.

Question 3

The area of a circle with radius 7 cm is (use π = 22/7):
(a) 154 cm²  (b) 44 cm²  (c) 38.5 cm²  (d) 22 cm²

Accepted Answer

Area = πr² = (22/7) × 7 × 7 = 22 × 7 = 154 cm².

Answer: (a) 154 cm². Trap (b) 44 cm² comes from computing the CIRCUMFERENCE (2πr=2×22/7×7=44) instead of the area — a wrong-quantity trap. Trap (c) 38.5 cm² comes from wrongly treating the given 7 cm as the diameter rather than the radius, using r=3.5: (22/7)×3.5²=38.5. Trap (d) 22 cm² comes from forgetting to square the radius, computing πr instead of πr²: (22/7)×7=22.

Question 4

A parallelogram has base 10 cm, height 6 cm, and a slant side of 8 cm. Its area is:
(a) 60 cm²  (b) 80 cm²  (c) 36 cm²  (d) 30 cm²

Accepted Answer

Area = base × height = 10 × 6 = 60 cm² (the slant side of 8 cm is not needed).

Answer: (a) 60 cm². Trap (b) 80 cm² comes from wrongly using the slant side instead of the perpendicular height: 10×8=80. Trap (c) 36 cm² comes from confusing this with a perimeter-style calculation, 2×(10+8)=36. Trap (d) 30 cm² comes from wrongly applying the triangle formula (halving the product) instead of the parallelogram formula: ½×10×6=30.

Question 5

The total surface area of a cube with edge 5 cm is:
(a) 150 cm²  (b) 100 cm²  (c) 125 cm³  (d) 25 cm²

Accepted Answer

Total surface area = 6a² = 6×5² = 6×25 = 150 cm².

Answer: (a) 150 cm². Trap (b) 100 cm² comes from using 4a² (a lateral-surface-style count, forgetting the top and bottom faces) instead of 6a²: 4×25=100. Trap (c) 125 cm³ comes from computing the VOLUME (a³=125) instead of the surface area — flagged further by the wrong (cubic) unit. Trap (d) 25 cm² comes from forgetting the factor of 6 altogether, reporting the area of just one face.

Question 6

The volume of a cylinder with radius 7 cm and height 10 cm is (use π = 22/7):
(a) 1540 cm³  (b) 440 cm²  (c) 220 cm³  (d) 748 cm²

Accepted Answer

Volume = πr²h = (22/7) × 49 × 10 = 22×7×10 = 1540 cm³.

Answer: (a) 1540 cm³. Trap (b) 440 cm² comes from computing the curved surface area (2πrh=2×22/7×7×10=440) instead of the volume — flagged by the wrong (square) unit. Trap (c) 220 cm³ comes from forgetting to square the radius, using πrh instead of πr²h: (22/7)×7×10=220. Trap (d) 748 cm² comes from computing the total surface area (2πr(r+h)=2×22/7×7×17=748) instead of the volume.

Question 7

Two rectangles both have a perimeter of 24 cm: rectangle P measures 8 cm × 4 cm, and rectangle Q measures 6 cm × 6 cm. Which statement is correct?
(a) Both rectangles have the same perimeter (24 cm) but different areas — P has area 32 cm² and Q has area 36 cm²
(b) Since both rectangles have the same perimeter, they must also have the same area
(c) Rectangle P has both a larger perimeter and a larger area than rectangle Q
(d) A rectangle's area can be determined directly from its perimeter alone, without knowing its individual side lengths

Accepted Answer

P: perimeter=2(8+4)=24 cm, area=8×4=32 cm². Q: perimeter=2(6+6)=24 cm, area=6×6=36 cm². The perimeters are equal but the areas differ, with the square (Q) having the larger area.

Answer: (a). Trap (b) states the core misconception this question is built to test — assuming equal perimeter forces equal area, which is false. Trap (c) is wrong on both counts: the perimeters are actually equal (not P larger), and Q has the larger area, not P. Trap (d) overgeneralises the same false belief into a claim that perimeter alone determines area, which it does not — the same perimeter can pair with many different areas depending on the actual side lengths.

Question 8

A classroom floor has an area of 6 m². Expressed in cm², this is:
(a) 60,000 cm²  (b) 600 cm²  (c) 6,000,000 cm²  (d) 6 cm²

Accepted Answer

Since 1 m = 100 cm, 1 m² = 100×100 = 10,000 cm². So 6 m² = 6 × 10,000 = 60,000 cm².

Answer: (a) 60,000 cm². Trap (b) 600 cm² comes from forgetting to square the conversion factor, using 1 m²=100 cm² instead of 10,000 cm²: 6×100=600. Trap (c) 6,000,000 cm² comes from wrongly cubing the conversion factor (as if converting a volume), using 100³=1,000,000: 6×1,000,000=6,000,000. Trap (d) 6 cm² comes from assuming the numeral doesn't change at all across units, just the unit label.

Question 9

When using the graph-paper grid-counting method (as recommended in the NCERT Class VII Mensuration chapter) to estimate the area of an irregular closed figure, which convention is correctly applied?
(a) Count each fully covered square as 1, each square covered more than half as 1, and ignore (count as 0) each square covered less than half
(b) Count only the fully covered squares; ignore all partially covered squares entirely
(c) Count every square that the boundary touches, whether fully, partially, or barely covered, as 1 full unit
(d) Multiply the number of squares along the length by the number of squares along the width, exactly as done for a rectangle

Accepted Answer

The standard NCERT convention counts a fully covered square as 1, a more-than-half-covered square as 1, an exactly-half-covered square as ½, and ignores (counts as 0) any square covered less than half.

Answer: (a). Trap (b) ignores all partial squares, badly underestimating the area and defeating the purpose of the grid method for irregular shapes. Trap (c) overcounts by treating every touched square as full, greatly overestimating the area. Trap (d) wrongly applies the rectangle area formula (length × width) to an irregular shape that isn't a rectangle at all — exactly the premature-formula error the grid method exists to avoid.

Question 10

A teacher wants students to discover the area formula for a parallelogram (rather than simply stating 'Area = base × height') through a hands-on activity. Which approach best reflects the NCERT-recommended decomposition method?
(a) Cut a right triangle from one slanted end of the parallelogram and reattach it to the other end, transforming the parallelogram into a rectangle of the same base and height, whose area students already know how to find
(b) Have students memorise the formula and verify it by substituting numbers from the textbook's solved examples
(c) Ask students to measure only the two slant sides and multiply them together
(d) Tell students that a parallelogram's area always equals half the product of its diagonals, matching the rhombus formula

Accepted Answer

The classic cut-and-rearrange activity transforms a parallelogram into an equal-area rectangle by relocating a triangular piece from one end to the other, letting students see WHY base×height works rather than being told to accept it.

Answer: (a). Trap (b) is exactly the premature, rote-memorisation approach this chapter's pedagogy content explicitly contrasts against hands-on discovery — a plausible trap because it's dressed up as a normal instructional step. Trap (c) multiplies the wrong dimensions entirely (slant sides aren't height) and doesn't correspond to any correct formula. Trap (d) incorrectly imports the rhombus diagonal formula (½d₁d₂, valid only for shapes with perpendicular diagonals) into the general parallelogram case, where it does not hold.

Question 11

A Class VII student converts a room's floor area from 5 m² to cm² by writing '5 × 100 = 500 cm²'. What is the most likely source of this error, and what should the teacher clarify?
(a) The student applied the linear conversion factor (1 m = 100 cm) directly to an area quantity, without squaring it — the teacher should clarify that converting an area unit requires squaring the linear conversion factor, since area itself is a product of two lengths
(b) The student made a simple arithmetic mistake in multiplying 5 by 100, and the teacher should only reteach multiplication facts
(c) The student confused metres with centimetres as the larger unit, and the teacher should clarify which unit is bigger
(d) The error is actually correct, since 1 m² does equal 100 cm²

Accepted Answer

The student correctly recalled the linear conversion factor (1 m = 100 cm) but applied it directly to an area quantity without squaring it — the actual conversion is 1 m² = 10,000 cm².

Answer: (a) precisely names the squaring-omission error central to this chapter's unit-conversion theme. Trap (b) misdiagnoses a conceptual unit-conversion error as a mere arithmetic slip, which would lead the teacher to reteach the wrong skill entirely — the multiplication 5×100=500 is arithmetically correct, only the underlying conversion factor is wrong. Trap (c) also misdiagnoses the issue: the student clearly knows a metre is bigger than a centimetre (that's why they multiplied by 100 rather than dividing) — the actual gap is about squaring, not about unit-size ordering. Trap (d) simply asserts the incorrect conversion is correct, reinforcing the exact misconception under examination.

Practice — "Mathematics — Mensuration"

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