CBSE Class 8 Mathematics Area MCQs Set 11

Answer: (b) the number of non-overlapping unit squares that exactly cover it

Explanation:
1. Area is the count of non-overlapping unit squares (fractions of a square allowed) that cover the region exactly.
2. The length of the boundary, or the sum of the sides, is the perimeter, not the area.

Teacher's Note:
1. Area = space covered; perimeter = distance around.
2. Unit squares must not overlap and must leave no gaps.

Answer: (c) l × w

Explanation:
1. The rectangle holds w rows of l unit squares each.
2. So the number of unit squares is l × w.
3. Option (b), 2(l + w), is the perimeter, not the area.

Teacher's Note:
1. Area of a rectangle = length × width.
2. Do not mix up the area formula with the perimeter formula.

Answer: (d) 54

Explanation:
1. Each row along the 9 cm side holds 9 unit squares.
2. There are 6 such rows.
3. Number of unit squares = 9 × 6 = 54, so the area is 54 cm².

Teacher's Note:
1. 15 is 9 + 6, and 30 is the perimeter; neither counts squares.
2. Multiply the two sides to count unit squares.

Answer: (b) \( \frac{1}{2} \) × base × height

Explanation:
1. A triangle is exactly half of the rectangle drawn on its base with the same height.
2. So area of a triangle = \( \frac{1}{2} \) × base × height.

Teacher's Note:
1. The height must be perpendicular to the base.
2. Remember the factor one-half for triangles.

Answer: (d) 20 cm²

Explanation:
1. Base BC = 8 cm and height AD = 5 cm.
2. Area = \( \frac{1}{2} \) × 8 × 5 = 4 × 5 = 20 cm².
3. Option (b), 40 cm², forgets the factor \( \frac{1}{2} \); 13 is just 8 + 5.

Teacher's Note:
1. Read the base and the perpendicular height from the figure.
2. Area of a triangle = ½ × base × height.

Answer: (a) base × corresponding height

Explanation:
1. Cut off the right triangle at one end and fit it at the other end; the parallelogram becomes a rectangle.
2. That rectangle has the base and the height as its sides, so area = base × corresponding height.
3. The product of two adjacent sides is too big, because the height is shorter than the slant side.

Teacher's Note:
1. The height is the perpendicular distance between the base and its opposite side.
2. No factor ½ for a parallelogram.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A diagonal cuts a rectangle into two congruent triangles, so R is true.
2. Congruent triangles cover equal regions, so they have equal areas; A is true.
3. R is exactly why A holds, so R is the correct explanation of A.

Teacher's Note:
1. Congruent figures always have equal areas.
2. Each triangle has half the area of the rectangle.

Answer: (b) Only Statement II is true.

Explanation:
1. A 1 cm × 10 cm rectangle has perimeter 22 cm and area 10 cm².
2. A 4 cm × 5 cm rectangle has the smaller perimeter 18 cm but the larger area 20 cm².
3. So Statement II is true. Also 3 cm × 3 cm and 1 cm × 5 cm both have perimeter 12 cm but areas 9 cm² and 5 cm², so Statement I is false.

Teacher's Note:
1. Perimeter does not decide area.
2. One counter-example is enough to show a statement is false.

Answer: (d) \( \frac{1}{2} \) × product of its diagonals

Explanation:
1. The diagonals of a rhombus are perpendicular bisectors of each other.
2. The rhombus can be dissected into a rectangle of sides d₁ and \( \frac{d_2}{2} \).
3. So area = \( \frac{1}{2} \) × d₁ × d₂, half the product of the diagonals.

Teacher's Note:
1. A rhombus is half of the rectangle drawn around its diagonals.
2. Multiply the diagonals, then halve.

Answer: (a) (i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)

Explanation:
1. Rectangle → length × width (r).
2. Triangle → ½ × base × height (p).
3. Parallelogram → base × height (s).
4. Rhombus → ½ × product of diagonals (q).

Teacher's Note:
1. Triangle and rhombus formulas carry the factor ½.
2. Parallelogram and rectangle formulas do not.

Answer: (c) \( \frac{1}{2} \) × height × sum of the parallel sides

Explanation:
1. Split the trapezium into a rectangle and two triangles, or join two copies to make a parallelogram.
2. Either way, area = \( \frac{1}{2} \)h(a + b).
3. That is, \( \frac{1}{2} \) × height × sum of the parallel sides.

Teacher's Note:
1. Add the parallel sides, multiply by the height, then halve.
2. Two copies of a trapezium make a parallelogram.

Answer: (c) A is true, but R is false.

Explanation:
1. A median goes to the midpoint, so the two triangles have equal bases and the same height from the vertex. A is true.
2. But their other sides are usually different, so they are generally not congruent. R is false.

Teacher's Note:
1. Equal area does not mean congruent.
2. Equal base + same height ⇒ equal area.

Answer: (d) 126 cm²

Explanation:
1. Area = \( \frac{1}{2} \) × d₁ × d₂ = \( \frac{1}{2} \) × 18 × 14.
2. = 9 × 14 = 126 cm².
3. 252 cm² is the full product, without halving; 32 is just the sum.

Teacher's Note:
1. Halve the product of the diagonals.
2. Halve 18 first to make the arithmetic easy.

Answer: (a) Only Statement I is true.

Explanation:
1. Each side can act as a base with its own height, and all three pairs give the same area. Statement I is true.
2. Since base × height is the same (twice the area) for all pairs, a longer base goes with a shorter height.
3. So the longest side has the shortest height, and Statement II is false.

Teacher's Note:
1. Fixed product: if one factor grows, the other shrinks.
2. Example: 12-16-20 triangle has heights 16, 12 and 9.6 cm.

Answer: (c) The two triangles have equal areas.

Explanation:
1. Both triangles stand on the same base BC.
2. Their apexes A and P lie on l, which is parallel to m, so both heights equal the distance between l and m.
3. Same base and same height give equal areas.

Teacher's Note:
1. Triangles on the same base between the same parallels have equal areas.
2. Their shapes and perimeters may still differ.

Answer: (a) dissection

Explanation:
1. Cutting a figure into pieces and re-arranging them into a new figure is called dissection.
2. The chapter uses it to turn a parallelogram, a rhombus or a trapezium into a rectangle of the same area.

Teacher's Note:
1. Dissection keeps the area unchanged.
2. Reflection and rotation move a whole figure; they do not cut it.

Answer: (b) 9 cm

Explanation:
1. \( \frac{1}{2} \) × 12 × h = 54, so 6h = 54.
2. h = 54 ÷ 6 = 9 cm.
3. Check: \( \frac{1}{2} \) × 12 × 9 = 54 cm².

Teacher's Note:
1. Put the known values in the formula and solve for h.
2. 4.5 cm forgets the ½ in the formula.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The rectangle has area 5 × 4 = 20 cm².
2. In a slanting parallelogram, the 4 cm side is slanted, so the perpendicular height on the 5 cm base is less than 4 cm. R is true.
3. So its area = 5 × (height) is less than 20 cm². A is true, and R is the reason.

Teacher's Note:
1. The perpendicular is the shortest distance between two lines.
2. The more it slants, the smaller the area.

Answer: (b) (i)-(q), (ii)-(p), (iii)-(r), (iv)-(s)

Explanation:
1. (i) \( \frac{1}{2} \) × 10 × 7 = 35 cm² → (q).
2. (ii) 8 × 6 = 48 cm² → (p).
3. (iii) \( \frac{1}{2} \) × 12 × 10 = 60 cm² → (r).
4. (iv) \( \frac{1}{2} \) × 5 × (9 + 11) = \( \frac{1}{2} \) × 5 × 20 = 50 cm² → (s).

Teacher's Note:
1. Work out each area before looking at the options.
2. Use the correct formula for each shape.

Answer: (b) Only Statement II is true.

Explanation:
1. The correct formula is \( \frac{1}{2} \)h(a + b), not height × longer side. Statement I is false.
2. Two copies, one rotated, join along a slant side to form a parallelogram of base (a + b) and height h.
3. Its area h(a + b) is twice the trapezium's area, so Statement II is true.

Teacher's Note:
1. The trapezium formula uses both parallel sides.
2. Two copies give a parallelogram; halve it for one trapezium.

Answer: (c) 103.2256 cm²

Explanation:
1. Area = 4 × 4 = 16 in².
2. 16 × 6.4516 = 103.2256 cm².
3. 10.16 (= 4 × 2.54) converts only the side length, not the area.

Teacher's Note:
1. Find the area first, then convert square units.
2. 1 in² = 2.54 × 2.54 = 6.4516 cm².

Answer: (d) 68 cm²

Explanation:
1. Complete the L-shape to a 10 cm × 8 cm rectangle: area = 80 cm².
2. The missing top-right corner is 4 cm × 3 cm = 12 cm².
3. Area of the tile = 80 − 12 = 68 cm².
4. Check by splitting: top part 6 × 3 = 18 cm², bottom part 10 × 5 = 50 cm², total 68 cm².

Teacher's Note:
1. Complete-and-subtract or split-and-add; both must agree.
2. 80 cm² forgets to remove the missing corner.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Any polygon can be cut into triangles by its diagonals, and their areas added. So A is true.
2. A triangle has 3 sides, the fewest possible for a polygon. So R is true.
3. But A holds because of the splitting into triangles, not because a triangle has the fewest sides. R does not explain A.

Teacher's Note:
1. Check each statement first, then check the link.
2. Polygon area = sum of areas of its triangles.

Answer: (b) the length of AC and the perpendicular distances of B and D from AC

Explanation:
1. The diagonal AC splits ABCD into △ABC and △ACD, both with base AC.
2. Their heights are the perpendicular distances of B and D from AC.
3. So we need AC and those two perpendicular distances.

Teacher's Note:
1. Area = ½ × AC × (sum of the two perpendiculars).
2. Side lengths alone do not fix the area of a quadrilateral.

Answer: (c) A is true, but R is false.

Explanation:
1. All these triangles have the same base and the same height (the distance between the parallel lines), so A is true.
2. As the vertex slides far away, the two slant sides get longer, so the perimeters differ. R is false.
3. The smallest perimeter occurs for only one position, found by the mirror-reflection argument.

Teacher's Note:
1. Equal area does not mean equal perimeter.
2. Area depends on base and height only.

Answer: (b) 216 m²

Explanation:
1. Outer rectangle: (30 + 2 + 2) m by (20 + 2 + 2) m = 34 m by 24 m.
2. Outer area = 34 × 24 = 816 m²; park area = 30 × 20 = 600 m².
3. Path area = 816 − 600 = 216 m².

Teacher's Note:
1. Add the path width on both sides.
2. Path area = outer area − inner area.

Answer: (a) Only Statement I is true.

Explanation:
1. Join NY. In △XYZ, YN is a median, so area △XYN = \( \frac{1}{2} \) area △XYZ.
2. In △XYN, NM is a median, so area △XMN = \( \frac{1}{2} \) area △XYN.
3. Hence area △XMN = \( \frac{1}{4} \) area △XYZ. Statement I is true and Statement II is false.

Teacher's Note:
1. Use the median property twice.
2. Each median halves the area.

Answer: (d) (i)-(r), (ii)-(s), (iii)-(q), (iv)-(p)

Explanation:
1. 1 in = 2.54 cm (r).
2. 1 ft = 12 in (s).
3. 1 in² = 2.54 × 2.54 = 6.4516 cm² (q).
4. 1 acre = 43,560 ft² (p).

Teacher's Note:
1. A length unit matches a length; an area unit matches an area.
2. Square the conversion factor for square units.

Answer: (d) A is false, but R is true.

Explanation:
1. The diagonals of a rectangle bisect each other, so R is true.
2. The four triangles have equal areas, but only opposite pairs are congruent (unless the rectangle is a square). So A is false.
3. R is in fact what makes the four areas equal.

Teacher's Note:
1. Equal area is weaker than congruent.
2. In a non-square rectangle, adjacent triangles have different shapes.

Answer: (a) Doubling the base of a triangle while keeping its height unchanged doubles its area.

Explanation:
1. Area of a triangle = \( \frac{1}{2} \) × base × height, so doubling the base with the same height doubles the area. (a) is true.
2. Doubling the side of a square makes the area (2s)² = 4s², four times, so (b) is false.
3. Equal perimeters do not force equal areas, so (c) is false; a triangle is half only of the rectangle on its own base and height, so (d) is false.

Teacher's Note:
1. Area is proportional to the base when the height is fixed.
2. Squaring makes 'double' become 'four times'.