CBSE Class 8 Mathematics Area Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. How do we measure the area of a flat region? Answer in one sentence.
Answer:
We count how many non-overlapping unit squares (of side 1 cm, 1 m, etc.) cover the region exactly; this count, which may include fractions of a square, is its area.

Teacher's Note:
1. Unit squares must not overlap or leave gaps.
2. Area is written in square units such as cm².

Question 2. Write the formula for the area of a rectangle in terms of its length and width.
Answer:
Area of a rectangle = length × width.

Teacher's Note:
1. It counts rows × squares in each row.
2. Do not confuse it with perimeter = 2(length + width).

Question 3. Find the area of a rectangular notice board of length 12 cm and width 5 cm.
Answer:
Area = 12 × 5 = 60 cm².

Teacher's Note:
1. Multiply length by width.
2. Write the answer in square units.

Question 4. Write the formula for the area of a triangle.
Answer:
Area of a triangle = \( \frac{1}{2} \) × base × height, where the height is the perpendicular distance from the opposite vertex to the base.

Teacher's Note:
1. The height must be perpendicular to the chosen base.
2. A triangle is half of a rectangle with the same base and height.

Question 5. Find the area of a triangle whose base is 14 cm and whose corresponding height is 6 cm.
Answer:
Area = \( \frac{1}{2} \) × 14 × 6 = 7 × 6 = 42 cm².

Teacher's Note:
1. Halve the even number first.
2. Remember the factor ½.

Question 6. A square flower bed has a side of 9 m. Find its area.
Answer:
Area = side × side = 9 × 9 = 81 m².

Teacher's Note:
1. A square is a rectangle with equal sides.
2. Area of a square = side².

Question 7. State the formula for the area of a parallelogram.
Answer:
Area of a parallelogram = base × corresponding height; any side may be taken as the base, with the height drawn to that side.

Teacher's Note:
1. Use the perpendicular height, not the slant side.
2. No ½ in this formula.

Question 8. Find the area of a parallelogram with base 11 cm and corresponding height 4 cm.
Answer:
Area = 11 × 4 = 44 cm².

Teacher's Note:
1. Base × height for a parallelogram.
2. Units: cm × cm = cm².

Question 9. Write the formula for the area of a rhombus in terms of its diagonals.
Answer:
Area of a rhombus = \( \frac{1}{2} \) × d₁ × d₂, that is, half the product of its diagonals.

Teacher's Note:
1. The diagonals of a rhombus are perpendicular bisectors of each other.
2. Halve the product, not the sum.

Question 10. The diagonals of a rhombus are 16 cm and 9 cm. Find its area.
Answer:
Area = \( \frac{1}{2} \) × 16 × 9 = 8 × 9 = 72 cm².

Teacher's Note:
1. Halve 16 first.
2. Area of a rhombus = ½ d₁d₂.

Question 11. Write the formula for the area of a trapezium.
Answer:
Area of a trapezium = \( \frac{1}{2} \) × height × (sum of the parallel sides).

Teacher's Note:
1. Add the two parallel sides first.
2. The height is the distance between the parallel sides.

Question 12. A trapezium has parallel sides 7 cm and 13 cm and height 6 cm. Find its area.
Answer:
Area = \( \frac{1}{2} \) × 6 × (7 + 13) = 3 × 20 = 60 cm².

Teacher's Note:
1. Sum of parallel sides = 20 cm.
2. Half of the height 6 is 3.

Question 13. A rectangular card is 7 cm by 4 cm. It is cut along a diagonal. What is the area of each piece?
Answer:
A diagonal makes two congruent triangles, so each piece has area \( \frac{1}{2} \) × 7 × 4 = 14 cm².

Teacher's Note:
1. Each triangle is half of the rectangle.
2. Rectangle area 28 cm² ÷ 2 = 14 cm².

Question 14. How many base-height pairs does a triangle have, and what can you say about the areas they give?
Answer:
A triangle has three base-height pairs, one for each side taken as base, and all three give the same area.

Teacher's Note:
1. A longer base goes with a shorter height.
2. This helps find an unknown altitude.

Question 15. The area of a triangle is 48 cm² and its base is 16 cm. Find the corresponding height.
Answer:
\( \frac{1}{2} \) × 16 × h = 48, so 8h = 48 and h = 6 cm.

Teacher's Note:
1. Substitute and solve for h.
2. Check: ½ × 16 × 6 = 48.

Question 16. The area of a triangle is 30 cm² and one of its heights is 5 cm. Find the base corresponding to this height.
Answer:
\( \frac{1}{2} \) × b × 5 = 30, so 2.5b = 30 and b = 12 cm.

Teacher's Note:
1. The base and height must belong to each other.
2. Check: ½ × 12 × 5 = 30.

Question 17. In △ABC, AD is a median and the area of △ABC is 36 cm². Find the area of △ABD.
Answer:
A median divides a triangle into two triangles of equal area, so area △ABD = \( \frac{36}{2} \) = 18 cm².

Teacher's Note:
1. A median ends at the midpoint of the opposite side.
2. Equal bases and the same height give equal areas.

Question 18. Two triangles stand on the same base and their third vertices lie on a line parallel to that base. What can you say about their areas?
Answer:
Their areas are equal, because they have the same base and the same height (the distance between the two parallel lines).

Teacher's Note:
1. Area depends only on base and height.
2. Their perimeters may still be different.

Question 19. What is meant by dissection of a figure?
Answer:
Dissection means cutting a figure into pieces and rearranging the pieces to form a different figure of the same area.

Teacher's Note:
1. No area is lost or gained in a dissection.
2. A parallelogram can be dissected into a rectangle.

Question 20. A parallelogram has area 72 cm² and base 12 cm. Find the height corresponding to this base.
Answer:
12 × h = 72, so h = 72 ÷ 12 = 6 cm.

Teacher's Note:
1. Area of parallelogram = base × height.
2. Divide the area by the base.

Question 21. The area of a rhombus is 84 cm² and one of its diagonals is 12 cm. Find the other diagonal.
Answer:
\( \frac{1}{2} \) × 12 × d = 84, so 6d = 84 and d = 14 cm.

Teacher's Note:
1. Use area = ½ d₁d₂.
2. Check: ½ × 12 × 14 = 84.

Question 22. In quadrilateral ABCD, the diagonal AC = 20 cm, and the perpendicular distances of B and D from AC are 5 cm and 3 cm. Find the area of ABCD.
Answer:
Area = \( \frac{1}{2} \) × 20 × 5 + \( \frac{1}{2} \) × 20 × 3 = 50 + 30 = 80 cm².

Teacher's Note:
1. The diagonal splits the quadrilateral into two triangles.
2. Shortcut: ½ × 20 × (5 + 3) = 80.

Question 23. Express a length of 5 in in centimetres.
Answer:
1 in = 2.54 cm, so 5 in = 5 × 2.54 = 12.7 cm.

Teacher's Note:
1. Multiply by 2.54 to change inches to cm.
2. This is a length, so no squaring.

Question 24. How many square centimetres make 1 in²? Show the step.
Answer:
A square of side 1 in is a square of side 2.54 cm, so 1 in² = 2.54 × 2.54 = 6.4516 cm².

Teacher's Note:
1. Square the length conversion for areas.
2. 1 in² is not 2.54 cm².

Question 25. Give an example of two rectangles in which the one with the larger perimeter has the smaller area.
Answer:
A 1 cm × 10 cm rectangle: perimeter 22 cm, area 10 cm². A 4 cm × 5 cm rectangle: perimeter 18 cm, area 20 cm².
The first has the larger perimeter but the smaller area.

Teacher's Note:
1. Long thin shapes have large perimeters but small areas.
2. Perimeter is not a measure of area.

Question 26. If the side of a square is doubled, how many times does its area become? Give a reason.
Answer:
It becomes 4 times, because a square of side 2s has area (2s)² = 4s², while the original area is s².

Teacher's Note:
1. Both length and width double, so the area is 2 × 2 = 4 times.
2. Not just 2 times.

Question 27. In △XYZ, M and N are the midpoints of XY and XZ. What fraction of the area of △XYZ is the area of △XMN?
Answer:
It is \( \frac{1}{4} \) of the area of △XYZ: YN is a median of △XYZ and NM is a median of △XYN, each halving the area.

Teacher's Note:
1. Join YN first.
2. ½ of ½ = ¼.

Question 28. A path 2 m wide is laid all around a rectangular lawn measuring 15 m by 10 m, on the outside. Find the area of the path.
Answer:
Outer rectangle = 19 m × 14 m = 266 m²; lawn = 15 × 10 = 150 m².
Area of the path = 266 − 150 = 116 m².

Teacher's Note:
1. Add 2 m on each side: 15 + 4 and 10 + 4.
2. Path = outer − inner.

Question 29. The area of a trapezium is 90 cm², its height is 9 cm and one parallel side is 8 cm. Find the other parallel side.
Answer:
\( \frac{1}{2} \) × 9 × (8 + b) = 90, so 8 + b = 20 and b = 12 cm.

Teacher's Note:
1. Multiply both sides by 2 and divide by 9.
2. Check: ½ × 9 × 20 = 90.

Question 30. A rectangular field measures 220 ft by 198 ft. Taking 1 acre = 43,560 ft², how many acres is the field?
Answer:
Area = 220 × 198 = 43,560 ft², which is exactly 1 acre.

Teacher's Note:
1. Find the area in ft² first.
2. Then divide by 43,560 to get acres.