CBSE Class 8 Mathematics Area Short Answer Questions with Answers

Short Answer Questions

Question 1. A rangoli design is drawn on a rectangular slab that is 9 cm long and 6 cm wide. The slab is completely packed with non-overlapping unit squares of side 1 cm.
(i) How many unit squares are needed?
(ii) Write the area of the slab in cm².
Answer:
(i) Along the length, one row holds 9 unit squares, and 6 such rows fit one above the other. So 9 × 6 = 54 unit squares are needed.
(ii) Area = length × width = 9 × 6 = 54 cm².
The number of unit squares packed in a region is its area, so both answers are the same number.
In simple words: 54 squares of 1 cm² fit, so the area is 54 cm².

Teacher's Note:
1. Rows × squares per row = length × width.
2. Each unit square has area 1 cm².

Question 2. The sail of a toy boat is triangular. Its base measures 12 cm and the height drawn on that base measures 7 cm. Write the formula for the area of a triangle and use it to find the area of the sail.
Answer:
Formula: Area of a triangle = \( \frac{1}{2} \) × base × height.
Here base = 12 cm and height = 7 cm.
Area = \( \frac{1}{2} \) × 12 × 7 = 6 × 7 = 42 cm².
So the sail uses 42 cm² of cloth.
In simple words: Half of 12 × 7 gives 42 cm².

Teacher's Note:
1. State the formula before substituting.
2. Halve 12 first to keep the numbers small.

Question 3. Rectangle A measures 10 m × 1 m and Rectangle B measures 4 m × 4 m.
(i) Find the perimeter and the area of each rectangle.
(ii) What does your answer tell you about using perimeter as a measure of area?
Answer:
(i) Rectangle A: perimeter = 2(10 + 1) = 22 m, area = 10 × 1 = 10 m².
Rectangle B: perimeter = 2(4 + 4) = 16 m, area = 4 × 4 = 16 m².
(ii) A has the larger perimeter (22 m > 16 m) but the smaller area (10 m² < 16 m²).
So perimeter cannot be used as a measure of area; area must be found by counting unit squares.
In simple words: The longer boundary encloses less space here, so perimeter does not tell area.

Teacher's Note:
1. Perimeter is a length (m); area is in square units (m²).
2. Long, thin shapes have big perimeters but small areas.

Question 4. A rectangular sheet of card measuring 16 cm by 9 cm is cut along one of its diagonals. Find the area of each triangular piece and give the reason for your answer.
Answer:
Area of the whole sheet = 16 × 9 = 144 cm².
A diagonal divides a rectangle into two congruent triangles, so the two pieces have equal areas.
Area of each piece = \( \frac{144}{2} \) = 72 cm².
The triangle formula gives the same: \( \frac{1}{2} \) × 16 × 9 = 72 cm².
In simple words: Each triangle is half of 144 cm², i.e. 72 cm².

Teacher's Note:
1. Congruent pieces have equal area.
2. Check with ½ × base × height.

Question 5. A floor tile is shaped like a parallelogram with base 8 cm and corresponding height 5.5 cm. Find the area of the tile.
Answer:
Area of a parallelogram = base × corresponding height.
Area = 8 × 5.5 = 44 cm².
The height is the perpendicular distance between the base and the opposite side, not the slant side of the tile.
In simple words: 8 × 5.5 = 44 cm².

Teacher's Note:
1. 8 × 5.5 = 8 × 5 + 8 × 0.5 = 40 + 4.
2. Use the perpendicular height.

Question 6. The two diagonals of a rhombus-shaped window grill measure 18 cm and 14 cm. State the formula for the area of a rhombus and find the area of the grill.
Answer:
Formula: Area of a rhombus = \( \frac{1}{2} \) × product of its diagonals = \( \frac{1}{2} \) × d₁ × d₂.
Area = \( \frac{1}{2} \) × 18 × 14 = 9 × 14 = 126 cm².
So the grill covers 126 cm².
In simple words: Half of 18 × 14 gives 126 cm².

Teacher's Note:
1. Halve the product of the diagonals.
2. 252 would be the answer without halving, which is wrong.

Question 7. A name plate is shaped like a trapezium. Its parallel sides are 11 cm and 7 cm long, and the distance between them is 6 cm. Find the area of the name plate.
Answer:
Area of a trapezium = \( \frac{1}{2} \) × height × sum of the parallel sides.
Sum of parallel sides = 11 + 7 = 18 cm; height = 6 cm.
Area = \( \frac{1}{2} \) × 6 × 18 = 3 × 18 = 54 cm².
In simple words: Half of 6 × 18 = 54 cm².

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

Question 8. In △PQR, M is the midpoint of QR and PM is joined. If the area of △PQR is 48 cm², write the area of △PQM and give the reason.
Answer:
Area of △PQM = 24 cm².
Reason: PM joins the vertex P to the midpoint M of QR, so PM is a median.
△PQM and △PRM have equal bases (QM = MR) and the same height (the perpendicular from P to QR).
Equal bases and equal heights give equal areas, so each part is half of 48 cm², that is 24 cm².
In simple words: A median cuts the triangle into two equal-area halves, so 48 ÷ 2 = 24 cm².

Teacher's Note:
1. A median always halves the area.
2. The two halves need not be congruent.

Question 9. A triangular flower bed has an area of 54 m² and its base measures 12 m.
(i) Find the height of the bed on that base.
(ii) Check your answer by substituting it back into the area formula.
Answer:
(i) \( \frac{1}{2} \) × base × height = area ⇒ \( \frac{1}{2} \) × 12 × h = 54 ⇒ 6h = 54 ⇒ h = 9 m.
(ii) Check: \( \frac{1}{2} \) × 12 × 9 = 6 × 9 = 54 m², which is the given area. So the height is correct.
In simple words: The height is 9 m, and ½ × 12 × 9 gives back 54 m².

Teacher's Note:
1. Solve for the unknown step by step.
2. Always substitute back to check.

Question 10. A rectangular park is 30 m long and 20 m wide. A uniform path 2 m wide is laid all around the park on the outside. Find the area of the path.
Answer:
The path adds 2 m on each of the four sides.
Outer length = 30 + 2 + 2 = 34 m; outer width = 20 + 2 + 2 = 24 m.
Area of the outer rectangle = 34 × 24 = 816 m².
Area of the park = 30 × 20 = 600 m².
Area of the path = 816 − 600 = 216 m².
In simple words: Big rectangle minus park: 816 − 600 = 216 m².

Teacher's Note:
1. Add the width twice to each side.
2. Path = outer area − inner area.

Question 11. The figure shows an L-shaped plot made by removing a 6 m × 4 m corner from a 14 m × 10 m rectangle. Find the area of the plot in two different ways.

Answer:
Method 1 (subtracting): Full rectangle = 14 × 10 = 140 m²; removed corner = 6 × 4 = 24 m². Area of the plot = 140 − 24 = 116 m².
Method 2 (splitting): Cut the plot into a lower strip 14 m × 6 m = 84 m² and an upper part 8 m × 4 m = 32 m² (8 = 14 − 6). Total = 84 + 32 = 116 m².
Both methods give the same area, 116 m².
In simple words: Either 140 − 24 or 84 + 32 gives 116 m².

Teacher's Note:
1. Two methods agreeing is a good check.
2. Find missing lengths: 14 − 6 = 8 and 10 − 4 = 6.

Question 12. A line l is drawn parallel to BC, and P, Q, R are three different points on l.
(i) Explain why △PBC, △QBC and △RBC have equal areas.
(ii) If BC = 10 cm and the distance between l and BC is 6 cm, find this common area.
Answer:
(i) All three triangles have the same base BC. Their third vertices lie on l, and l ∥ BC, so the perpendicular distance from each of P, Q, R to BC is the same. Equal base and equal height give equal areas.
(ii) Common area = \( \frac{1}{2} \) × 10 × 6 = 30 cm².
In simple words: Same base and same height, so each triangle has area 30 cm².

Teacher's Note:
1. The distance between parallel lines is the same everywhere.
2. Triangles between the same parallels on the same base have equal areas.

Question 13. In △ABC, AB = 10 cm and the altitude drawn from C to AB is 6 cm long.
(i) Find the area of △ABC.
(ii) If AC = 12 cm, find the length of the altitude drawn from B to AC.
Answer:
(i) Taking AB as base with the altitude from C as height: area = \( \frac{1}{2} \) × 10 × 6 = 30 cm².
(ii) Now take AC = 12 cm as base and let the altitude from B be h. The area is still 30 cm², so
\( \frac{1}{2} \) × 12 × h = 30 ⇒ 6h = 30 ⇒ h = 5 cm.
In simple words: Area is 30 cm², so on base 12 cm the height must be 5 cm.

Teacher's Note:
1. Every base-height pair gives the same area.
2. This is how an unknown altitude is found.

Question 14. A square flower bed of side 6 m is laid at the centre of a rectangular lawn measuring 15 m by 10 m. Find the area of the lawn left outside the flower bed, and express it as a fraction of the whole lawn in its simplest form.
Answer:
Area of the lawn = 15 × 10 = 150 m².
Area of the flower bed = 6 × 6 = 36 m².
Area left outside = 150 − 36 = 114 m².
Fraction = \( \frac{114}{150} \) = \( \frac{19}{25} \) (dividing both by 6).
In simple words: 114 m² is left, which is 19/25 of the lawn.

Teacher's Note:
1. Subtract the inner area from the outer area.
2. Divide by the HCF, 6, to simplify.

Question 15. One side of a parallelogram is 12 cm and the height drawn on it is 5 cm.
(i) Find the area of the parallelogram.
(ii) If the adjacent side measures 10 cm, find the height corresponding to that side.
Answer:
(i) Area = base × height = 12 × 5 = 60 cm².
(ii) Any side can be the base with its own height; the area stays 60 cm². With the 10 cm side as base: 10 × h = 60 ⇒ h = 6 cm.
In simple words: The area is 60 cm², so on the 10 cm side the height is 6 cm.

Teacher's Note:
1. Same area, different base-height pairs.
2. Check: 10 × 6 = 12 × 5.

Question 16. A tabletop measures 30 in by 20 in. Taking 1 in = 2.54 cm, find the area of the tabletop (i) in in² and (ii) in cm².
Answer:
(i) Area = 30 × 20 = 600 in².
(ii) 1 in² = 2.54 × 2.54 = 6.4516 cm², so area = 600 × 6.4516 = 3870.96 cm².
Check by converting the sides first: 30 in = 76.2 cm, 20 in = 50.8 cm, and 76.2 × 50.8 = 3870.96 cm².
In simple words: 600 in², which is 3870.96 cm².

Teacher's Note:
1. Square the factor 2.54 for areas.
2. Converting sides first must give the same answer.

Question 17. Among all rectangles of perimeter 24 cm whose side lengths are whole numbers of centimetres, find the one with the greatest area and the one with the least area. Hence explain why equal perimeters do not force equal areas.
Answer:
Perimeter 24 cm means length + width = 12 cm.
Pairs and areas: 1 and 11 → 11 cm²; 2 and 10 → 20 cm²; 3 and 9 → 27 cm²; 4 and 8 → 32 cm²; 5 and 7 → 35 cm²; 6 and 6 → 36 cm².
Greatest area: the square 6 cm × 6 cm, 36 cm². Least area: 1 cm × 11 cm, 11 cm².
All have the same perimeter, yet the areas range from 11 cm² to 36 cm². Perimeter measures the boundary and area measures the surface, so one does not decide the other.
In simple words: Same 24 cm boundary, but areas go from 11 cm² to 36 cm².

Teacher's Note:
1. List the cases in order so none is missed.
2. The square gives the largest area for a fixed perimeter.

Question 18. In △PQR, D is the midpoint of PQ and E is the midpoint of PR. Join DR. Using the property that a line from a vertex to the midpoint of the opposite side halves the area, find what fraction of the area of △PQR is the area of △PDE. Justify each step.
Answer:
Step 1. In △PQR, R is joined to D, the midpoint of the opposite side PQ. So RD is a median and area △PDR = \( \frac{1}{2} \) area △PQR.
Step 2. In △PDR, E is the midpoint of PR, and DE joins vertex D to it. So DE is a median of △PDR and area △PDE = \( \frac{1}{2} \) area △PDR.
Step 3. Combining: area △PDE = \( \frac{1}{2} \) × \( \frac{1}{2} \) area △PQR = \( \frac{1}{4} \) area △PQR.
In simple words: Halving twice gives one-fourth.

Teacher's Note:
1. Use the median property in two different triangles.
2. ½ × ½ = ¼.

Question 19. A trapezium has an area of 144 cm², a height of 12 cm, and one parallel side of length 14 cm.
(i) Find the length of the other parallel side.
(ii) If the height is doubled while both parallel sides are halved, what happens to the area? Justify with the formula.
Answer:
(i) Let the other side be b. \( \frac{1}{2} \) × 12 × (14 + b) = 144 ⇒ 6(14 + b) = 144 ⇒ 14 + b = 24 ⇒ b = 10 cm.
(ii) New height = 24 cm; new parallel sides = 7 cm and 5 cm. New area = \( \frac{1}{2} \) × 24 × (7 + 5) = 12 × 12 = 144 cm².
The area is unchanged: doubling the height doubles the area, halving both sides halves their sum, and 2 × \( \frac{1}{2} \) = 1.
In simple words: The other side is 10 cm, and the area stays 144 cm².

Teacher's Note:
1. Divide 144 by 6 to get the sum of parallel sides.
2. A ×2 change and a ×½ change cancel.

Question 20. A rhombus has diagonals of lengths 24 cm and 18 cm, and each of its sides measures 15 cm.
(i) Find its area using the diagonals.
(ii) A rhombus is also a parallelogram. Use this to find the perpendicular distance between a pair of its opposite sides.
Answer:
(i) Area = \( \frac{1}{2} \) × 24 × 18 = 12 × 18 = 216 cm².
(ii) As a parallelogram, area = base × height. Taking a 15 cm side as base and the distance between that pair of opposite sides as h: 15 × h = 216 ⇒ h = \( \frac{216}{15} \) = 14.4 cm.
Check: 15 × 14.4 = 216 cm², matching part (i).
In simple words: The area is 216 cm², so the distance between opposite sides is 216 ÷ 15 = 14.4 cm.

Teacher's Note:
1. One shape, two formulas, same area.
2. The height 14.4 cm is less than the side 15 cm, as it should be.