16+ Area Questions for CAT with Solutions

Q1. Find the maximum distance between two points on the perimeter of a rectangular garden whose length and breadth are 100 m and 50 m.

Answer. 111.8 m.

Detailed Solution. Clearly, the two points which are maximum distance apart are the end-points of a diagonal.

∴  Reqd. distance = Length of the diagonal = √(100)² + (50)² m

= √10000 + 2500 m = √12500 m

= 50√5 m = (50 × 2.236) = 111.8 m.

Q2. One side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the area of the field.

Answer. 120 m².

Detailed Solution. Other side = √(17)² - (15)² = √289 - 225 = √64 = 8 m.

∴  Area = (15 × 8) m² = 120 m².

Q3. A lawn is in the form of a rectangle having its sides in the ratio 2 : 3. The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn.

Answer. 50 m.

Detailed Solution.Let length = 2x metres and breadth = 3x metres.

Now, area = (1/6 × 1000) m² + (5000/3) m².

So, 2x × 3x = 5000/3 <=> x² <=> 2500/9 <=> x = (50/3)

∴  Length = 2x = 100/3 m = 33.1/3 m and Breadth = 3x = (3 × 50)/3 m = 50 m.

Q4. Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of ` 12.40 per square metre.

Answer.  ₹ 1934.40.

Detailed Solution. Area of the carpet = Area of the room = (13 × 9) m² = 117 m².

Length of the carpet = (Area/Width) = (117 ÷ 4)/3 m = 156 m.

∴  Cost of carpeting = ₹ (156 × 12.40) = ₹ 1934.40.

Q5. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq. cm. Find the length of the rectangle.

Answer.  20 cm.

Detailed Solution. Let breadth = x. Then, length = 2x. Then,

(2x – 5) (x + 5) – 2x × x = 75 ⇔ 5x – 25 = 75 ⇔ x = 20

∴  Length of the rectangle = 20 cm.

Q6. A rectangular carpet has an area of 120 sq. metres and a perimeter of 46 metres. Find the length of its diagonal.

Answer. 17 m.

Detailed Solution. Let the length and breadth of the rectangle be l and b metres respectively.

Then, 2 (l + b) = 46 ⇒ l + b = 23 ⇒ b = (23 – l).

And, lb = 120 ⇒ l(23 - l) = 120 ⇒ 23l - l² = 120 ⇒ l² -23l + 120 = 0

⇒ l² - 15l - 8l + 120 = 0

⇒ l (l-15) - 8 (l - 15) = 0

⇒ (l - 15) (l - 8) = 0 ⇒ l = 15.

So, l = 15 and b = 8.

∴ Length of diagonal = √(l² + b²) = √(15² + 8²m) = √289 m = 17 m.

Q7. The length of a rectangle is increased by 30%. By what percent would the breadth have to be decreased to maintain the same area?

Answer. 37.5%.

Detailed Solution. Let the length and breadth of the rectangle be l and b units respectively.
Then, area of rectangle = (lb) sq. units.
New length = 160% of l = 8l/5 units.

Desired breadth = ^Area⁄_{New length} = ^lb⁄_{(8l ⁄5)} = ^5b⁄₈ units.

Decrease in breadth = [ b - ^5b⁄₈ ] units = ^3b⁄₈ units.

∴ Decrease% = ( ^3b⁄₈ × ¹⁄_b × 100 )% = ⁷⁵⁄₂% = 37.5%.

Q8. In measuring the sides of a rectangular plot, one side is taken 5% in excess and the other 6% in deficit. Find the error percent in area calculated, of the plot.

Answer.  1.3%.

Detailed Solution. Let the length and breadth of the rectangle be l and b units respectively.

Then, correct area = (lb) sq. units.

Calculated area = ((105 l × 100) / (94 l × 100)) sq. units. = (987 lb / 1000) sq. units.

Error in measurement = (lb - (987/1000).lb) sq. units = (13lb/1000) sq. units.

∴ Error% = (13lb/1000 × 1/lb × 100)%a = 1.3%.

Q9. Instead of walking along two adjacent sides of a rectangular field, a boy took a shor-cut along the diagonal of the field and saved a distance equal to half of the longer side. Find the ratio of the shorter side of the rectangle to the longer side

Answer.  3 : 4.

Detailed Solution. Let the length of the longer side of the field be l and that of the shorter side be b.

Then, diagonal = √l² + b².

∴ (l + b) - √l² + b² = (1/2).l → √l² + b² = 1/2 + b

→ 2√l² + b² = l + 2b → 4(l² + b²) = l² + 4b² + 4lb

→ 3l² = 4lb → 3l = 4b → ^b/_l = 3/4

Hence, required ratio = 3 : 4.

Q10. A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. metre.

Answer.  Cost of gravelling the path =   ₹ (850 × 80/100) × ₹ 680.

Detailed Solution. Area of the plot = (110 × 65) m² = 7150 m²

Area of the plot excluding the path = [(110 - 5) × (65 - 5)] m² = 6300 m²

∴  Area of the path = (7150 - 6300) m² = 850 m²

Cost of gravelling the path =   ₹ (850 × 80/100) × ₹ 680.

Q11. The diagonal of a rectangular field is 15 m and its area is 108 sq. m. What will be the total expenditure in fencing the field at the rate of ` 5 per metre?

Answer. ₹ 210.

Detailed Solution. Let the length and breadth of the rectangle be l and b metres respectively.

Then, √l² + b² = 15 and lb = 108 ⇒ l² + b² = 225 and lb = 108

⇒ (l + b)² = (l² + b²) + 2lb = 225 + 216 = 441

⇒ l + b = √441 = 21.

∴  Perimeter of the field = 2 (l + b) = (2 × 21) m = 42 m.

Hence, cost of fencing = ₹ (42 × 5) = ₹ 210.

Q12. The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.

Answer. 24 cm.

Detailed Solution. Side of first square = (40/4) cm ≈ 10 cm; Side of second square = (32/4) cm ≈ 8 cm.

Area of third square = [(10)² - (8)² ] cm² = (100 - 64) cm² = 36 cm²

Side of third square = √36 cm ≈ 6 cm.

∴ Required perimeter = (6 × 4) cm = 24 cm.

Q13. The length of a rectangle R is 10% more than the side of a square S. The width of the rectangle R is 10% less than the side of the square S. What is the ratio of the area of R to that of S?

Answer. 99:100.

Detailed Solution. Let each side of the square S be x units.

Then, length of rectangle R = 110% of x = (11x/10) units.

And, width of rectangle R = 90% of x = (9x/10) units.

∴ Ratio of areas of R and S = ((11x × 10x) x (9x/10)) : x² = (99x²/100) : x² = 99:100.

Q14. Find the largest size of a bamboo that can be placed in a square of area 100 sq. m.

Answer. 14.14 m.

Detailed Solution. Side of the square = √100 m = 10 m.

Largest size of bamboo = Length of diagonal of the square

= 10 √2 m. = (10 × 1.414) m = 14.14 m.

Q15. A rectangular courtyard, 3.78 m long and 5.25 m broad, is to be paved exactly with square tiles, all of the same size. Find the least number of square tiles covered.

Answer. 450.

Detailed Solution. Area of the room = (378 × 525) cm².

Size of largest square tile = H.C.F. of 378 cm and 525 cm = 21 cm.

Area of 1 tile = (21 × 21) cm².

∴ Number of tiles required = ((378 × 525) / (21 × 21)) ≈ 450.

Q16. Find the area of a square, one of whose diagonals is 3.8 m long.

Answer. 7.22 m²

Detailed Solution. Area of the square = ½ × (diagonal)² = (½ × 3.8 × 3.8) m² = 7.22 m².

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