Area is a formula chapter, so learn the formulas and half the work is done. But the marks in the paper are not lost on the formula, they are lost on two other things. The first is units. If the length is in metres and the breadth is in centimetres, convert before you multiply, and remember that 1 square metre is 10000 square centimetres, not 100. The second is the percentage change question, where the paper says the side is increased by 20% and asks what happens to the area.
Do not calculate with the actual numbers, use the successive change formula and you will finish in ten seconds. One more habit worth building, draw the figure even when the question looks simple, especially for paths, roads and rooms with borders. The diagram will show you at once whether the path is inside or outside, and that single fact decides the answer.
Area Formulas
Rectangle
Area = length × breadth
Perimeter = 2 × (length + breadth)
Diagonal = √(l² + b²)
Square
Area = side², or Area = ½ × diagonal²
Perimeter = 4 × side
Diagonal = side × √2
Triangle
Area = ½ × base × height
Heron’s formula: Area = √[s (s − a) (s − b) (s − c)], where s = (a + b + c) ÷ 2
Equilateral triangle: Area = (√3 ÷ 4) × side², Height = (√3 ÷ 2) × side
Right angled triangle: Hypotenuse = √(base² + perpendicular²)
Triplets: 3, 4, 5 and 5, 12, 13 and 8, 15, 17 and 7, 24, 25
Parallelogram, rhombus, trapezium
Parallelogram: Area = base × height
Rhombus: Area = ½ × d₁ × d₂, and Side = ½ × √(d₁² + d₂²)
Trapezium: Area = ½ × (sum of the parallel sides) × distance between them
Circle
Area = πr²
Circumference = 2πr
Take π as 22/7 when the radius is a multiple of 7, and as 3.14 otherwise
Parts of a circle
Length of an arc = (θ ÷ 360) × 2πr
Area of a sector = (θ ÷ 360) × πr²
Area of a ring = π (R² − r²)
Polygons
Sum of the interior angles = (n − 2) × 180°
Each interior angle of a regular polygon = (n − 2) × 180° ÷ n
Sum of the exterior angles is always 360°
Number of diagonals = n (n − 3) ÷ 2
Paths and roads
Path outside a rectangle of width w: Area = 2w (l + b + 2w)
Path inside a rectangle of width w: Area = 2w (l + b − 2w)
Two roads crossing at the centre, each of width w: Area = w (l + b − w)
The last term is subtracted because the crossing square gets counted twice. Forgetting that is the classic mistake.
Percentage change in area
Length changes by a% and breadth by b%, the area changes by a + b + (a × b ÷ 100) percent
Side of a square or radius of a circle changes by x%, the area changes by 2x + (x² ÷ 100) percent
So a 20% rise in the side gives a 44% rise in the area, without touching the actual numbers.
Unit conversions
1 m = 100 cm, so 1 m² = 10000 cm²
1 km² = 1000000 m²
1 hectare = 10000 m²
Convert everything to one unit before you begin, not at the end.
60 Area Aptitude Questions and Answers (Solved MCQs)
Question 1. The length and breadth of a rectangle are in the ratio 5:3. If its perimeter is 128 metres, what is its area?
a) 840 m²
b) 900 m²
c) 960 m²
d) 1,020 m²
Answer:
c) 960 m² — Let the length and breadth be 5x and 3x. The perimeter is 2(5x+3x) = 16x. Therefore, 16x = 128, giving x = 8. The dimensions are 40 metres and 24 metres, so the area is 40×24 = 960 m².
Question 2. The diagonal of a square is 14√2 centimetres. What is its area?
a) 144 cm²
b) 169 cm²
c) 196 cm²
d) 224 cm²
Answer:
c) 196 cm² — If the side of a square is s, its diagonal is s√2. Therefore, s√2 = 14√2, giving s = 14 cm. Hence, the area is 14² = 196 cm².
Question 3. A triangular field has a base of 20 metres and a perpendicular height of 15 metres. What is its area?
a) 120 m²
b) 140 m²
c) 150 m²
d) 180 m²
Answer:
c) 150 m² — The area of a triangle is 1/2×base×height. Therefore, the area is 1/2×20×15 = 150 m².
Question 4. The area of a parallelogram is 252 cm². If its base is 18 cm, what is the corresponding height?
a) 12 cm
b) 14 cm
c) 16 cm
d) 18 cm
Answer:
b) 14 cm — Area of a parallelogram = base×height. Therefore, 18×height = 252, giving height = 252/18 = 14 cm.
Question 5. The diagonals of a rhombus are 24 cm and 10 cm. What is its area?
a) 100 cm²
b) 110 cm²
c) 120 cm²
d) 140 cm²
Answer:
c) 120 cm² — The area of a rhombus is 1/2×product of its diagonals. Therefore, the area is 1/2×24×10 = 120 cm².
Question 6. The parallel sides of a trapezium are 18 cm and 32 cm. If the perpendicular distance between them is 12 cm, what is its area?
a) 280 cm²
b) 300 cm²
c) 320 cm²
d) 360 cm²
Answer:
b) 300 cm² — The area of a trapezium is 1/2×sum of parallel sides×height. Therefore, the area is 1/2×(18+32)×12 = 1/2×50×12 = 300 cm².
Question 7. The circumference of a circular field is 88 metres. What is its area? Use π = 22/7.
a) 544 m²
b) 588 m²
c) 616 m²
d) 644 m²
Answer:
c) 616 m² — The circumference is 2πr. Therefore, 2×22/7×r = 88, giving r = 14 metres. The area is πr² = 22/7×14² = 616 m².
Question 8. The perimeter of a semicircular region, including its diameter, is 72 cm. What is its area? Use π = 22/7.
a) 264 cm²
b) 288 cm²
c) 308 cm²
d) 336 cm²
Answer:
c) 308 cm² — The perimeter of a semicircle including its diameter is πr+2r. Therefore, r(22/7+2) = 72, or 36r/7 = 72, giving r = 14 cm. The area is 1/2×πr² = 1/2×22/7×14² = 308 cm².
Question 9. What is the area of a quadrant of radius 14 cm? Use π = 22/7.
a) 144 cm²
b) 154 cm²
c) 164 cm²
d) 176 cm²
Answer:
b) 154 cm² — A quadrant is one-fourth of a circle. Therefore, its area is 1/4×πr² = 1/4×22/7×14² = 154 cm².
Question 10. The length of a rectangle is increased by 25%, while its breadth is decreased by 20%. What is the resulting percentage change in its area?
a) 5% increase
b) 5% decrease
c) No change
d) 10% decrease
Answer:
c) No change — The new length is 125% of the original and the new breadth is 80% of the original. Therefore, the new area is 1.25×0.80 = 1 times the original area. Hence, there is no change in the area.
Question 11. A square has a side of 18 cm. A rectangle has the same perimeter as the square, and its length and breadth are in the ratio 5:4. What is the area of the rectangle?
a) 288 cm²
b) 300 cm²
c) 320 cm²
d) 324 cm²
Answer:
c) 320 cm² — The perimeter of the square is 4×18 = 72 cm. Let the rectangle’s length and breadth be 5x and 4x. Then 2(5x+4x) = 72, so 18x = 72 and x = 4. The dimensions are 20 cm and 16 cm. Therefore, the area is 20×16 = 320 cm².
Question 12. One perpendicular side of a right-angled triangle is 7 cm, and its hypotenuse is 25 cm. What is its area?
a) 72 cm²
b) 84 cm²
c) 96 cm²
d) 108 cm²
Answer:
b) 84 cm² — Let the other perpendicular side be x. By the Pythagorean theorem, x²+7² = 25². Therefore, x² = 625−49 = 576, giving x = 24 cm. The area is 1/2×7×24 = 84 cm².
Question 13. What is the area of an equilateral triangle whose side is 12 cm?
a) 24√3 cm²
b) 30√3 cm²
c) 36√3 cm²
d) 48√3 cm²
Answer:
c) 36√3 cm² — The area of an equilateral triangle is √3/4×side². Therefore, the area is √3/4×12² = √3/4×144 = 36√3 cm².
Question 14. A plot of land has an area of 2.5 hectares. What is its area in square metres?
a) 2,500 m²
b) 10,000 m²
c) 20,000 m²
d) 25,000 m²
Answer:
d) 25,000 m² — One hectare equals 10,000 m². Therefore, 2.5 hectares = 2.5×10,000 = 25,000 m².
Question 15. A circle is inscribed inside a square of side 14 cm. What is the area inside the square but outside the circle? Use π = 22/7.
a) 36 cm²
b) 42 cm²
c) 48 cm²
d) 56 cm²
Answer:
b) 42 cm² — The diameter of the inscribed circle equals the side of the square, so its radius is 7 cm. The square’s area is 14² = 196 cm². The circle’s area is 22/7×7² = 154 cm². Therefore, the required area is 196−154 = 42 cm².
Question 16. A rectangular lawn is 40 metres long and 30 metres wide. A path 2 metres wide is constructed outside it. What is the area of the path?
a) 280 m²
b) 288 m²
c) 296 m²
d) 304 m²
Answer:
c) 296 m² — The outer dimensions are 40+4 = 44 metres and 30+4 = 34 metres. The outer area is 44×34 = 1,496 m², while the lawn’s area is 40×30 = 1,200 m². Therefore, the area of the path is 1,496−1,200 = 296 m².
Question 17. A path 3 metres wide runs inside a rectangular field measuring 50 metres by 36 metres, along all four sides. What is the area of the path?
a) 456 m²
b) 468 m²
c) 480 m²
d) 492 m²
Answer:
c) 480 m² — The inner dimensions are 50−6 = 44 metres and 36−6 = 30 metres. The total field area is 50×36 = 1,800 m², and the inner area is 44×30 = 1,320 m². Therefore, the path’s area is 1,800−1,320 = 480 m².
Question 18. A uniform border 2 cm wide is added outside a square photograph of side 20 cm. What is the area of the border?
a) 160 cm²
b) 168 cm²
c) 176 cm²
d) 184 cm²
Answer:
c) 176 cm² — The side of the outer square is 20+2+2 = 24 cm. The outer area is 24² = 576 cm², and the photograph’s area is 20² = 400 cm². Therefore, the border’s area is 576−400 = 176 cm².
Question 19. Two concentric circles have radii 14 cm and 7 cm. What is the area of the ring between them? Use π = 22/7.
a) 420 cm²
b) 441 cm²
c) 462 cm²
d) 484 cm²
Answer:
c) 462 cm² — The area of the ring is π(R²−r²). Therefore, the area is 22/7×(14²−7²) = 22/7×(196−49) = 22/7×147 = 462 cm².
Question 20. From each corner of a square sheet of side 28 cm, a quadrant of radius 7 cm is removed. What is the area of the remaining sheet? Use π = 22/7.
a) 602 cm²
b) 616 cm²
c) 630 cm²
d) 644 cm²
Answer:
c) 630 cm² — The four removed quadrants together form one complete circle of radius 7 cm. The square’s area is 28² = 784 cm², and the removed area is 22/7×7² = 154 cm². Therefore, the remaining area is 784−154 = 630 cm².
Question 21. A stadium consists of a rectangular region 70 metres long and 28 metres wide, with a semicircle attached at each shorter end. What is its total area? Use π = 22/7.
a) 2,464 m²
b) 2,520 m²
c) 2,576 m²
d) 2,604 m²
Answer:
c) 2,576 m² — The two semicircles form one circle with diameter 28 metres and radius 14 metres. The rectangular area is 70×28 = 1,960 m². The circular area is 22/7×14² = 616 m². Therefore, the total area is 1,960+616 = 2,576 m².
Question 22. A room measures 12 metres by 9 metres. An 8-metre by 6-metre carpet is placed at its centre. What area of the floor remains uncovered?
a) 54 m²
b) 60 m²
c) 64 m²
d) 72 m²
Answer:
b) 60 m² — The room’s area is 12×9 = 108 m². The carpet’s area is 8×6 = 48 m². Therefore, the uncovered area is 108−48 = 60 m².
Question 23. How many square tiles of side 40 cm are required to cover a floor having an area of 48 m²?
a) 250
b) 280
c) 300
d) 320
Answer:
c) 300 — The side of each tile is 40 cm = 0.4 metre. Its area is 0.4×0.4 = 0.16 m². Therefore, the number of tiles required is 48/0.16 = 300.
Question 24. A square is inscribed in a circle whose diameter is 28 cm. What is the area inside the circle but outside the square? Use π = 22/7.
a) 196 cm²
b) 210 cm²
c) 224 cm²
d) 238 cm²
Answer:
c) 224 cm² — The circle has radius 14 cm, so its area is 22/7×14² = 616 cm². The diagonal of the inscribed square equals the circle’s diameter, which is 28 cm. The square’s area is diagonal²/2 = 28²/2 = 392 cm². Therefore, the required area is 616−392 = 224 cm².
Question 25. Two rectangular regions measure 12 metres by 8 metres and 10 metres by 6 metres. They overlap over a rectangular region measuring 4 metres by 3 metres. What is the total area covered by the two rectangles?
a) 132 m²
b) 138 m²
c) 144 m²
d) 156 m²
Answer:
c) 144 m² — The two rectangular areas are 12×8 = 96 m² and 10×6 = 60 m². Their overlapping area is 4×3 = 12 m². Since the overlap is counted twice in the sum, the total covered area is 96+60−12 = 144 m².
Question 26. A circular field has a radius of 21 metres. A path 7 metres wide is constructed outside it. What is the area of the path? Use π = 22/7.
a) 1,012 m²
b) 1,056 m²
c) 1,078 m²
d) 1,100 m²
Answer:
c) 1,078 m² — The outer radius is 21+7 = 28 metres. The path’s area is π(28²−21²) = 22/7×(784−441) = 22/7×343 = 1,078 m².
Question 27. Two paths, each 4 metres wide, run through the centre of a rectangular lawn measuring 60 metres by 40 metres. One path is parallel to the length and the other to the breadth. What is the total area occupied by the paths?
a) 368 m²
b) 376 m²
c) 384 m²
d) 400 m²
Answer:
c) 384 m² — The path parallel to the length has an area of 60×4 = 240 m². The other path has an area of 40×4 = 160 m². Their central overlap of 4×4 = 16 m² is counted twice. Therefore, the total area is 240+160−16 = 384 m².
Question 28. A square plot has an area of 1,600 m². A uniform path 5 metres wide runs inside the plot along all four sides. What is the area of the path?
a) 600 m²
b) 650 m²
c) 700 m²
d) 750 m²
Answer:
c) 700 m² — The side of the square is √1,600 = 40 metres. After allowing for the 5-metre path on both sides, the side of the inner square is 40−10 = 30 metres. Therefore, the path’s area is 40²−30² = 1,600−900 = 700 m².
Question 29. A semicircle is attached externally to one side of a square of side 14 cm, with that side serving as the semicircle’s diameter. What is the total area of the resulting figure? Use π = 22/7.
a) 252 cm²
b) 266 cm²
c) 273 cm²
d) 280 cm²
Answer:
c) 273 cm² — The square’s area is 14² = 196 cm². The semicircle has radius 7 cm, so its area is 1/2×22/7×7² = 77 cm². Since the semicircle is attached externally without overlapping the square, the total area is 196+77 = 273 cm².
Question 30. Two concentric semicircles have radii 14 cm and 7 cm. What is the area between their arcs? Use π = 22/7.
a) 210 cm²
b) 220 cm²
c) 231 cm²
d) 242 cm²
Answer:
c) 231 cm² — The required region is half the difference between the areas of the two complete circles. Therefore, its area is 1/2×π(14²−7²) = 1/2×22/7×147 = 231 cm².
Question 31. A rectangular plot is 120 metres long and 80 metres wide. If the plot costs ₹45 per square metre, what is its total cost?
a) ₹4,12,000
b) ₹4,20,000
c) ₹4,32,000
d) ₹4,50,000
Answer:
c) ₹4,32,000 — The area of the plot is 120×80 = 9,600 m². At ₹45 per square metre, the total cost is 9,600×45 = ₹4,32,000.
Question 32. A triangular wall has a base of 18 metres and a perpendicular height of 8 metres. What will it cost to paint the wall at ₹35 per square metre?
a) ₹2,340
b) ₹2,420
c) ₹2,520
d) ₹2,640
Answer:
c) ₹2,520 — The area of the wall is 1/2×18×8 = 72 m². Therefore, the painting cost is 72×35 = ₹2,520.
Question 33. The sides of two square plots are in the ratio 5:7. What is the ratio of their areas?
a) 5:7
b) 10:14
c) 25:49
d) 125:343
Answer:
c) 25:49 — The areas of similar figures are proportional to the squares of their corresponding dimensions. Therefore, the ratio of the areas is 5²:7² = 25:49.
Question 34. The length of a rectangle is increased by 20%. If its area increases by only 8%, what is the percentage change in its breadth?
a) 8% decrease
b) 10% decrease
c) 12% decrease
d) 15% decrease
Answer:
b) 10% decrease — Let the original area be LB. The new area is 1.08LB, while the new length is 1.20L. Therefore, the new breadth is 1.08LB/(1.20L) = 0.90B. Hence, the breadth decreases by 10%.
Question 35. What is the area of a triangular field whose sides are 13 metres, 14 metres and 15 metres?
a) 72 m²
b) 78 m²
c) 84 m²
d) 90 m²
Answer:
c) 84 m² — Using Heron’s formula, the semiperimeter is s = (13+14+15)/2 = 21 metres. The area is √[21(21−13)(21−14)(21−15)] = √(21×8×7×6) = √7,056 = 84 m².
Question 36. What is the area of the triangle whose vertices are (2,3), (8,3) and (5,11)?
a) 20 square units
b) 24 square units
c) 28 square units
d) 32 square units
Answer:
b) 24 square units — The first two points form a horizontal base of length 8−2 = 6 units. The perpendicular distance of (5,11) from the line y = 3 is 11−3 = 8 units. Therefore, the area is 1/2×6×8 = 24 square units.
Question 37. A circular field has an area of 1,386 m². What will it cost to fence the field at ₹75 per metre? Use π = 22/7.
a) ₹9,240
b) ₹9,600
c) ₹9,900
d) ₹10,200
Answer:
c) ₹9,900 — From πr² = 1,386, we get 22/7×r² = 1,386. Thus, r² = 441 and r = 21 metres. The circumference is 2×22/7×21 = 132 metres. Therefore, the fencing cost is 132×75 = ₹9,900.
Question 38. The length of a rectangle is reduced by 20%. By what percentage must its breadth be increased to keep the area unchanged?
a) 20%
b) 22.5%
c) 25%
d) 30%
Answer:
c) 25% — After a 20% reduction, the new length is 80% or 0.8 times the original. To preserve the area, the breadth must be multiplied by 1/0.8 = 1.25. Therefore, it must be increased by 25%.
Question 39. The parallel sides of a trapezium are in the ratio 3:2. Its height is 12 cm and its area is 540 cm². What is the length of the larger parallel side?
a) 48 cm
b) 50 cm
c) 54 cm
d) 60 cm
Answer:
c) 54 cm — Let the parallel sides be 3x and 2x. The area is 1/2×(3x+2x)×12 = 30x. Therefore, 30x = 540, giving x = 18. The larger side is 3×18 = 54 cm.
Question 40. A triangular agricultural field has an area of 3 hectares and a base of 250 metres. What is its perpendicular height?
a) 200 metres
b) 220 metres
c) 240 metres
d) 250 metres
Answer:
c) 240 metres — Three hectares equal 30,000 m². Using area = 1/2×base×height, we get 30,000 = 1/2×250×height. Therefore, the height is 60,000/250 = 240 metres.
Question 41. The area of a square is increased by 44%. By what percentage is its side increased?
a) 18%
b) 20%
c) 22%
d) 24%
Answer:
b) 20% — If the original side is s, the new area is 1.44s². Therefore, the new side is √1.44×s = 1.2s. Hence, the side increases by 20%.
Question 42. The radius of a circle is increased by 10%. By what percentage does its area increase?
a) 10%
b) 20%
c) 21%
d) 22%
Answer:
c) 21% — The new radius is 1.1 times the original radius. Since area is proportional to the square of the radius, the new area is (1.1)² = 1.21 times the original. Therefore, the area increases by 21%.
Question 43. A rectangular plot measuring 60 metres by 40 metres contains a triangular pond with a base of 20 metres and a perpendicular height of 12 metres. What area of the plot remains available for cultivation?
a) 2,160 m²
b) 2,240 m²
c) 2,280 m²
d) 2,320 m²
Answer:
c) 2,280 m² — The rectangular plot’s area is 60×40 = 2,400 m². The pond’s area is 1/2×20×12 = 120 m². Therefore, the area available for cultivation is 2,400−120 = 2,280 m².
Question 44. A hall measures 15 metres by 12 metres. What is the total flooring cost at ₹125 per square metre?
a) ₹21,600
b) ₹22,000
c) ₹22,500
d) ₹23,400
Answer:
c) ₹22,500 — The hall’s area is 15×12 = 180 m². At ₹125 per square metre, the total cost is 180×125 = ₹22,500.
Question 45. A triangle and a parallelogram stand on the same base and between the same parallel lines. If the parallelogram’s area is 480 cm², what is the triangle’s area?
a) 200 cm²
b) 220 cm²
c) 240 cm²
d) 260 cm²
Answer:
c) 240 cm² — The triangle and parallelogram have the same base and perpendicular height. A triangle’s area is half the area of a parallelogram with the same base and height. Therefore, the triangle’s area is 480/2 = 240 cm².
Question 46. Among all rectangles having a perimeter of 100 metres, what is the maximum possible area?
a) 600 m²
b) 625 m²
c) 650 m²
d) 676 m²
Answer:
b) 625 m² — If the length and breadth are l and b, then l+b = 50. For a fixed sum, the product lb is maximum when l = b. Therefore, l = b = 25 metres, and the maximum area is 25×25 = 625 m².
Question 47. The area of a rectangle is 600 cm², and its length is 10 cm greater than its breadth. What is the length of its diagonal?
a) 10√10 cm
b) 10√11 cm
c) 10√12 cm
d) 10√13 cm
Answer:
d) 10√13 cm — Let the breadth be x cm, so the length is x+10 cm. Then x(x+10) = 600, giving x²+10x−600 = 0. Factorising gives (x−20)(x+30) = 0, so the breadth is 20 cm and the length is 30 cm. The diagonal is √(20²+30²) = √1,300 = 10√13 cm.
Question 48. A square has a side of 20 cm. A second square is formed by joining the midpoints of its sides, and a third square is formed by joining the midpoints of the second square’s sides. What is the area of the third square?
a) 80 cm²
b) 100 cm²
c) 120 cm²
d) 200 cm²
Answer:
b) 100 cm² — Joining the midpoints of a square’s sides produces a new square with half the area of the original. The first square has an area of 20² = 400 cm². The second square has an area of 200 cm², and the third square has an area of 100 cm².
Question 49. An equilateral triangle of side 14√3 cm is inscribed in a circle. What is the area of the circle? Use π = 22/7.
a) 462 cm²
b) 528 cm²
c) 616 cm²
d) 686 cm²
Answer:
c) 616 cm² — The circumradius of an equilateral triangle of side a is a/√3. Therefore, the radius is 14√3/√3 = 14 cm. The circle’s area is 22/7×14² = 616 cm².
Question 50. The length of a rectangle is increased by 30%. By what percentage must its breadth be decreased so that its area decreases by 9%?
a) 25%
b) 28%
c) 30%
d) 35%
Answer:
c) 30% — Let the original area be LB. The new area must be 91% of the original. If the proportional decrease in breadth is d, then 1.30L×(1−d)B = 0.91LB. Thus, 1−d = 0.91/1.30 = 0.70. Therefore, d = 0.30, so the breadth must decrease by 30%.
Question 51. A circular field has a radius of 35 metres. A path 7 metres wide runs inside the field along its boundary. What is the area of the path? Use π = 22/7.
a) 1,232 m²
b) 1,320 m²
c) 1,386 m²
d) 1,408 m²
Answer:
c) 1,386 m² — The inner radius is 35−7 = 28 metres. The path’s area is π(35²−28²) = 22/7×(1,225−784) = 22/7×441 = 1,386 m².
Question 52. A square and a circle have equal perimeters. If the square’s side is 11 cm, how much greater is the circle’s area than the square’s area? Use π = 22/7.
a) 22 cm²
b) 28 cm²
c) 33 cm²
d) 44 cm²
Answer:
c) 33 cm² — The square’s perimeter is 4×11 = 44 cm. Therefore, the circle’s circumference is also 44 cm. From 2×22/7×r = 44, we get r = 7 cm. The circle’s area is 154 cm², while the square’s area is 121 cm². The difference is 154−121 = 33 cm².
Question 53. A rectangle has an area of 168 cm² and a diagonal of 25 cm. What is its perimeter?
a) 56 cm
b) 60 cm
c) 62 cm
d) 64 cm
Answer:
c) 62 cm — Let the length and breadth be l and b. Then lb = 168 and l²+b² = 25² = 625. Therefore, (l+b)² = l²+b²+2lb = 625+336 = 961. Thus, l+b = 31, and the perimeter is 2(l+b) = 62 cm.
Question 54. An isosceles triangle has two equal sides of 13 cm each and a base of 10 cm. What is its area?
a) 48 cm²
b) 54 cm²
c) 60 cm²
d) 65 cm²
Answer:
c) 60 cm² — The altitude from the vertex bisects the base into two segments of 5 cm each. Using the Pythagorean theorem, the height is √(13²−5²) = √144 = 12 cm. Therefore, the area is 1/2×10×12 = 60 cm².
Question 55. What is the area of a sector with radius 21 cm and central angle 120°? Use π = 22/7.
a) 420 cm²
b) 441 cm²
c) 462 cm²
d) 484 cm²
Answer:
c) 462 cm² — The sector represents 120/360 = 1/3 of the complete circle. Therefore, its area is 1/3×22/7×21² = 1/3×1,386 = 462 cm².
Question 56. What is the area of a regular hexagon whose side is 8 cm?
a) 64√3 cm²
b) 80√3 cm²
c) 96√3 cm²
d) 108√3 cm²
Answer:
c) 96√3 cm² — A regular hexagon can be divided into six equilateral triangles of side 8 cm. The area of each triangle is √3/4×8² = 16√3 cm². Therefore, the hexagon’s area is 6×16√3 = 96√3 cm².
Question 57. The medians of a triangle meet at its centroid and divide the triangle into six smaller triangles of equal area. If the original triangle’s area is 180 cm², what is the combined area of the two smaller triangles adjacent to one side?
a) 30 cm²
b) 45 cm²
c) 60 cm²
d) 90 cm²
Answer:
c) 60 cm² — The three medians divide the triangle into six smaller triangles of equal area. Each small triangle has an area of 180/6 = 30 cm². Therefore, two such triangles have a combined area of 2×30 = 60 cm².
Question 58. The vertices of a quadrilateral are (0,0), (6,2), (4,8) and (−2,6), taken in order. What is its area?
a) 32 square units
b) 36 square units
c) 40 square units
d) 48 square units
Answer:
c) 40 square units — Using the shoelace formula, the sum of the downward products is 0×2+6×8+4×6+(−2)×0 = 72. The sum of the upward products is 0×6+2×4+8×(−2)+6×0 = −8. Therefore, the area is 1/2|72−(−8)| = 1/2×80 = 40 square units.
Question 59. A rectangular field measures 80 metres by 60 metres. A uniform path is constructed outside it, increasing the total area by 1,184 m². What is the width of the path?
a) 3 metres
b) 4 metres
c) 5 metres
d) 6 metres
Answer:
b) 4 metres — Let the path’s width be x metres. The outer dimensions are 80+2x and 60+2x. Therefore, (80+2x)(60+2x)−4,800 = 1,184. Simplifying gives 4x²+280x−1,184 = 0, or x²+70x−296 = 0. Factorising gives (x−4)(x+74) = 0. Since width must be positive, x = 4 metres.
Question 60. A semicircle has a diameter of 28 cm. A triangle inside it has the diameter as its base and the radius as its perpendicular height. What is the area inside the semicircle but outside the triangle? Use π = 22/7.
a) 98 cm²
b) 105 cm²
c) 112 cm²
d) 126 cm²
Answer:
c) 112 cm² — The semicircle has radius 14 cm, so its area is 1/2×22/7×14² = 308 cm². The triangle’s area is 1/2×28×14 = 196 cm². Therefore, the area inside the semicircle but outside the triangle is 308−196 = 112 cm².
