IUC AcademyHome881. The areas of two equilateral triangles are in the ratio 25 : 36. Their altitudes will be in the ratio
- A. 25 : 36 / 25: 36
- B. 36 : 25 / 36: 25
- C. 5 : 6 / 5: 6
- D. 5: 6
Answer: 5 : 6 / 5: 6
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882. The altitude of an equilateral triangle of side 2 3 cm is
- A. 1 cm 2
- B. 3 cm 4
- C. 3 2 cm
- D. 3 cm
Answer: 3 cm
883. If the area of an equilateral triangle is 24 3 sq. cm, then its perimeter is
- A. 2 6 cm
- B. 4 6 cm
- C. 12 6 cm
- D. 96 cm
Answer: 12 6 cm
884. The perimeter of a triangle is 30 cm and its area is 30 cm2. If the largest side measures 13 cm, then what is the length of the smallest side of the triangle?
- A. 3 cm
- B. 4 cm
- C. 5 cm
- D. 6 cm
Answer: 5 cm
885. In an isosceles triangle, the measure of each of the equal sides is 10 cm and the angle between them is 45º. The area of the triangle is
- A. 25 cm2
- B. 25 2 2 cm2
- C. 25 2 cm2
- D. 25 3 cm2
Answer: 25 2 cm2
886. The perimeter of an isosceles right-angled triangle having an area of 162 cm2 is
- A. 40 cm
- B. 56.5 cm
- C. 61.38 cm
- D. 68.2 cm
Answer: 61.38 cm
887. If the perimeter of an isosceles right triangle is (6 3 2 ) + m, then the area of the triangle is
- A. 4.5 m2
- B. 5.4 m2
- C. 9 m2
- D. 81 m2
Answer: 4.5 m2
888. If the perimeter of a right-angled isosceles triangle is (4 24 + ) cm, the length of the hypotenuse is
- A. 4 cm
- B. 6 cm
- C. 8 cm
- D. 10 cm
Answer: 4 cm
889. The perimeter of a right-angled triangle is 60 cm. Its hypotenuse is 26 cm. The area of the triangle is
- A. 120 cm2
- B. 240 cm2
- C. 390 cm2
- D. 780 cm2
Answer: 120 cm2
890. If the hypotenuse of a right-angled triangle is 41 cm and the area of the triangle is 180 sq. cm, then the difference between the lengths of the legs of the triangle must be
- A. 22 cm
- B. 25 cm
- C. 27 cm
- D. 31 cm
Answer: 31 cm
891. The base and altitude of a right-angled triangle are 12 cm and 5 cm respectively. The perpendicular distance of its hypotenuse from the opposite vertex is
- A. 4 4 13 cm
- B. 8 413 cm
- C. 5 cm
- D. 7 cm
Answer: 8 413 cm
892. The area of a right-angled triangle is 20 sq. cm and one of the sides containing the right angle is 4 cm. The altitude on the hypotenuse is
- A. 41 34 cm
- B. 41 40 cm
- C. 29 20 cm
- D. 20 29 cm
Answer: 20 29 cm
893. One side of a right-angled triangle is twice the other, and the hypotenuse is 10 cm. The area of the triangle is
- A. 20 cm2
- B. 1 33 3 cm2
- C. 40 cm2
- D. 50 cm2
Answer: 20 cm2
894. The hypotenuse of a right-angled isosceles triangle is 5 cm. The area of the triangle is
- A. 5 cm2
- B. 6.25 cm2
- C. 6.5 cm2
- D. 12.5 cm2
Answer: 6.25 cm2
895. The areas of two similar triangles are 12 cm2 and 48 cm2. If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is
- A. 0.525 cm
- B. 4.2 cm
- C. 4.41 cm
- D. 8.4 cm
Answer: 4.2 cm
Explanation: Note : The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. Let the length of the required altitude be x cm. Then, 2 2 2 12 (2.1) (4.41 4) 2.1 2 4.2 cm. 48 x x x = ⇒ = × ⇒= ×=
Explanation: Note : The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. Let the length of the required altitude be x cm. Then, 2 2 2 12 (2.1) (4.41 4) 2.1 2 4.2 cm. 48 x x x = ⇒ = × ⇒= ×=
896. The sides of a triangle are 5 cm, 6 cm, and 7 cm. One more triangle is formed by joining the mid-points of the sides. The perimeter of the second triangle in cm is
- A. 6
- B. 9
- C. 12
- D. 18
Answer: 9
897. If D, E and F are the mid-points of the sides of a DABC, the ratio of the areas of the triangles DEF and DCE is
- A. 1.1 : 1 / 1.1: 1
- B. 1 : 1.1 / 1: 1.1
- C. 1 : 1 / 1: 1
- D. 0.9 : 1 / 0.9: 1
Answer: 1 : 1 / 1: 1
Explanation: Keeping the above note in mind, we have: Required ratio = 1 : 1
Explanation: Keeping the above note in mind, we have: Required ratio = 1 : 1
898. The sides of a triangle are 3 cm, 4 cm and 5 cm. The area (in cm2) of the triangle formed by joining the mid-points of the sides of this triangle is
- A. 3 . 4 / 3. 4
- B. 32
- C. 3
- D. 6
Answer: 32
899. If three sides of a triangle are 6 cm, 8 cm and 10 cm, then the altitude of the triangle, using the largest side as its base, will be
- A. 4.4 cm
- B. 4.8 cm
- C. 6 cm
- D. 8 cm
Answer: 4.8 cm
900. The area of a triangle is 216 cm2 and its sides are in the ratio 3 : 4 : 5. The perimeter of the triangle is
- A. 6 cm
- B. 12 cm
- C. 36 cm
- D. 72 cm
Answer: 72 cm
901. The sides of a triangle are consecutive integers. The perimeter of the triangle is 120 cm. Find the length of the greatest side.
- A. 39 cm
- B. 40 cm
- C. 41 cm
- D. 42 cm
Answer: 41 cm
Explanation: Let the sides of the triangle be x cm, (x + 1) cm and (x + 2) cm respectively. Then, x + (x + 1) + (x + 2) = 120 ⇒ 3x + 3 = 120 ⇒ 3x = 117 ⇒ x = 39. ∴ Length of greatest side = (39 + 2) cm = 41 cm.
Explanation: Let the sides of the triangle be x cm, (x + 1) cm and (x + 2) cm respectively. Then, x + (x + 1) + (x + 2) = 120 ⇒ 3x + 3 = 120 ⇒ 3x = 117 ⇒ x = 39. ∴ Length of greatest side = (39 + 2) cm = 41 cm.
902. The sides of a triangle are in the ratio of 111 ::. 234 If the perimeter is 52 cm, then the length of the smallest side is
- A. 9 cm
- B. 10 cm
- C. 11 cm
- D. 12 cm
Answer: 12 cm
903. The perimeter of an isosceles triangle is equal to 14 cm and the lateral side is to the base in the ratio 5 : 4. The area of the triangle is
- A. 21 cm2
- B. 0.5 21 cm2
- C. 1.5 21 cm2
- D. 2 21 cm2
Answer: 2 21 cm2
904. The three sides of a triangular field are 20 metres, 21 metres and 29 metres long respectively. The area of the field is
- A. 210 sq. m
- B. 215 sq. m
- C. 230 sq. m
- D. None of these
Answer: 210 sq. m
905. The area of a triangle whose sides are of lengths 3 cm, 4 cm and 5 cm is
- A. 8 cm2
- B. 6 cm2
- C. 10 cm2
- D. None of these
Answer: 6 cm2
906. If the area of a triangle is 1176 cm2 and base : cor-responding altitude is 3 : 4, then the altitude of the triangle is
- A. 42 cm
- B. 52 cm
- C. 54 cm
- D. 56 cm
Answer: 56 cm
907. Out of a square of side 8 cm, a triangle is drawn with base as one side of the square and third vertex at any point on the opposite side of the square. What is the area of the remaining portion of the square if the triangle is taken out?
- A. 16 sq. cm
- B. 32 sq. cm
- C. 64 sq. cm
- D. Cannot be determined
Answer: 32 sq. cm
908. What is the area of DPQR, shown in Q. 162?
- A. 2 154 sq. cm
- B. 3 154 sq. cm
- C. 4 308 sq. cm
- D. Cannot be determined
Answer: 3 154 sq. cm
909. In DPQR, side PQ = 32 cm and side PR = 25 cm. What is the measure of side QR ?
- A. 4 154 cm
- B. 2 308 cm
- C. 4 308 cm
- D. None of these
Answer: None of these
910. The area of a triangle is equal to the area of a square whose each side is 60 metres. The height of the triangle is 90 metres. The base of the triangle will be
- A. 65 m
- B. 75 m
- C. 80 m
- D. 85 m
Answer: 80 m
911. ABCD is a rectangle and ABE is a triangle whose vertex E lies on CD. If AB = 5 cm and the area of the triangle is 10 sq. cm, then the perimeter of the rectangle is
- A. 14 cm
- B. 15 cm
- C. 18 cm
- D. 20 cm
Answer: 18 cm
912. The ratio of the areas of a square of side 6 cm and an equilateral triangle of side 6 cm is
- A. 3: 3
- B. 8 : 3 / 8: 3
- C. 6 : 3 / 6: 3
- D. 4 : 3 / 4: 3
Answer: 4 : 3 / 4: 3
913. The area of a right-angled triangle is 40 times its base. What is its height?
- A. 45 cm
- B. 60 cm
- C. 80 cm
- D. Data inadequate
Answer: 80 cm
914. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is
- A. 8 cm
- B. 9 cm
- C. 12.5 cm
- D. 18 cm
Answer: 18 cm
915. Consider the following: I. 9 4 II. 6 6 III. 8 9 Which one of the following conclusions can be drawn form these figures?
- A. The areas of the three figures are all different.
- B. The areas of all the three figures are equal.
- C. The perimeters of the three figures are equal.
- D. The perimeters of figures I and II are equal.
Answer: The areas of all the three figures are equal.
916. The length, breadth and height of the room are in the ratio 3 : 2 : 1. The breadth and height of the room are halved and length of the room is doubled. The area of the four walls of the room will
- A. decrease by 13.64%
- B. decrease by 15%
- C. decrease by 18.75%
- D. decrease by 30%
Answer: decrease by 30%
917. The ratio of the height of a room to its semi-perimeter is 2 : 5. It costs ₹ 260 to paper the walls of the room with paper 50 cm wide at ₹ 2 per metre allowing an area of 15 sq. m for doors and windows. The height of the room is
- A. 2.6 m
- B. 3.9 m
- C. 4 m
- D. 4.2 m
Answer: 4 m
918. The cost of papering the four walls of a room is ₹ 475. Each one of the length, breadth and height of another room is double that of this room. The cost of papering the walls of this new room is
- A. ₹ 712.50
- B. ₹ 950
- C. ₹ 1425
- D. ₹ 1900
Answer: ₹ 1900
919. A hall, whose length is 16 m and the breadth is twice its height, takes 168 m of paper with 2 m as its width to cover its four walls. The area of the floor is
- A. 96 m2
- B. 190 m2
- C. 192 m2
- D. 216 m2
Answer: 192 m2
Explanation: Let the height of the room be x metres. Then, breadth of the room = (2x) metres. Area of 4 walls = [2(16 + 2x) × x] m2 = (32x + 4x2) m2. ∴ 32x + 4x2 = 168 × 2 ⇒ x2 + 8x – 84 = 0 ⇒ x2 + 14x – 6x – 84 = 0 ⇒ x(x + 14) – 6(x + 14) = 0 ⇒ (x + 14) (x – 6) = 0 ⇒ x = 6. Area of the floor = (16 × 12) m2 = 192 m2
Explanation: Let the height of the room be x metres. Then, breadth of the room = (2x) metres. Area of 4 walls = [2(16 + 2x) × x] m2 = (32x + 4x2) m2. ∴ 32x + 4x2 = 168 × 2 ⇒ x2 + 8x – 84 = 0 ⇒ x2 + 14x – 6x – 84 = 0 ⇒ x(x + 14) – 6(x + 14) = 0 ⇒ (x + 14) (x – 6) = 0 ⇒ x = 6. Area of the floor = (16 × 12) m2 = 192 m2
920. A room measures 12.5 m × 9 m × 7 m and has 2 doors of 2.5 m × 1.2 m and 4 windows of 1.5 m × 1 m. Find the cost of painting its walls at ₹ 3.50 per square metre.
- A. ₹ 1050.50
- B. ₹ 1011.50
- C. ₹ 1101.50
- D. Cannot be determined
Answer: ₹ 1011.50
Explanation: Area of 4 walls = 2 (l + b) × h = [2(12.5 + 9) × 7] m2 = 301 m2. Area of 2 doors and 4 windows = [2(2.5 × 1.2) + 4 (1.5 × 1)] m2 = 12 m2. ∴ Area to be painted = (301 – 12) m2 = 289 m2. Cost of painting = ₹ (289 × 3.50) = ₹ 1011.50.
Explanation: Area of 4 walls = 2 (l + b) × h = [2(12.5 + 9) × 7] m2 = 301 m2. Area of 2 doors and 4 windows = [2(2.5 × 1.2) + 4 (1.5 × 1)] m2 = 12 m2. ∴ Area to be painted = (301 – 12) m2 = 289 m2. Cost of painting = ₹ (289 × 3.50) = ₹ 1011.50.
921. The length of a room is double its breadth. The cost of colouring the ceiling at ₹ 25 per sq. m is ₹ 5000 and the cost of painting the four walls at ₹ 240 per sq. m is ₹ 64800. Find the height of the room.
- A. 3.5 m
- B. 4 m
- C. 4.5 m
- D. 5 m
Answer: 4.5 m
922. A tank is 25 m long, 12 m wide and 6 m deep. What will it cost to plaster its walls and floor at 75 paise per sq. m?
- A. ₹ 456
- B. ₹ 458
- C. ₹ 558
- D. ₹ 568
Answer: ₹ 558
923. The area of the four walls of a room is 120 m2 and the length is twice the breadth. If the height of the room is 4 m, then the area of the floor is
- A. 48 m2
- B. 49 m2
- C. 50 m2
- D. 52 m2
Answer: 50 m2
Explanation: Let breadth = x metres and length = (2x) metres. Area of 4 walls = [2 (2x + x) × 4] m2 = (24x) m2. ∴ 24x = 120 ⇒ x = 5. So, length = 10 m, breadth = 5 m. Area of the floor = (10 × 5) m2 = 50 m2.
Explanation: Let breadth = x metres and length = (2x) metres. Area of 4 walls = [2 (2x + x) × 4] m2 = (24x) m2. ∴ 24x = 120 ⇒ x = 5. So, length = 10 m, breadth = 5 m. Area of the floor = (10 × 5) m2 = 50 m2.
924. What will be the area of 4 metre high wall on all four sides of a rectangular hall having perimeter 64 m?
- A. 256 m2
- B. 328 m2
- C. 384 m2
- D. Cannot be determined (d) None of these
Answer: 256 m2
925. Twenty-nine times the area of a square is one square metre less than six times the area of the second square and nine times its side exceeds the perimeter of other square by 1 metre. The difference in the sides of these squares is
- A. 5 m
- B. 54 11 m
- C. 6 m
- D. 11 m
Answer: 6 m
926. The area of a square is twice that of a rectangle. The perimeter of the rectangle is 10 cm. If its length and breadth each is increased by 1 cm, the area of the rectangle becomes equal to the area of the square. The length of side of the square is
- A. 2 3 cm
- B. 3 2 cm
- C. 4 3 cm
- D. 12 cm
Answer: 2 3 cm
Explanation: Let the length and breadth of the rectangle be l cm and b cm respectively. Then, 2(l + b) = 10 ⇒ l + b = 5 ⇒ b = (5 – l) cm. Area of the rectangle = l(5 – l) cm2 = (5l – l 2) cm2. Area of the square = 2 (5l – l 2) cm2 = (10l – 2l2) cm2. ∴ (l + 1) (6 – l) = (10l – 2l 2) ⇒ l2 – 5l + 6 = 0 ⇒ (l – 3) (l – 2) = 0 ⇒ l = 3. Area of the square = (10 × 3 – 2 × 9) cm2 = 12 cm2. ∴ Side of the square = 2 cm 2 3 cm. =
Explanation: Let the length and breadth of the rectangle be l cm and b cm respectively. Then, 2(l + b) = 10 ⇒ l + b = 5 ⇒ b = (5 – l) cm. Area of the rectangle = l(5 – l) cm2 = (5l – l 2) cm2. Area of the square = 2 (5l – l 2) cm2 = (10l – 2l2) cm2. ∴ (l + 1) (6 – l) = (10l – 2l 2) ⇒ l2 – 5l + 6 = 0 ⇒ (l – 3) (l – 2) = 0 ⇒ l = 3. Area of the square = (10 × 3 – 2 × 9) cm2 = 12 cm2. ∴ Side of the square = 2 cm 2 3 cm. =
927. The areas of a square and a rectangle are equal. The length of the rectangle is greater than the length of any side of the square by 5 cm and the breadth is less by 3 cm. Find the perimeter of the rectangle.
- A. 17 cm
- B. 26 cm
- C. 30 cm
- D. 34 cm
Answer: 34 cm
928. The difference of the areas of two squares drawn on two line segments of different lengths is 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2 cm.
- A. 7 cm
- B. 9 cm
- C. 11 cm
- D. 16 cm
Answer: 9 cm
Explanation: Let the lengths of the line segments be x cm and (x + 2) cm. Then, (x + 2)2 – x2 = 32 ⇔ x2 + 4x + 4 – x2 = 32 ⇔ 4x = 28 ⇔ x = 7. ∴ Length of longer line segment = (7 + 2) cm = 9 cm.
Explanation: Let the lengths of the line segments be x cm and (x + 2) cm. Then, (x + 2)2 – x2 = 32 ⇔ x2 + 4x + 4 – x2 = 32 ⇔ 4x = 28 ⇔ x = 7. ∴ Length of longer line segment = (7 + 2) cm = 9 cm.
929. If the side of a square is increased by 5 cm, the area increases by 165 sq. cm. The side of the square is
- A. 12 cm
- B. 13 cm
- C. 14 cm
- D. 15 cm
Answer: 14 cm
Explanation: Let original side = x cm. Then, new side = (x + 5) cm. ∴ (x + 5)2 – x2 = 165 ⇔ x2 + 10x + 25 – x2 = 165 ⇔ 10x = 140 ⇔ x = 14. Hence, the side of the square is 14 cm.
Explanation: Let original side = x cm. Then, new side = (x + 5) cm. ∴ (x + 5)2 – x2 = 165 ⇔ x2 + 10x + 25 – x2 = 165 ⇔ 10x = 140 ⇔ x = 14. Hence, the side of the square is 14 cm.
930. A rectangle becomes a square when its length is reduced by 10 units and its breadth is increased by 5 units. But by this process the area of the rectangle is reduced by 210 sq. units. The area of the rectangle (a) in square units is
- A. 2950 > A < 2900
- B. 2900 > A > 2875
- C. 2925 < A > 2875
- D. 2925 > A >2900
Answer: 2925 > A >2900
Explanation: Let the length and breadth of the rectangle be l and b units respectively. Then, l – 10 = b + 5 ⇒ l – b = 15 ...(i) And, lb – (l – 10) (b + 5) = 210 ⇒ lb – (lb + 5l – 10b – 50) = 210 ⇒ – 5l + 10b = 160 ⇒ – l + 2b = 32 ...(ii) Adding (i) and (ii), we get: b = 47. Putting b = 47 in (i), we get: l = 62. Hence, area of the rectangle = lb = (62 × 47) sq. units = 2914 sq.units. Clearly, 2925 > A > 2900.
Explanation: Let the length and breadth of the rectangle be l and b units respectively. Then, l – 10 = b + 5 ⇒ l – b = 15 ...(i) And, lb – (l – 10) (b + 5) = 210 ⇒ lb – (lb + 5l – 10b – 50) = 210 ⇒ – 5l + 10b = 160 ⇒ – l + 2b = 32 ...(ii) Adding (i) and (ii), we get: b = 47. Putting b = 47 in (i), we get: l = 62. Hence, area of the rectangle = lb = (62 × 47) sq. units = 2914 sq.units. Clearly, 2925 > A > 2900.
931. If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. The perimeter of the original rectangle (in cm) is
- A. 44
- B. 46
- C. 48
- D. 50
Answer: 50
Explanation: Let the length and width of the rectangle be l cm and b cm respectively. Then, (l – 4) (b + 3) = lb ⇒ lb + 3l – 4b – 12 = lb ⇒ 3l – 4b = 12 ...(i) And, l – 4 = b + 3 ⇒ l – b = 7 ...(ii) Multiplying (ii) by 4 and subtracting (i) from it, we get : l = 16. Putting l = 16 in (ii), we get : b = 9. ∴ Perimeter of the original rectangle = 2(l + b) = [2(16 + 9)] cm = 50 cm.
Explanation: Let the length and width of the rectangle be l cm and b cm respectively. Then, (l – 4) (b + 3) = lb ⇒ lb + 3l – 4b – 12 = lb ⇒ 3l – 4b = 12 ...(i) And, l – 4 = b + 3 ⇒ l – b = 7 ...(ii) Multiplying (ii) by 4 and subtracting (i) from it, we get : l = 16. Putting l = 16 in (ii), we get : b = 9. ∴ Perimeter of the original rectangle = 2(l + b) = [2(16 + 9)] cm = 50 cm.
932. The length of one pair of opposite sides of a square is increased by 5 cm on each side; the ratio of the length and the breadth of the newly formed rectangle becomes 3 : 2. What is the area of the original square?
- A. 25 sq. cm
- B. 81 sq. cm
- C. 100 sq. cm
- D. 225 sq. cm
Answer: 225 sq. cm
933. The length and breadth of a square are increased by 40% and 30% respectively. The area of the resulting rectangle exceeds the area of the square by
- A. 35%
- B. 42%
- C. 62%
- D. 82%
Answer: 82%
934. If the diagonal of a square is made 1.5 times, then the ratio of the areas of two squares is
- A. 4 : 3 / 4: 3
- B. 4 : 5 / 4: 5
- C. 4 : 7 / 4: 7
- D. 4 : 9 / 4: 9
Answer: 4 : 9 / 4: 9
935. If the area of a square increases by 69%, then the side of the square increases by
- A. 13%
- B. 30%
- C. 39%
- D. 69%
Answer: 30%
936. An error of 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is
- A. 2%
- B. 2.02%
- C. 4%
- D. 4.04%
Answer: 4.04%
937. If the sides of a square be doubled find the increase of percentage in area.
- A. 100%
- B. 200%
- C. 300%
- D. 400%
Answer: 300%
938. What happens to the area of a square when its side is halved? Its area will
- A. remain same
- B. become half
- C. become one-fourth
- D. become double
Answer: become one-fourth
939. If each side of a square is increased by 50%, the ratio of the area of the resulting square to that of the given square is
- A. 4 : 5 / 4: 5
- B. 5 : 4 / 5: 4
- C. 4 : 9 / 4: 9
- D. 9 : 4 / 9: 4
Answer: 9 : 4 / 9: 4
940. If each side of a square is increased by 10%, its area will be increased by
- A. 10%
- B. 21%
- C. 44%
- D. 100%
Answer: 21%
941. Of the two square fields, the area of one is 1 hectare while the other one is broader by 1%. The difference in their areas is
- A. 100 m2
- B. 101 m2
- C. 200 m2
- D. 201 m2
Answer: 201 m2
Explanation: Area = 1 hect. = 10000 sq. m. ⇒ side = 10000 m 100 m. = Side of the other square = 101 m. Difference in their areas = [(101)2 – (100)2] m2 = [(101 + 100) (101 – 100)] m2 = 201 m2.
Explanation: Area = 1 hect. = 10000 sq. m. ⇒ side = 10000 m 100 m. = Side of the other square = 101 m. Difference in their areas = [(101)2 – (100)2] m2 = [(101 + 100) (101 – 100)] m2 = 201 m2.
942. If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeters is
- A. 225 : 256 / 225: 256
- B. 256 : 225 / 256: 225
- C. 15 : 16 / 15: 16
- D. 16 : 15 / 16: 15
Answer: 15 : 16 / 15: 16
943. The ratio of the areas of two squares, one having its diagonal double than the other, is
- A. 2 : 1 / 2: 1
- B. 2 : 3 / 2: 3
- C. 3 : 1 / 3: 1
- D. 4 : 1 / 4: 1
Answer: 4 : 1 / 4: 1
944. A square S1 encloses another square S2 in such a manner that each corner of S2 is at the mid-point of the side of S1. If A1 is the area of S1 and A2 is the area of S2, then
- A. A1 = 4 A2
- B. A1 = 2 A2
- C. A2 = 2 A1
- D. A1 = A2
Answer: A1 = 2 A2
945. The diagonal of a square is 42 cm. The diagonal of another square whose area is double that of the first square, is
- A. 8 cm
- B. 82 cm
- C. 42 cm
- D. 16 cm
Answer: 8 cm
946. If the perimeters of a square and a rectangle are the same, then the areas A and B enclosed by them would satisfy the condition
- A. A < B
- B. A ≤ B
- C. A > B
- D. A ≥ B
Answer: A > B
Explanation: Take a square of side 4 cm and a rectangle having l = 6 cm, b = 2 cm. Then, perimeter of square = perimeter of rectangle. Area of square = 16 cm2, area of rectangle = 12 cm2. ∴ A > B.
Explanation: Take a square of side 4 cm and a rectangle having l = 6 cm, b = 2 cm. Then, perimeter of square = perimeter of rectangle. Area of square = 16 cm2, area of rectangle = 12 cm2. ∴ A > B.
947. A square and a rectangle have equal areas. If their perimeters are p1 and p2 respectively, then
- A. p1 < p2
- B. p1 = p2
- C. p1 > p2
- D. None of these
Answer: p1 < p2
Explanation: A square and a rectangle with equal areas will satisfy the relation p1 < p2.
Explanation: A square and a rectangle with equal areas will satisfy the relation p1 < p2.
948. The length of a rectangle is 20% more than its breadth. What will be the ratio of the area of a rectangle to that of a square whose side is equal to the breadth of the rectangle?
- A. 2 : 1 / 2: 1
- B. 5 : 6 / 5: 6
- C. 6 : 5 / 6: 5
- D. Data inadequate
Answer: 6 : 5 / 6: 5
949. Area of a square natural lake is 50 sq. kms. A diver wishing to cross the lake diagonally, will have to swim a distance of
- A. 10 miles
- B. 12 miles
- C. 15 miles
- D. None of these
Answer: None of these
950. The area of a square field is 0.5 hectare. Its diagonal would be
- A. 50 m
- B. 50 2 m
- C. 100 m
- D. 250 m
Answer: 100 m
Explanation: Area = 0.5 hectare = (0.5 × 10000) m2 = 5000 m2. ∴ 1 2 (diagonal) 5000 diagonal = 10000 100 m. 2 =⇒ =
Explanation: Area = 0.5 hectare = (0.5 × 10000) m2 = 5000 m2. ∴ 1 2 (diagonal) 5000 diagonal = 10000 100 m. 2 =⇒ =
951. What will be the length of the diagonal of that square plot whose area is equal to the area of a rectangular plot of length 45 metres and breadth 40 metres?
- A. 42.5 metres
- B. 60 metres
- C. 75 metres
- D. Data inadequate
Answer: 60 metres
Explanation: Area = (45 × 40) m2 1 2 (diagonal) 2 ⇔× = 1800 diagonal 60 m. ⇔=
Explanation: Area = (45 × 40) m2 1 2 (diagonal) 2 ⇔× = 1800 diagonal 60 m. ⇔=
952. The area of a square field is 69696 cm2. Its diagonal will be equal to
- A. 313.296 m
- B. 353.296 m
- C. 373.296 m
- D. 393.296 m
Answer: 373.296 m
953. If the length of the diagonal of a square is 20 cm, then its perimeter must be
- A. 10 2 cm
- B. 40 cm
- C. 40 2 cm
- D. 200 cm
Answer: 40 2 cm
954. A man walking at 4 kmph takes 3 minutes to cross a square field diagonally. Find the area of the field.
- A. 18000 m2
- B. 19000 m2
- C. 20000 m2
- D. 25000 m2
Answer: 20000 m2
955. If the length of diagonal AC of a square ABCD is 5.2 cm, then the area of the square is
- A. 10.52 sq.cm
- B. 11.52 sq.cm
- C. 12.52 sq.cm
- D. 13.52 sq.cm
Answer: 13.52 sq.cm
956. A man cut across a square plot along its diagonal instead of walking along two edges. Roughly what percentage of the distance did he save?
- A. 20
- B. 24
- C. 30
- D. 33
Answer: 30
957. A park square in shape has a 3 metre wide road inside it running along its sides. The area occupied by the road is 1764 square metres. What is the perimeter along the outer edge of the road?
- A. 576 metres
- B. 600 metres
- C. 640 metres
- D. Data inadequate
Answer: 600 metres
958. Fencing a square field at ₹ 20 per metre costs ₹ 10,080. What will it cost to lay a 3 m wide pavement inside along the fence at ₹ 50 per sq. metre?
- A. ₹ 37,350
- B. ₹ 73,800
- C. ₹ 77,400
- D. None of these
Answer: ₹ 73,800
959. The area of a rectangle is four times the area of a square. The length of the rectangle is 90 cm and the 5 c m 5cm 2 c m 2cm breadth of the rectangle is 2 rd 3 the side of the square. What is the side of the square?
- A. 9 cm
- B. 10 cm
- C. 20 cm
- D. None of these
Answer: None of these
960. The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq. m. The length of the rectangle is
- A. 30 m
- B. 40 m
- C. 50 m
- D. 60 m
Answer: 30 m