IUC AcademyHome161. The side length of a cube is 5.6 cm. What is the volume of the largest sphere that can be taken out of the cube.
- A. 91.98 cm3
- B. 96.98 cm3
- C. 90.69 cm3
- D. 99.96 cm3
Answer: 91.98 cm3
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162. A cube of side 5cm is assembled using 120 cubes of edge 1cm. One cube is removed from the center of each of the face of the cube. What is the surface area of (in cm 2) the remaining solid?
- A. 150
- B. 174
- C. 155
- D. 144
Answer: 174
Explanation: Total surface area of cube = 6a 2 = 6 × 5 2 = 150 cm2 The overall surface area of the remaining solid is = 150 - 6 × 1 + 5 × (1)2 × 6 = 150 - 6 + 30 174= cm2
Explanation: Total surface area of cube = 6a 2 = 6 × 5 2 = 150 cm2 The overall surface area of the remaining solid is = 150 - 6 × 1 + 5 × (1)2 × 6 = 150 - 6 + 30 174= cm2
163. A solid cube, whose each edge of length 48 cm, is melted. Identical solid cubes, each of volume 64 cm 3, are made out of this molten cube, without any wastage. How many such small cubes are obtained ?
- A. 1738
- B. 1748
- C. 1718
- D. 1728
Answer: 1728
164. If the surface area of a cube is 5046 cm 2, then what will be the volume of the cube?
- A. 26189 cm 3
- B. 24389 cm 3
- C. 22812 cm 3
- D. 25145 cm 3
Answer: 24389 cm 3
Explanation: Total surface area of cube = 6 (side)2 6 (side)2 = 5046 (side)2 = 841 side = 29cm Volume of cube = (side)3 = (29)3 = 24389 cm 3
Explanation: Total surface area of cube = 6 (side)2 6 (side)2 = 5046 (side)2 = 841 side = 29cm Volume of cube = (side)3 = (29)3 = 24389 cm 3
165. If the volume of a cube is 729 cm 3. Find the total surface area of the cube.
- A. 488 cm2
- B. 486 cm2
- C. 484 cm2
- D. 482 cm2
Answer: 486 cm2
Explanation: Volume of cube = (side)3 (Side)3 = 729 side = 9cm Total surface area of cube = 6 × (side)2 = 6 × (9)2 = 486 cm 2
Explanation: Volume of cube = (side)3 (Side)3 = 729 side = 9cm Total surface area of cube = 6 × (side)2 = 6 × (9)2 = 486 cm 2
166. The base of a right prism is a triangle with s ides 20 cm, 21 cm and 29 cm. If its volume is 7560 cm3, then its lateral surface area (in cm2) is
- A. 2448
- B. 2556
- C. 2520
- D. 2484
Answer: 2520
Explanation: Volume of prism = Area of base × height 17560 20 21 H2= × × × 7560 2H 3620 21×= = × Perimeter = 20 + 21 + 29 = 70 cm Lateral surface area = perimeter × height = 70 × 36 = 2520 cm 2
Explanation: Volume of prism = Area of base × height 17560 20 21 H2= × × × 7560 2H 3620 21×= = × Perimeter = 20 + 21 + 29 = 70 cm Lateral surface area = perimeter × height = 70 × 36 = 2520 cm 2
167. The length of two parallel sides of a trapezium are 30 cm and 40 cm. If the area of the trapezium is 350 cm 2, then what is the value (in cm) of its height?
- A. 8
- B. 10
- C. 15
- D. 12
Answer: 8
Explanation: (b) Area of trapezium = 12 (Sum of sides) × height According to the question, ∴ 1350 (30 40)2= × + × height Height = 350 270× = 10 cm
Explanation: (b) Area of trapezium = 12 (Sum of sides) × height According to the question, ∴ 1350 (30 40)2= × + × height Height = 350 270× = 10 cm
168. Given that the area of a triangle is A. Its perimeter, inradius, circumradius and the average of medians are p, r, R and d respectively. The ratio A : p will be equal to
- A. (R - r)2 : r
- B. r : 1
- C. (R + r)2 : d
- D. r : 2
Answer: r : 2
169. A regular octagon ABCDEFGH is inscribed in a circle with centre O. What will be the ratio of ∠OAB to ∠AOB?
- A. 4 : 3
- B. 3 : 1
- C. 8 : 3
- D. 3 : 2
Answer: 3 : 2
170. The area of a triangle is 15 sq cm and the radius of its incircle is 3 cm. Its perimeter is equal to
- A. 20 cm
- B. 10 cm
- C. 12 cm
- D. 5 cm
Answer: 10 cm
171. The perimeter of a square is 64 cm. The area of a rectangle is 16 cm 2 less than the area of the square. If the length of the rectangle is 20 cm. What is the perimeter of the rectangle?
- A. 18 cm
- B. 64 cm
- C. 12 cm
- D. 15 cm
Answer: 64 cm
Explanation: Given that- Perimeter of square (4a) = 64 cm then side = 644 = 16 cm Area of square = (side)2 = (16)2 = 256 cm2 According to the question, Area of rectangle = 256 - 16 = 240 cm Length of rectangle (l) = 20 cm Area of rectangle = l × b 240 = 20 × b b = 12 cm Hence, the perimeter of rectangle = 2 (l + b) = 2 ( 20+12) = 2 × 32 = 64 cm
Explanation: Given that- Perimeter of square (4a) = 64 cm then side = 644 = 16 cm Area of square = (side)2 = (16)2 = 256 cm2 According to the question, Area of rectangle = 256 - 16 = 240 cm Length of rectangle (l) = 20 cm Area of rectangle = l × b 240 = 20 × b b = 12 cm Hence, the perimeter of rectangle = 2 (l + b) = 2 ( 20+12) = 2 × 32 = 64 cm
172. What is the Circumference of the largest circl e which can be inscribed in a square of side 14 cm?
- A. 66 cm
- B. 88 cm
- C. 22 cm
- D. 44 cm
Answer: 44 cm
Explanation: ∵ Side of square = 14 cm ∴ Diameter of circle = 14 cm Radius of circle 14 72= = cm then circumference of circle 2 r= π 222 7 7= × × 44cm=
Explanation: ∵ Side of square = 14 cm ∴ Diameter of circle = 14 cm Radius of circle 14 72= = cm then circumference of circle 2 r= π 222 7 7= × × 44cm=
173. The lengths of two diagonals of a rhombus are 15 cm and 20 cm. What is the area (in cm 2) of the rhombus?
- A. 120
- B. 150
- C. 190
- D. 180
Answer: 120
Explanation: (b) Given that- Diagonal (d1) = 15 cm (d 2) = 20 cm Area of rhombus 1 2d dA 2×= 15 202×= A = 150 cm2
Explanation: (b) Given that- Diagonal (d1) = 15 cm (d 2) = 20 cm Area of rhombus 1 2d dA 2×= 15 202×= A = 150 cm2
174. The length of the diagonals of a rhombus are 48 cm and 20 cm. What is the perimeter of the rhombus?
- A. 100 cm
- B. 200 cm
- C. 120 cm
- D. 104 cm
Answer: 104 cm
Explanation: Given, Length of diagonals of rhombus = 48 cm and 20 cm The diagonals of the rhombus bisect each other at r ight angles. BO = 24 cm, CO = 10 cm In right angle triangle ∆BOC BC2 = BO2 + CO2 BC 2 = (24)2 + (10)2 BC 2 = 576 + 100 BC = 676 BC = 26cm Hence, length of each side of rhombus = 26cm Hence perimeter = 4 × 26cm = 104cm
Explanation: Given, Length of diagonals of rhombus = 48 cm and 20 cm The diagonals of the rhombus bisect each other at r ight angles. BO = 24 cm, CO = 10 cm In right angle triangle ∆BOC BC2 = BO2 + CO2 BC 2 = (24)2 + (10)2 BC 2 = 576 + 100 BC = 676 BC = 26cm Hence, length of each side of rhombus = 26cm Hence perimeter = 4 × 26cm = 104cm
175. The area of the largest triangle that can be inscribed in a semi-circle of radius 6 cm is
- A. 34 cm2
- B. 36 cm2
- C. 38 cm2
- D. 35 cm2
Answer: 36 cm2
176. The area of the largest triangle that can be inscribed in a semi-circle of radius 4 cm (in cm2) is
- A. 18 cm2
- B. 12 cm2
- C. 14 cm2
- D. 16 cm2
Answer: 16 cm2
Explanation: Radius of semi-circle = 4 cm Diameter of semi-circle (AB) = 8 cm The largest triangle that can be inscribed in a sem i- circle will have a base equal to the diameter and Height of triangle is equal to radius of semi-circl e. Hence the Area of largest triangle = 1 8 42 × × = 16 cm2
Explanation: Radius of semi-circle = 4 cm Diameter of semi-circle (AB) = 8 cm The largest triangle that can be inscribed in a sem i- circle will have a base equal to the diameter and Height of triangle is equal to radius of semi-circl e. Hence the Area of largest triangle = 1 8 42 × × = 16 cm2
177. ABCDEF is a regular hexagon. What is the ratio of the area of triangle ACE and area of triangle AEF?
- A. 6 : 1
- B. 4 : 1
- C. 3 : 1
- D. 5 : 1
Answer: 3 : 1
Explanation: A regular hexagon is made up of six equilateral triangles. ∆OAB, ∆OBC, ∆OCD, ∆ODE, ∆OEF and ∆OAF are equilateral triangles. Area of ∆ACE 12= × Area of six equilateral triangles Area of ∆AEF 16= × Area of six equilateral triangles ∴ ar (∆ACE) : ar (∆AEF) = 3 : 1
Explanation: A regular hexagon is made up of six equilateral triangles. ∆OAB, ∆OBC, ∆OCD, ∆ODE, ∆OEF and ∆OAF are equilateral triangles. Area of ∆ACE 12= × Area of six equilateral triangles Area of ∆AEF 16= × Area of six equilateral triangles ∴ ar (∆ACE) : ar (∆AEF) = 3 : 1
178. The diagonal of rectangle is 15 cm and length is 12 cm. Find the area of the rectangle.
- A. 108 cm2
- B. 112 cm2
- C. 114 cm2
- D. 116 cm2
Answer: 108 cm2
179. The breadth of a rectangle is four times of it s length. If the area of the rectangle is 1764, then what is the breadth of the rectangle?
- A. 21 cm
- B. 84 cm
- C. 44 cm
- D. 56 cm
Answer: 84 cm
180. The length of a rectangle is 16 cm. If the length of the diagonal is 20 cm, then what will be the breadth of the rectangle?
- A. 8 cm.
- B. 12 cm
- C. 10 cm.
- D. 14 cm.
Answer: 12 cm
181. The length and breadth of a rectangle are in the ratio 8 : 3. If the perimeter of the rectangle is 220 cm, then what is the length (in cm) of the rectangle?
- A. 60
- B. 70
- C. 80
- D. 85
Answer: 80
Explanation: Let length of the rectangle = 8x cm. and breadth = 3x cm. Perimeter of the rectangle = 2 (l + b) 220 = 2(8x + 3x) 220 = 2 × 11x x = 10 Hence length of the rectangle = 8 × 10 = 80cm.
Explanation: Let length of the rectangle = 8x cm. and breadth = 3x cm. Perimeter of the rectangle = 2 (l + b) 220 = 2(8x + 3x) 220 = 2 × 11x x = 10 Hence length of the rectangle = 8 × 10 = 80cm.
182. The perimeter of a rectangle is 80cm and its area is 375cm 2. What is the difference between the length and the breadth of the rectangle?
- A. 20cm
- B. 10cm
- C. 16cm
- D. 12cm
Answer: 10cm
Explanation: Let the length and breadth of the rectangle are l and b respectively. Area of the circle (lb) = 375cm2 Perimeter of rectangle = 2(l + b) = 80 ∴ l + b = 40 ∴ (l - b)2 + 4lb = (l + b)2 or (l - b)2 + 4lb = (40)2 ( l - b)2 + 4 × 375 = 1600 ( l - b)2 = 1600 - 1500 l - b = 100 l - b = 10cm
Explanation: Let the length and breadth of the rectangle are l and b respectively. Area of the circle (lb) = 375cm2 Perimeter of rectangle = 2(l + b) = 80 ∴ l + b = 40 ∴ (l - b)2 + 4lb = (l + b)2 or (l - b)2 + 4lb = (40)2 ( l - b)2 + 4 × 375 = 1600 ( l - b)2 = 1600 - 1500 l - b = 100 l - b = 10cm
183. The ratio of the length and the perimeter of a rectangle is 2 : 7. What is the ratio of the length and breadth of the rectangle?
- A. 5 : 4
- B. 4 : 3
- C. 4 : 5
- D. 5 : 3
Answer: 4 : 3
184. If the perimeter of a certain rectangle is 50 units and its area is 150 sq. units, then how many units is the length of its shorter side?
- A. 15
- B. 9
- C. 10
- D. 12
Answer: 10
Explanation: Let the sides of the rectangle are l and b respectively 2 (l+b) = 50 l + b = 25 ......... (1) l.b = 150 ............ (2) (l-b)2 = (l + b)2 - 4lb = 625-600 l - b = 5 ......... (3) on solving both equations (1) and (3), 2b = 20 b = 10 Hence, shorter side = 10 units
Explanation: Let the sides of the rectangle are l and b respectively 2 (l+b) = 50 l + b = 25 ......... (1) l.b = 150 ............ (2) (l-b)2 = (l + b)2 - 4lb = 625-600 l - b = 5 ......... (3) on solving both equations (1) and (3), 2b = 20 b = 10 Hence, shorter side = 10 units
185. A circle is inscribed in a right-angled triangle. The lengths of the two sides containing the right angle are 15 cm and 8 cm. What is the radius of the in-circle?
- A. 4.5 cm
- B. 3 cm
- C. 3.75 cm
- D. 4 cm
Answer: 3 cm
186. The perimeter of a rectangle is 86 cm. The numbers representing its area and breadth are in the ratio of 9:1, respectively. The breadth of the rectangle is
- A. 34 cm
- B. 32 cm
- C. 30 cm
- D. 36 cm
Answer: 34 cm
Explanation: Given, ∵ Perimeter of rectangle = 86 cm. ∴ 2(l+b) = 86 ......[where l = length and b = breadth of rectangle] l+b = 862 = 43 ........ (i) Let the area and breadth of rectangle is 9x and x respectively. Then, Area 9x 9cmBreadth x= =l = On putting the value of l in equation (i) 9+b = 43 b = 43 - 9 = 34 cm
Explanation: Given, ∵ Perimeter of rectangle = 86 cm. ∴ 2(l+b) = 86 ......[where l = length and b = breadth of rectangle] l+b = 862 = 43 ........ (i) Let the area and breadth of rectangle is 9x and x respectively. Then, Area 9x 9cmBreadth x= =l = On putting the value of l in equation (i) 9+b = 43 b = 43 - 9 = 34 cm
187. The length of a rectangle is 10 cm more than the side of a square and is 8 cm less than the side of the square. If the areas of both the rectangle and square are equal then, what will be the perimeter (in cm) of the rectangle?
- A. 160
- B. 156
- C. 164
- D. 144
Answer: 164
Explanation: Let each side of the square be a cm. According to the question, (a + 10) (a - 8) = a 2 ⇒ a2 - 8a + 10a - 80 = a 2 ⇒ 2a = 80, ⇒ a = 40 cm ∴ Length of the rectangle = 40 + 10 = 50 cm Breadth of the rectangle = 40 - 8 = 32cm Hence, Perimeter of the rectangle = 2 (50 + 32) = 164 cm
Explanation: Let each side of the square be a cm. According to the question, (a + 10) (a - 8) = a 2 ⇒ a2 - 8a + 10a - 80 = a 2 ⇒ 2a = 80, ⇒ a = 40 cm ∴ Length of the rectangle = 40 + 10 = 50 cm Breadth of the rectangle = 40 - 8 = 32cm Hence, Perimeter of the rectangle = 2 (50 + 32) = 164 cm
188. If the adjacent sides of a rectangle whose perimeter is 60 cm are in the ratio 3:2, then what will be the area of the rectangle?
- A. 300 cm2
- B. 216 cm2
- C. 60 cm2
- D. 864 cm2
Answer: 216 cm2
Explanation: Let the sides of rectangle are 3x and 2x respectively. So, 2(3x + 2x) = 60 ⇒ 2(5x) = 60 ⇒ 10x = 60 ⇒ x = 6 Sides of rectangle are 18 cm and 12 cm ∴ Area of the rectangle = 18 × 12 = 216 cm 2
Explanation: Let the sides of rectangle are 3x and 2x respectively. So, 2(3x + 2x) = 60 ⇒ 2(5x) = 60 ⇒ 10x = 60 ⇒ x = 6 Sides of rectangle are 18 cm and 12 cm ∴ Area of the rectangle = 18 × 12 = 216 cm 2
189. The area of a square and rectangle are equal. The length of the rectangle is greater than the length of a side of the square by 10 cm and the breadth is less than 5 cm. The perimeter (in cm) of the rectangle is
- A. 40
- B. 50
- C. 100
- D. 80
Answer: 50
Explanation: Let the length of a side of the square = x cm Length of the rectangle = (x+10) cm Breadth = (x-5) cm Area of the square = Area of the rectangle x2 = (x+10) (x-5) x2 = x2 - 5x + 10x - 50 50 = 5x x = 10 Perimeter of the rectangle = 2(x+10+x-5) = 2 (2x+5) = 2×25 = 50 cm
Explanation: Let the length of a side of the square = x cm Length of the rectangle = (x+10) cm Breadth = (x-5) cm Area of the square = Area of the rectangle x2 = (x+10) (x-5) x2 = x2 - 5x + 10x - 50 50 = 5x x = 10 Perimeter of the rectangle = 2(x+10+x-5) = 2 (2x+5) = 2×25 = 50 cm
190. The area of a triangular field whose sides are 65 m, 72 m, and 97 m is equal to the area of a rectangular park whose sides are in the ratio of 5 : 13. What is the perimeter (in m) of the rectangular park?
- A. 108
- B. 180
- C. 216
- D. 144
Answer: 216
191. The perimeter of rectangular field is 386 m and the difference between its two adjacent sides is 95 m. The side of a square field, having the same area as that of the rectangle, is
- A. 64 m
- B. 84 m
- C. 74 m
- D. 82 m
Answer: 84 m
192. The length of the diagonal of rectangle is 26 cm and one side 10 cm. Find the area of the rectangle.
- A. 240 cm2
- B. 260 cm2
- C. 65 cm2
- D. 130 cm2
Answer: 240 cm2
193. A boy walked along two adjacent sides of a rectangular field. If he had walked along the diagonal, then he would have saved a distance equal to one-fourth of the larger side. The ratio of the larger to the smaller side is
- A. 7 : 24
- B. 11 : 18
- C. 24 : 7
- D. 18 : 11
Answer: 24 : 7
194. A rectangular LCD has an area of 480 cm 2 and a perimeter of 92 cm. If the size of the LCD is defined in terms of the length of the diagonal, then what is the size of LCD?
- A. 35 cm
- B. 31 cm
- C. 39 cm
- D. 34 cm
Answer: 34 cm
Explanation: Let the length and breadth be l and b respectively According to the question Area of rectangular LCD = 480 cm 2 l.b = 480 cm2 ……………(i) And perimeter = 2(l + b) 92 = 2( l + b) l + b = 46 cm ……………(ii) ∵ (l + b)2 = l2 + b2 + 2lb (46) 2 = l2 + b2 + 2 × 480 [from eq n (i) & (ii)] 2116 - 960 = l 2 + b2 l2 + b2 = 1156 ∴Diagonal of rectangular LCD = 2 2 b+l 1156= = 34 cm
Explanation: Let the length and breadth be l and b respectively According to the question Area of rectangular LCD = 480 cm 2 l.b = 480 cm2 ……………(i) And perimeter = 2(l + b) 92 = 2( l + b) l + b = 46 cm ……………(ii) ∵ (l + b)2 = l2 + b2 + 2lb (46) 2 = l2 + b2 + 2 × 480 [from eq n (i) & (ii)] 2116 - 960 = l 2 + b2 l2 + b2 = 1156 ∴Diagonal of rectangular LCD = 2 2 b+l 1156= = 34 cm
195. A rectangular park is 120 long and 104 m wide. A 1-m wide path runs along the boundary of the park, remaining completely inside the park area. Thus, the outside edges of the path run along the boundary wall of the park. The inside edges of the path are marked with a white line of negligible thickness. If it costs ₹2.50 to mark each metre with the white line, then how much would it cost (in ₹) to fully mark the inside edges of the path?
- A. 1090
- B. 1080
- C. 1120
- D. 1100
Answer: 1100
Explanation: Perimeter of inner edge of the path = 2(l+b) = 2(118+102) = 440 m Cost to mark each metre with white line = ₹2.50 Hence the required cost to fully mark the inside ed ges of the path = 440 × 2.50 = ₹1100
Explanation: Perimeter of inner edge of the path = 2(l+b) = 2(118+102) = 440 m Cost to mark each metre with white line = ₹2.50 Hence the required cost to fully mark the inside ed ges of the path = 440 × 2.50 = ₹1100
196. The length of rectangle is 5 times of its breadth. If breadth is 10 cm, then what is the perimeter of this rectangle?
- A. 95 cm
- B. 60 cm
- C. 110 cm
- D. 120 cm
Answer: 120 cm
Explanation: According to the question, Breadth of rectangle = 10 cm Then the length of rectangle = 5 × 10 = 50 cm Hence, Perimeter of rectangle = 2 (l + b) = 2 (50 + 10) = 2 × 60 = 120cm
Explanation: According to the question, Breadth of rectangle = 10 cm Then the length of rectangle = 5 × 10 = 50 cm Hence, Perimeter of rectangle = 2 (l + b) = 2 (50 + 10) = 2 × 60 = 120cm
197. The diagonal of a square A is (a + b) units. What is the area (in square units) of the square drawn on the diagonal of square B whose area is twice the area of A?
- A. 2(a+b) 2
- B. 4(a+b) 2
- C. 8(a+b) 2
- D. (a+b) 2
Answer: 2(a+b) 2
198. The perimeter of a square is 64 cm. Its area will be
- A. 256 cm 2
- B. 32 cm 2
- C. 8 cm 2
- D. 128 cm 2
Answer: 256 cm 2
Explanation: The perimeter of a square = 64 cm 4a = 64 a = 16 cm Area of square = a2 = 256 cm2
Explanation: The perimeter of a square = 64 cm 4a = 64 a = 16 cm Area of square = a2 = 256 cm2
199. If the width of the path around a square field is 4.5m and the area of path is 252m 2 then the length of the side of the field is
- A. 9.5 m
- B. 9 m
- C. 8 m
- D. 8.5
Answer: 9.5 m
Explanation: Let the side of square field = a Side of square field along the path (a+4.5+4.5) = (a+9) Area of square = a2 Now, According to the question, 252 = (a+9)2-a2 252 = a 2+81+18a-a2 252 -81=18a 171 a18 = 9.5m = a ∴ a = 9.5m
Explanation: Let the side of square field = a Side of square field along the path (a+4.5+4.5) = (a+9) Area of square = a2 Now, According to the question, 252 = (a+9)2-a2 252 = a 2+81+18a-a2 252 -81=18a 171 a18 = 9.5m = a ∴ a = 9.5m
200. The perimeter of a square is 64 cm. What will be its area?
- A. 256 cm 2
- B. 32 cm 2
- C. 8 cm 2
- D. 128 cm 2
Answer: 256 cm 2
Explanation: Perimeter of square= 64 cm 4a = 64 a = 16 cm Hence the area of square = (a 2) = 256 cm2
Explanation: Perimeter of square= 64 cm 4a = 64 a = 16 cm Hence the area of square = (a 2) = 256 cm2
201. The area of a square ABC is 36 cm 2. Find the area of the square obtained by joining the mid points of the side of the square ABCD.
- A. 18
- B. 27
- C. 24
- D. 21
Answer: 18
202. The perimeter of a square and a circle are same. If the area of the circle is 1386 cm 2, then what will be the area of the square?
- A. 1089 cm 2
- B. 841 cm 2
- C. 1024 cm 2
- D. 1225 cm 2
Answer: 1089 cm 2
Explanation: Area of the circle = 1386 cm2 πr2 = 1386 222 r 13867 = 2 1386 7r 63 7 22×= = × r = 21cm Perimeter of a square = circumference of circle 4× sides 222 217= × × Side = 33 cm. Area of the square = (33)2 =1089
Explanation: Area of the circle = 1386 cm2 πr2 = 1386 222 r 13867 = 2 1386 7r 63 7 22×= = × r = 21cm Perimeter of a square = circumference of circle 4× sides 222 217= × × Side = 33 cm. Area of the square = (33)2 =1089
203. If the area of circle is 154 cm 2, then ratio of circumference of this circle to circumference of another circle whose radius is 21 cm.
- A. 1 : 3
- B. 2 : 3
- C. 2 : 1
- D. 1 : 2
Answer: 1 : 3
204. Two small circular grounds of diameters 42 m and 26 m are to be replaced by a bigger circular ground. What would be the radius of the new ground if the new ground has the same area as the two small grounds?
- A. 25.01 m
- B. 24.69 m
- C. 25 m
- D. 23 m
Answer: 24.69 m
205. What is the area of the largest square which can be inscribed in a circle of radius 28 cm?
- A. 3136 cm2
- B. 1568 cm2
- C. 784 cm2
- D. 196 cm2
Answer: 1568 cm2
Explanation: QDiagonal of square = Diameter of circle 2a 2R 2 28 56= = × = Where, R = radius of circle a = side of square 56a cm 2 = ∴ Area of square 2 56 56(a) 28 562×= = = × = 1568 cm2
Explanation: QDiagonal of square = Diameter of circle 2a 2R 2 28 56= = × = Where, R = radius of circle a = side of square 56a cm 2 = ∴ Area of square 2 56 56(a) 28 562×= = = × = 1568 cm2
206. The measure of central angle of a sector of circle of radius 30 cm is 210º. What is the area of the sector (in cm 2)?
- A. 1650
- B. 1645
- C. 1649
- D. 1647
Answer: 1650
207. Two concentric circles form a ring. The inner and outer circumferences of the ring are 22 cm and 44 cm respectively. The width of the ring is : (take 22π = 7 )
- A. 1.5 cm
- B. 2.5 cm
- C. 3 cm
- D. 3.5 cm
Answer: 1.5 cm
Explanation: (d) Outer circumferences - inner circumferences = 44 - 22 2πR - 2πr = 22 2 π (R-r) = 22 2× 227 (R-r) = 22 (R-r) = 72 = 3.5 cm Hence, the width of the ring = 3.5 cm
Explanation: (d) Outer circumferences - inner circumferences = 44 - 22 2πR - 2πr = 22 2 π (R-r) = 22 2× 227 (R-r) = 22 (R-r) = 72 = 3.5 cm Hence, the width of the ring = 3.5 cm
208. The diameter of wheel of a car has 210 cm. How many revolutions per minute must the wheel make so that the speed of the car is kept at 120 km/h?
- A. 245
- B. 326.42
- C. 303.03
- D. 289
Answer: 303.03
Explanation: Let the required number of revolutions = n According to the question, Distance covered by car in 1 hour = 120 km Distance covered by car in 1 minute 120 km60= 120 22 1052 n60 7 100 1000= × 120 1000 20000n 6 22 3 66×= = × × n = 303.03
Explanation: Let the required number of revolutions = n According to the question, Distance covered by car in 1 hour = 120 km Distance covered by car in 1 minute 120 km60= 120 22 1052 n60 7 100 1000= × 120 1000 20000n 6 22 3 66×= = × × n = 303.03
209. The area of the quadrant of a circle whose circumference is 22 cm, will be
- A. 38.5 cm 2
- B. 10 cm 2
- C. 3.5 cm 2
- D. 9.625 cm 2
Answer: 9.625 cm 2
210. The radii of two circles are 20 cm and 13 cm, respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
- A. 30 cm
- B. 33 cm
- C. 28 cm
- D. 32 cm
Answer: 33 cm
Explanation: Let the radius of two circles are r 1 and r 2 respectively And the radius of other circle = R ∴ Circumference of radius of larger circle = Sum of circumference of both smaller circles 1 22 R 2 r 2 rπ = π + π R = 20 + 13 R = 33 cm
Explanation: Let the radius of two circles are r 1 and r 2 respectively And the radius of other circle = R ∴ Circumference of radius of larger circle = Sum of circumference of both smaller circles 1 22 R 2 r 2 rπ = π + π R = 20 + 13 R = 33 cm
211. The area of the largest triangle that can be inscribed in a semi-circle of radius 4 cm in square centimetres is
- A. 18 cm2
- B. 12 cm2
- C. 14 cm2
- D. 16 cm2
Answer: 16 cm2
Explanation: Radius of semi-circle = 4 cm Diameter of semi-circle (AB) = 8 cm QThe height of triangle will be equal to the radius of the semi-circle. Area of area largest triangle = 1 8 42 × × = 16 cm2
Explanation: Radius of semi-circle = 4 cm Diameter of semi-circle (AB) = 8 cm QThe height of triangle will be equal to the radius of the semi-circle. Area of area largest triangle = 1 8 42 × × = 16 cm2
212. A = Area of the largest circle drawn inside a square of side 1 cm. B = Sum of areas of 4 identical (largest possible) circles drawn inside a square of side 1 cm. C = Sum of areas of 9 identical circle (largest possible) drawn inside a square of side 1 cm. D = Sum of areas of 16 identical circles (largest possible) drawn inside a square of side 1 cm. Which of the following is TRUE about A, B, C and D ?
- A. A > B > C > D
- B. A < B < C < D
- C. A > B = C > D
- D. No option is correct
Answer: No option is correct
213. A rectangular sheet of length 42 cm and breadt h 14 cm is cut from a circular sheet. What is the minimum area (in cm2) of circular sheet ?
- A. 3080
- B. 1540
- C. 770
- D. 1030
Answer: 1540
214. The base of right prism is a trapezium whose parallel sides are 11 cm and 15 cm and the distance between them is 9 cm. If the volume of the prism is 1731.6 cm 3, then the height (in cm) of the prism will be
- A. 14.2
- B. 14.8
- C. 15.6
- D. 15.2
Answer: 14.8
215. Find the area and circumference of a circle if the radius is 14 cm (Take 22π = 7 )
- A. Area = 88 cm 2; Circumference = 616 cm
- B. Area = 44 cm 2; Circumference = 308 cm
- C. Area = 308 cm 2; Circumference = 44 cm
- D. Area = 616 cm 2; Circumference = 88 cm
Answer: Area = 616 cm 2; Circumference = 88 cm
Explanation: Area of circle 2 2 22r 14 14 616 cm7= π = × × = Circumference of the circle 222 r 2 14 88 cm7= π = × × =
Explanation: Area of circle 2 2 22r 14 14 616 cm7= π = × × = Circumference of the circle 222 r 2 14 88 cm7= π = × × =
216. The radius of a circular garden is 42m. The distance (in m) covered by running 8 rounds around it, is: (Take 227π = )
- A. 3248
- B. 2112
- C. 1124
- D. 4262
Answer: 2112
Explanation: Circumference of given circular garden =2πr 222 427= × × And the total distance covered by running 8 rounds around it = 8 × circumference of circular garden 228 2 42 7 = × × × = 2112 cm2
Explanation: Circumference of given circular garden =2πr 222 427= × × And the total distance covered by running 8 rounds around it = 8 × circumference of circular garden 228 2 42 7 = × × × = 2112 cm2
217. The radii of three concentric circles are in the ratio of 4 : 5 : 7. What is the ratio of the area between the two inner circles to that between the two outer circles?
- A. 4 : 5
- B. 3 : 8
- C. 5 : 9
- D. 4 : 7
Answer: 3 : 8
218. The area of a circle that is inscribed in a sq uare of area 2917 cm is :11
- A. 22 cm 2
- B. 28 cm 2
- C. 16 cm 2
- D. 14 cm 2
Answer: 14 cm 2
219. If the radius of a circle is equal to a diagon al of a square whose area is 12 cm 2, then the area of the circle is
- A. 24 π cm2
- B. 36 π cm2
- C. 28 π cm2
- D. 32 π cm2
Answer: 24 π cm2
220. The sum of radii of two circles is 286 cm and the area between the concentric circles is 50336 cm2. What will be the radii (in cm) of both the circles?
- A. 91 and 84
- B. 171 and 84
- C. 115 and 91
- D. 115 and 171
Answer: 91 and 84
221. A 35 cm high bucket in the form of a frustum is full of water. Radii of its lower and upper ends are 12cm and 18 cm, respectively. If water from this bucket is poured in a cylindrical drum, whose base radius is 20 cm, then what will be the height of water (in cm ) in the drum?
- A. 20.50
- B. 18.25
- C. 16.25
- D. 19.95
Answer: 19.95
222. The hour hand moves through 4 hours and has a length of 6 cm. Find the area (in cm 2, rounded off to two decimal places) of the sector covered by the hour hand.
- A. 35.75
- B. 30.67
- C. 37.71
- D. 32.69
Answer: 37.71
Explanation: The length of hour hand = radius of circle = 6cm Area of clock (circle) covered in 12 hours by hour hand 12 = πr2 The area of the sector covered in 4 hours = πr2 × 412 = 22 16 67 3× × × = 22 3621× = 37.71 cm2
Explanation: The length of hour hand = radius of circle = 6cm Area of clock (circle) covered in 12 hours by hour hand 12 = πr2 The area of the sector covered in 4 hours = πr2 × 412 = 22 16 67 3× × × = 22 3621× = 37.71 cm2
223. The circumference of the two circle is 264 cm and 396 cm respectively. What is the difference between their radii?
- A. 32 cm.
- B. 25 cm.
- C. 16 cm.
- D. 21 cm.
Answer: 21 cm.
Explanation: Circumference of first circle 2 πR1= 264 cm 1 222 R 2647× × = R1 = 42 cm Circumference of second circle = 2πR2 = 396 cm 2 222 R 3957× × = R 2 = 63 cm Hence the required difference = R 1 ∼ R2 = 63 - 42 = 21 cm
Explanation: Circumference of first circle 2 πR1= 264 cm 1 222 R 2647× × = R1 = 42 cm Circumference of second circle = 2πR2 = 396 cm 2 222 R 3957× × = R 2 = 63 cm Hence the required difference = R 1 ∼ R2 = 63 - 42 = 21 cm
224. The radius of a circle with centre at O is 6 cm and the central angle of a sector is 40º. Find the area of the sector.
- A. 8π cm2
- B. 6π cm2
- C. 5π cm2
- D. 4π cm2
Answer: 4π cm2
225. In quadrilateral PQRS, RM ⊥ QS, PN ⊥ QS and QS = 6cm. If RM = 3cm and PN = 2cm, then the area of PQRS is
- A. 15 cm 2
- B. 14 cm 2
- C. 13 cm 2
- D. 11 cm 2
Answer: 15 cm 2
226. The length of each side of a rhombus is 10 cm. If the length of one its diagonals is 16 cm, then what is the area of rhombus?
- A. 128 cm2
- B. 108 cm2
- C. 96 cm2
- D. 112 cm2
Answer: 96 cm2
Explanation: In right angle triangle ∆OBC (BC)2 = (OC)2 + (OB)2 (10) 2 = (OC)2 + (8)2 (OC) 2 = 100-64 (OC) 2 = 36 OC = 6 cm second diagonal (AC) = 2(OC) = 2 × 6 = 12 cm Area of rhombus 12= × First diagonal × second diagonal 1 16 122= × × = 96 cm2
Explanation: In right angle triangle ∆OBC (BC)2 = (OC)2 + (OB)2 (10) 2 = (OC)2 + (8)2 (OC) 2 = 100-64 (OC) 2 = 36 OC = 6 cm second diagonal (AC) = 2(OC) = 2 × 6 = 12 cm Area of rhombus 12= × First diagonal × second diagonal 1 16 122= × × = 96 cm2
227. How many bricks, of size 25 cm × 11.25 cm × 6 cm, will be needed to build a wall of 8 m × 6 m × 22.5 cm?
- A. 7020
- B. 6800
- C. 6400
- D. 5800
Answer: 6400
Explanation: We know that, จ Volume of cuboid = Length × Breadth × Height ∴ Volume of brick = 25 cm × 11.25 cm × 6 cm = 1687.5 cm 3 And volume of a wall = 800 × 600 × 22.5 = 10800000 cm 3 Hence, number of brick 10800000 64001687.5= =
Explanation: We know that, จ Volume of cuboid = Length × Breadth × Height ∴ Volume of brick = 25 cm × 11.25 cm × 6 cm = 1687.5 cm 3 And volume of a wall = 800 × 600 × 22.5 = 10800000 cm 3 Hence, number of brick 10800000 64001687.5= =
228. The length of the parallel sides of a trapezium are 10 cm and 20 cm. If the distance between these parallel lines is 5 cm, then what is the area of the trapezium?
- A. 60 cm2
- B. 50 cm2
- C. 75 cm2
- D. 45 cm2
Answer: 75 cm2
229. The ratio of the areas of two triangles is 1 : 2 and the ratio of their bases is 3 : 4. What will be the ratio of their height?
- A. 1 : 3
- B. 4 : 3
- C. 2 : 1
- D. 2 : 3
Answer: 2 : 3
Explanation: Let the height of first triangle is h 1 and second triangle is h 2. According to the question, 1 2 1 3 h1 2 12 4 h2 × × = × × ⇒ 1 2 h 4 h 6= ⇒ 1 2h : h 2 : 3=
Explanation: Let the height of first triangle is h 1 and second triangle is h 2. According to the question, 1 2 1 3 h1 2 12 4 h2 × × = × × ⇒ 1 2 h 4 h 6= ⇒ 1 2h : h 2 : 3=
230. Let ∆ABC and ∆ABD be on the same base AB and between the same parallels AB and CD. Then the relation between areas of ∆ABC and ∆ABD will be
- A. Area of (∆ABC) 12= Area of (∆ABD)
- B. Area of (∆ABC) = Area of (∆ABD)
- C. Area of (∆ABD) 13= Area of (∆ABC)
- D. Area of (∆ABD) 12= Area of (∆ABC)
Answer: Area of (∆ABC) = Area of (∆ABD)
Explanation: Area of triangles are same which drawn between two parallel line on same base. Area of (∆ABC) = Area of (∆ABD)
Explanation: Area of triangles are same which drawn between two parallel line on same base. Area of (∆ABC) = Area of (∆ABD)
231. What is the area of triangle having sides 35 cm , 84 cm and 91 cm?
- A. 2135 cm2
- B. 1530 cm2
- C. 1470 cm2
- D. 1880 cm2
Answer: 1470 cm2
Explanation: Triplet → (5, 12, 13) × 7 = 35, 84, 91 Hence, these are the sides of right angled triangle ∴ Area of triangle 21 35 84 1470cm2= × × =
Explanation: Triplet → (5, 12, 13) × 7 = 35, 84, 91 Hence, these are the sides of right angled triangle ∴ Area of triangle 21 35 84 1470cm2= × × =
232. The base and hypotenuse of a right angled triangle are 9cm and 41 cm respectively. What is the area of the triangle?
- A. 180 cm2
- B. 170 cm2
- C. 190 cm2
- D. 160 cm2
Answer: 180 cm2
233. The sides of an isosceles triangles are 10 cm, 10 cm and 12 cm. What is the area of the triangle?
- A. 60 cm2
- B. 48 cm2
- C. 40 cm2
- D. 44 cm2
Answer: 48 cm2
234. If one side of a triangle is 7 with its perimet er equal to 18, and area equal to 108 , then the other two sides are
- A. 3 and 8
- B. 7 and 4
- C. 3.5 and 7.5
- D. 6 and 5
Answer: 3 and 8
235. In an isosceles triangle ABC with AB = AC and AD is perpendicular to BC, if AD = 6 cm and the perimeter of ∆ABC is 36 cm, then the area of ∆ABC is
- A. 48 cm 2
- B. 45 cm 2
- C. 54 cm 2
- D. 64 cm 2
Answer: 48 cm 2
Explanation: (a) Let equal side of triangle is a cm. Perimeter = 2a+b = 36 cm ....(i) Height (AD) = 6 cm = 2 24a b 2 - On squaring both sides, 36×4 = 4a2 - b2 36×4 = (2a+b) (2a-b) 2a - b = 4 ....(ii) From equations (i) and (ii) a = 10 cm, b = 16 cm ∴ Area of an isosceles triangle = 2 2b 4a b4 - = 16 (2a b)(2a b)4 + - = 4 36 4× = 4×6×2 = 48 cm 2
Explanation: (a) Let equal side of triangle is a cm. Perimeter = 2a+b = 36 cm ....(i) Height (AD) = 6 cm = 2 24a b 2 - On squaring both sides, 36×4 = 4a2 - b2 36×4 = (2a+b) (2a-b) 2a - b = 4 ....(ii) From equations (i) and (ii) a = 10 cm, b = 16 cm ∴ Area of an isosceles triangle = 2 2b 4a b4 - = 16 (2a b)(2a b)4 + - = 4 36 4× = 4×6×2 = 48 cm 2
236. The perimeter of an isosceles triangle is 90 cm . If the base is 26 cm, then find the length of the equal sides.
- A. 42 cm
- B. 30 cm
- C. 32 cm
- D. 40 cm
Answer: 32 cm
Explanation: Let the sides on isosceles triangle are a, a and b respectively. As per the question. 2a + b = 90 (b = 26 cm) 2a + 26 = 90 2a = 90 - 26 ⇒ 2a = 64 a = 32 cm
Explanation: Let the sides on isosceles triangle are a, a and b respectively. As per the question. 2a + b = 90 (b = 26 cm) 2a + 26 = 90 2a = 90 - 26 ⇒ 2a = 64 a = 32 cm
237. The area of an isosceles right angle triangle i s 121 cm2. Find its hypotenuse.
- A. 21 cm
- B. 20 cm
- C. 23 cm
- D. 22 cm
Answer: 22 cm
Explanation: Let two equal sides of an isosceles right angle tri angle AB = BC = a cm ∵ Area of triangle = 12 Base × Height ∴ 121 = 1 a a2 × or 21 a 1212 = a = 11 2 ∴ Hypotenuse of right angle triangle = a 2 = 11 2 2 22× = cm
Explanation: Let two equal sides of an isosceles right angle tri angle AB = BC = a cm ∵ Area of triangle = 12 Base × Height ∴ 121 = 1 a a2 × or 21 a 1212 = a = 11 2 ∴ Hypotenuse of right angle triangle = a 2 = 11 2 2 22× = cm
238. A triangle has sides 25, 39, 34 units. If the a rea of a square exceeds the area of this triangle by 21 sq units, then the side of the square is
- A. 21 unit
- B. 22 unit
- C. 25 unit
- D. 18 unit
Answer: 21 unit
239. What is the area of a triangle whose sides measure 5 cm, 6 cm and 7 cm?
- A. 14.6969 cm 2
- B. 16.4545 cm 2
- C. 12.8484 cm 2
- D. 10.9797 cm 2
Answer: 14.6969 cm 2
240. In an isosceles triangle ABC, AB = AC and AD is perpendicular to BC at D. If AD = 8 cm and perimeter of ∆ABC is 64 cm, then the area of ∆ABC is
- A. 130 cm2
- B. 70 cm2
- C. 120 cm2
- D. 125 cm2
Answer: 120 cm2
Explanation: Let AB = AC = x cm And BC = 2y cm Then, BD = CD = y cm According to the question, Perimeter of triangle = 64 (2x + 2y) = 64 _____(i) In ∆ABD, x2 = y2 + 64 (x+y) (x-y) = 64 (x + y) (x - y) = 2(x+y) [From equation (i)] x-y = 2 ______(ii) On solving equations (i) and (ii), x = 17, y = 15 Hence, Area of ∆ABC = 1 2y 82 × × = 1 2 15 82 × × × = 120 cm 2
Explanation: Let AB = AC = x cm And BC = 2y cm Then, BD = CD = y cm According to the question, Perimeter of triangle = 64 (2x + 2y) = 64 _____(i) In ∆ABD, x2 = y2 + 64 (x+y) (x-y) = 64 (x + y) (x - y) = 2(x+y) [From equation (i)] x-y = 2 ______(ii) On solving equations (i) and (ii), x = 17, y = 15 Hence, Area of ∆ABC = 1 2y 82 × × = 1 2 15 82 × × × = 120 cm 2