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Q1. Surface area of a cube is 486 cm². Find the length of its side.

• 1) 12 cm
• 2) 6 cm
• 3) 8 cm
• 4) 9 cm

Solution

6 x side² = 486 side² = 81 side = 9 cm 

Q2. A rectangular tank is 3 m long, 2.5 m wide and 1.4 m deep. Calculate its capacity in liters.

Solution

Here, L = 3 m, B = 2.5 m and H = 1.4 m Volume of tank = L x B x H = 3 x 2.5 x 1.4 = 10.5 m³ Now, 1 m³ = 1000 liters Therefore, Capacity of tank = 10.5 x 1000 = 10500 liters.

Q3. A cuboidal box with sides 21 cm x 20 cm x 18 cm is to be filled with cuboidal boxes with sides 7 cm x 6 cm x 5 cm. How many cuboids can be placed in it? 

• 1) None
• 2) 26
• 3) 20
• 4) 22

Solution

number of cuboids = =  = 36 

Q4. Find the area of the quadrilateral.  

• 1) 250 cm²
• 2) 450 cm²
• 3) 400 cm²
• 4) 300 cm²

Solution

Area of a triangle = ½ × base × heightArea of a quadrilateral =     = 100 + 200= 300 cm²

Q5. In the figure given below, a circle is inscribed inside another circle. The radius of the outer circle is 12 cm and that of the inner circle is 7 cm. Find the area of the shaded portion between the circles.  

Solution

Given, Radius of the outer circle = 12 cm So, area of the outer circle = r² = 3.14 x 12² = 452.16 cm² Radius of the inner circle = 7 cm Area of the inner circle = r² = x 7² = 154 cm² Thus, the required area of the shaded portion = Area of the outer circle - Area of the inner circle = 452.16 - 154 = 298.16 cm²

Q6. In the given figure, ABCD is a trapezium in which parallel sides AB and DC are of length 24 m and 10 m. Also, CE||AD. If area of triangle CEB is 84 m². Find the area of given trapezium.

Solution

Now AECD is a parallelogram, therefore AE = DC = 10 m Therefore, EB = 24 - 10 = 14 m Now, area of triangle CEB = 84 m²

Q7. The volume of a cylinder is 1408 cm³ and its height is 7 cm. Find its total surface area.

Solution

Here, h = 7 cm Let r be the radius of cylinder. Now, Volume = 1408 cm³ Therefore,

Q8. Raju wants to fence the garden in his house which is in the shape of a rectangle having dimensions 25 m x 18 m. Find the cost of fencing at the rate of Rs 200 per metre.

Solution

The length of the fence required is the perimeter of the rectangular garden. Given that: Length (l) = 25 m ; Breadth (b) = 18 m i.e., Perimeter = 2 x (l + b) = 2 x (25 + 18) = 2 x 43 = 86 m So, the perimeter of the garden or the length of the fence required is 86 m. Cost of fencing 1 m = Rs 200 Thus, Cost of fencing 86 m = Rs 200 x 86 = Rs 17,200

Q9. A rectangular park is 38 m long and 15 m wide. A path 3.5 m wide is constructed outside the park. Find the perimeter of the path.

Solution

The above data can be shown in a figure as follows: Let PQRS represent the rectangular park and the shaded region represent the path 3.5 m wide. Thus, to find the length AB and breadth BC, we have to add 3.5 m to both sides of rectangular park whose dimensions are 38 x 15 So, the length and breadth of the path are: Length AB = (38 + 3.5 + 3.5) m = 45 m Breadth BC = (15 + 3.5 + 3.5) m = 22 m So, perimeter of the path = 2 x (l + b) = 2 x (45 + 22) = 2 x 67 = 134 m Thus, perimeter of the path is 134 m.

Q10. An iron pipe is 56 cm long with its outer diameter 5 cm and inner diameter 4 cm. If the metal weighs 12 grams per cubic cm, what is the weight of pipe?

Solution

Here, Height of pipe (h) = 56 cm Inner radius (r) =4/2 = 2 cm  Outer radius (R) = 5/2 cm Now, weight of pipe = 396 x 12 grams = 4752 grams.

Q11. How many 5 cm cubes can be obtained from a cube of side 30 cm.

Solution

Volume of cube of side 5 cm = (Side)³ =5³ = 125 cm³ Also, the volume of cube of side 30 cm = (Side)³ =30³ = 27000 cm³ Number of cubes of side 5 cm = Hence, 216 cubes are obtained.

Q12. A wire is in the shape of a square of side 22 cm. If the wire is re-bent into a circle, find its radius. Also find the area of circle.

Solution

Given that the side of the square = 22 cm Length of the wire = perimeter of the square = 4 x side = 4 x 22 = 88 cm Since the same wire is used to make the circle, both will have the same perimeter. So, perimeter of the circle = 88 Thus, radius of circle = 14 cm Now, area of circle =

Q13. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use = 3.14)

Solution

From the figure we have Height (h₁) of larger cylinder = 220cm Radius (r₁) of larger cylinder = = 12cm Height (h₂) of smaller cylinder = 60cm Radius (r₂) of smaller cylinder = 8cm   Mass of 1 cm³ iron = 8 gm Mass of 111532.8 cm³ iron = 111532.8 x 8 = 892262.4 gm = 892.262 kg

Q14. A closed cylindrical tank of radius 7 cm and height 5 cm is made from a metal sheet. If the breadth of the rectangular sheet is 10 cm, then find the length of the sheet.

Solution

Here, Radius (r) = 7 cm and height (h) = 5 cm Total surface area of cylindrical tank = Let the length of sheet required be x cm. Now, Area of sheet = Area of cylindrical tank Hence the length of sheet is 52.8 cm.

Q15. The surface area of a cuboidal wooden box is 470 cm². If its length and breadth is 15 cm and 8 cm respectively. Find its height.

Solution

Here, l = 15 cm and b = 8 cm Let the height of box be h cm. Now, Surface area of box = 2(lb + bh + hl) But surface area is given 470 cm². Therefore, we have

Q16. Find the surface area of the four walls of a cubical room whose side is 4m. Find the cost of whitewashing the room at the rate of Rs 5 per square metre.

Solution

Side of cube = 4 m Area of 4 walls = 4(Side)² = 4(4)² = 4 x 16 = 64 m². The whitewashing is done on 4 walls and 1 roof, So the surface area of 4 walls and 1 roof = 5(Side)² = 5(4)² = 5 x 16 = 80 m² Cost of whitewashing = Rs 5 x 80 = Rs 400.

Q17. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboids of dimensions?

Solution

Coins are cylindrical in shape. Height (h₁) of cylindrical coins = 2 mm = 0.2 cm Radius (r) of circular end of coins = Let n coins were melted to form the required cuboids. Volume of n coins = Volume of cuboids Hence, number of coins melted to form such cuboids is 400.

Q18. Find the area of the following quadrilateral ABCD, where the length of diagonal AC = 8 cm and the length of perpendiculars DP and BQ are 4.5 cm and 3.5 cm respectively.

Solution

Here, Diagonal AC = 8 cm Perpendicular DP = 4.5 cm And, Perpendicular BQ = 3.5 cm Therefore, sum of perpendicular = DP + BQ = 4.5 + 3.5 = 8 cm Now, Area of quadrilateral =

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