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A metallic solid cuboid of sides 44 cm, 32 cm and 36 cm melted and converted into some number of spheres of radius 12 cm. How many such sphere can be made with the metal (π = 22/7)?
1. 5
2. 6
3. 7
4. 8
Correct Answer - Option 3 : 7Given:The sides of the cuboid are 44 cm, 32 cm, and 36 cmThe radius of the sphere is 12 cmConcept Used:The volume of a cuboid of sides l, b and h = l × b × hThe volume of the sphere of radius r = (4/3)πr3Calculation:The volume of the metallic cuboid is (44 × 32 × 36) cm3Read more
Correct Answer – Option 3 : 7
Given:
The sides of the cuboid are 44 cm, 32 cm, and 36 cm
The radius of the sphere is 12 cm
Concept Used:
The volume of a cuboid of sides l, b and h = l × b × h
The volume of the sphere of radius r = (4/3)πr3
Calculation:
The volume of the metallic cuboid is (44 × 32 × 36) cm3
The volume of the sphere is (4/3) × π × 123
Let, the total number of such sphere is n
Accordingly,
44 × 32 × 36 = n × (4/3) × π × 123
⇒ 44 × 32 × 36 = n × (4/3) × (22/7) × 12 × 12 × 12
⇒ n = 44 × 32 × 36 × (3/4) × (7/22) × (1/12) × (1/12) × (1/12)
⇒ n = 7
∴ Such 7 spheres can be made by given metallic cuboid.
See lessAn agricultural field is in the form of a rectangle having length X1 meters and breadth X2 meters (X1 and X2 are variable). If X1 + X2 = 40 meters, then the area of the agricultural field will not exceed which one of the following values?
1. 400 sq m
2. 300 sq m
3. 200 sq m
4. 80 sq m
Correct Answer - Option 1 : 400 sq mGiven length of rectangle = X1Breadth of rectangle = X2Formula Used Area of rectangle = Length × Breadth Calculation ⇒ X1 + X2 = 40 ⇒ We know that, all the rectangle, a square has largest area ⇒ For given rectangle to be square X1 = X2⇒ X1 = X2 = 20 (for maximumRead more
Correct Answer – Option 1 : 400 sq m
Given
length of rectangle = X1
Breadth of rectangle = X2
Formula Used
Area of rectangle = Length × Breadth
Calculation
⇒ X1 + X2 = 40
⇒ We know that, all the rectangle, a square has largest area
⇒ For given rectangle to be square X1 = X2
⇒ X1 = X2 = 20 (for maximum area)
⇒ so Maximum Area = 20 × 20 = 400 sq m
See lessयदिम त्रिज्या वृत्त का क्षेत्रफल A त्रिजया `r` और परिधि C है तो इनमें से कौन सा कथन सत्य है?
A. `rC=2A`
B. `C/A=r/2`
C. `AC=(r^(2))/4`
D. `A/r=C`
Correct Answer - Aवृत्त का क्षेत्रफल `=A` वृत्त की त्रिज्या `=r` वृत्त की परिधि `=c` `pir^(2)=A`……………..i `2ir=c`……………..ii From i `-:` ii `(pir^(2))/(2pir)=A/C` `r/2=A/C` `rc=2A`
Correct Answer – A
See lessवृत्त का क्षेत्रफल `=A`
वृत्त की त्रिज्या `=r`
वृत्त की परिधि `=c`
`pir^(2)=A`……………..i
`2ir=c`……………..ii
From i `-:` ii
`(pir^(2))/(2pir)=A/C`
`r/2=A/C`
`rc=2A`
The ratio of length, width and height of a room is 3 : 2 : 1. If its volume is 3072 cubic meters, find its width.
A. 18 meters
B. 16 meters
C. 24 meters
D. 12 meters
1. B
2. C
3. D
4. A
Correct Answer - Option 1 : BGiven:The ratio of length, width and height = 3 : 2 : 1Volume = 3072 cubic metersFormula used:The volume of cuboid = length × width × heightCalculation:Let the length, width, and height be 3x, 2x, and x meters.The volume of room = length × width × height⇒ 3072 = 3x × 2x Read more
Correct Answer – Option 1 : B
Given:
The ratio of length, width and height = 3 : 2 : 1
Volume = 3072 cubic meters
Formula used:
The volume of cuboid = length × width × height
Calculation:
Let the length, width, and height be 3x, 2x, and x meters.
The volume of room = length × width × height
⇒ 3072 = 3x × 2x × x
⇒ 6x3 = 3072
⇒ x3 = 512
⇒ x = 8
Width = 2x
⇒ Width = 2 × 8
⇒ Width = 16 m
∴ The width of room is 16 meter.
See lessIf the area of a rhombus is 720 cm2 and the length of one of the two diagonals is 80 cm then find the length of each side of the rhombus.
1. 82 cm
2. 41 cm
3. 18 cm
4. 40 cm
Correct Answer - Option 2 : 41 cmGiven:Area of a rhombus = 720 cm2Length of first diagonal = 80 cmConcept used:The two diagonals bisect each other at 90° in a rhombus.Formula used:Area of rhombus = (1/2) × Product of the two diagonalsCalculations:Let the length of second diagonal be xArea of rhombusRead more
Correct Answer – Option 2 : 41 cm
Given:
Area of a rhombus = 720 cm2
Length of first diagonal = 80 cm
Concept used:
The two diagonals bisect each other at 90° in a rhombus.
Formula used:
Area of rhombus = (1/2) × Product of the two diagonals
Calculations:
Let the length of second diagonal be x
Area of rhombus = (1/2) × 80 × x
⇒ 720 = (1/2) × 80x
⇒ x = 18 cm
If the two diagonals of rhombus ABCD intersect at O then,
⇒ AO2 + BO2 = AB2
⇒ (80/2)2 + (18/2)2 = AB2
⇒ AB2 = 402 + 92
⇒ AB2 = 1600 + 81
⇒ AB2 = 1681
⇒ AB = 41 cm
∴ The length of each side of a rhombus is 41 cm.
See lessदो गोलों, के पृष्ठ क्षेत्रफलों का अनुपात `4:9` है। उनके आयतनों का अनुपात ज्ञात करें?
A. `2:3`
B. `4:9`
C. `8:27`
D. `64:729`
Correct Answer - Cगोलों के पृष्ठीय क्षेत्रफल का अनुपात `(4pir_(1)^(2))/(4pir_(2)^(2))=4/9` `(r_(1))/(r_(2))=2/3` आयतन का अनुपात `=(4/3pir_(1)^(3))/(4/3pir_(2)^(3))=((r_(1))/(r_(2)))^(3)=(2/3)^(3)=8/27` `=8:27`
Correct Answer – C
See lessगोलों के पृष्ठीय क्षेत्रफल का अनुपात
`(4pir_(1)^(2))/(4pir_(2)^(2))=4/9`
`(r_(1))/(r_(2))=2/3` आयतन का अनुपात
`=(4/3pir_(1)^(3))/(4/3pir_(2)^(3))=((r_(1))/(r_(2)))^(3)=(2/3)^(3)=8/27`
`=8:27`
A blacksmith bent a steel wire, in the form of a square, encloses an area of 484 sq. cm. The same wire he bent in the form of a circle. Find the area of circle (In sq.cm).
1. 520
2. 544
3. 660
4. 616
Correct Answer - Option 4 : 616Given:Area of square = 484 sq. cmSquare wire is converted into circleFormula used:Perimeter of circle = 2π r, where r = radius of circleArea of circle = π r2Perimeter of square = 4a, where a = side of the squareConcept used:When square shape bent into circle, perimeterRead more
Correct Answer – Option 4 : 616
Given:
Area of square = 484 sq. cm
Square wire is converted into circle
Formula used:
Perimeter of circle = 2π r, where r = radius of circle
Area of circle = π r2
Perimeter of square = 4a, where a = side of the square
Concept used:
When square shape bent into circle, perimeter of square becomes equal to perimeter of circle
Area of square = a2
∴ a = √area = √484 = 22 cm
∴ perimeter of square = 4a = 4 × 22 = 88 cm
Now, perimeter of circle = perimeter of square = 88 cm
⇒ 2π r = 88
∴ r = 88 / 2π = 88 × 7 / 22 × 1 / 2 = 14 cm
∴ Area of circle = π r2 = 22 / 7 × 142 = 22 / 7 × 196 = 616 sq.cm
Therefore, area of circle = 616 sq.cm
See lessA metal wire when bent in the form of a square encloses an area 121 cm2. If the same wire is bent in the form of a circle, then its area is:
1. 308 cm2
2. 254 cm2
3. 100 cm2
4. 154 cm2
Correct Answer - Option 4 : 154 cm2Given:A metal wire when bent in the form of a square encloses an area 121 cm2.Formula used:Area of circle = \(\pi \) × r2Radius = r , \(\pi \) = 22/7Calculation: Side of square = √ area⇒ √ 121⇒ 11 cmLength of the wire or circumference of the wire = 4 × side⇒ 4 × 11Read more
Correct Answer – Option 4 : 154 cm2
Given:
A metal wire when bent in the form of a square encloses an area 121 cm2.
Formula used:
Area of circle = \(\pi \) × r2
Radius = r , \(\pi \) = 22/7
Calculation:
Side of square = √ area
⇒ √ 121
⇒ 11 cm
Length of the wire or circumference of the wire = 4 × side
⇒ 4 × 11
⇒ 44 cm
Circumference of the circle = 2\(\pi \)r
⇒ 2\(\pi \)r = 44
⇒ r = (44 × 7)/(22 × 2)
⇒ r = 7 cm
Area of the circle = \(\pi \) × r2
⇒ (22/7) × 7 × 7
⇒ 154 cm2
∴ Area of the circle is 154 cm2.
See lessकिसी शंकु की त्रिज्या तथा ऊंचाई का अनुपात `4:3` है। शंकु के तिर्यक पृष्ठ क्षेत्रफल तथा सम्पूर्ण पृष्ठ क्षेत्रफल का अनुपात ज्ञात करें?
A. `5:9`
B. `3:7`
C. `5:4`
D. `16:9`
Correct Answer - ACone `implies` radius `:` height `4:3` Let `4x:3x` `:.`शंकु का पृष्ठीय क्षेत्रफल `implies pirl` `r=` radius `1=` slant height `1=sqrt(h^(2)+r^(2))` `=sqrt((4x)^(2)+(3x)^(2))=5x` `:.`पृष्ठीय क्षेत्रफल `impliespixx4x xx 5x` `implies 20pix^(2)` `:.` Total surface area `impliespirl+pirRead more
Correct Answer – A
See lessCone `implies` radius `:` height
`4:3`
Let `4x:3x`
`:.`शंकु का पृष्ठीय क्षेत्रफल
`implies pirl`
`r=` radius
`1=` slant height
`1=sqrt(h^(2)+r^(2))`
`=sqrt((4x)^(2)+(3x)^(2))=5x`
`:.`पृष्ठीय क्षेत्रफल
`impliespixx4x xx 5x`
`implies 20pix^(2)`
`:.` Total surface area
`impliespirl+pir^(2)`
`implies pir(l+r)`
`implies pixx 4x(5x+4x)`
`implies pixx4x xx 9x`
`implies 36pix^(2)`
`:.` Curved `:` Total area
`20pix^(2):36pix^(2)`
`5:9`
Applications of mensuration in real life
Mensuration is a subject or branch of Mathematics.Mensuration tells us about the lengths of sides, heights and perimeters, measures of angles, surface areas and volumes of 2-dimensional plates and 3-dimensional solids. Examples of different shapes are triangle, square, polygon, cylinder, cone, pyramRead more
Mensuration is a subject or branch of Mathematics.
Mensuration tells us about the lengths of sides, heights and perimeters, measures of angles, surface areas and volumes of 2-dimensional plates and 3-dimensional solids.
Examples of different shapes are triangle, square, polygon, cylinder, cone, pyramid, cuboid etc.
Some of the real life applications are:-
- Amount of paint required to cover a certain surface area
- Amount of carpet required for a particular room
- Fencing needed for the perimeter of a garden
- Pavement tessellation of a pathway
- Volume of soil needed to fill in a ditch
- The distance around a circular race track
- Travel and roadmap reading
- Amount of fuel needed for a given journey
- Gift wrapping
- Finding capacities of containers, tanks etc.
See less