Mensuration
- The breadth of a rectangular hall is three-fourth of its length. If the area of the floor is 768 sq. m., then the difference between the length and breadth of the hall is:
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Using Rule 9,
Let the length of rectangular hall = x–metre∴ Breadth = 
3 × x 
metre Area of rectangular √4 = Length × Breadth = x × 3 x sq.m. = 3 x² m.² 4 4
∴ According to question,= 3 x² = 768 4 x² = 768 × 4 3 or, x = √ 768 × 4 = 32m 3
∴ Length = 32 m and
Breadth = 24m
∴ Required difference = 32 – 24 = 8 mCorrect Option: A
Using Rule 9,
Let the length of rectangular hall = x–metre∴ Breadth = 3 × x metre Area of rectangular √4 = Length × Breadth = x × 3 x sq.m. = 3 x² m.² 4 4
∴ According to question,= 3 x² = 768 4 x² = 768 × 4 3 or, x = √ 768 × 4 = 32m 3
∴ Length = 32 m and
Breadth = 24m
∴ Required difference = 32 – 24 = 8 m
- The length of a rectangular garden is 12 metres and its breadth is 5 metres. Find the length of the diagonal of a square garden having the same area as that of the rectangular garden :
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Using Rule 9 and 10,
Area of the rectangular garden = 12 × 5 = 60 m²
= Area of the square garden
∴ Side of the square garden = 60² m
∴ Diagonal of the square garden = 2 × side
= √2 × √60 = √120 = √4 × 30 = √230mCorrect Option: A
Using Rule 9 and 10,
Area of the rectangular garden = 12 × 5 = 60 m²
= Area of the square garden
∴ Side of the square garden = 60² m
∴ Diagonal of the square garden = 2 × side
= √2 × √60 = √120 = √4 × 30 = √230m
- The length of a plot is five times its breadth. A playground measuring 245 square metres occupies half of the total area of the plot. What is the length of the plot?
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Using Rule 9,
Let breadth of plot = x m
∴ length = 5x m. According to question,5x² = 245 2 ⇒ x² = 245 × 2 = 98 5
⇒ x = 72m
∴ Length = 5 × 7 √2 = 35√2 m
Correct Option: A
Using Rule 9,
Let breadth of plot = x m
∴ length = 5x m. According to question,5x² = 245 2 ⇒ x² = 245 × 2 = 98 5
⇒ x = 72m
∴ Length = 5 × 7 √2 = 35√2 m
- A circular wire of diameter 42 cm is folded in the shape of a rectangle whose sides are in the ratio 6 : 5 . Find the area enclosed
by the rectangle. (Take p = 22/7 )
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Using Rule 9 and 14,
Radius of circular wire = 42 = 21cm 2
Circumference of wire = 2πr= 2 × 22 × 21 = 132cm 7
Let the length and breadth of rectangle be 6 x and 5 x cm respectively.
∴ Perimeter of rectangle = 2 (6x + 5 x) = 22 x
According to the question, 22x = 132⇒ x = 132 = 6 22
∴ Length of rectangle = 6x = 6 × 6 = 36 cm
Breadth of rectangle = 5x = 5 × 6 = 30 cm
∴ Area = 36 × 30 = 1080 cm²Correct Option: B
Using Rule 9 and 14,
Radius of circular wire = 42 = 21cm 2
Circumference of wire = 2πr= 2 × 22 × 21 = 132cm 7
Let the length and breadth of rectangle be 6 x and 5 x cm respectively.
∴ Perimeter of rectangle = 2 (6x + 5 x) = 22 x
According to the question, 22x = 132⇒ x = 132 = 6 22
∴ Length of rectangle = 6x = 6 × 6 = 36 cm
Breadth of rectangle = 5x = 5 × 6 = 30 cm
∴ Area = 36 × 30 = 1080 cm²
- A kite in the shape of a square with a diagonal 32 cm attached to an equilateral triangle of the base 8 cm. Approximately how much paper has been used to make it? (Use √3 = 1.732)
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Using Rule 6 and 10,
Area of paper = Area of square + Area of equilateral triangle= 1 (diagonal)² + √3 × (side)² 2 4 = 1 32 × 32 + √3 × 8 × 8 2 4
= 512 + 16 × 1.732
= 512 + 27.712 = 539.712 cm²
[Note : Diagonal of a square = √2 side]Correct Option: A
Using Rule 6 and 10,
Area of paper = Area of square + Area of equilateral triangle= 1 (diagonal)² + √3 × (side)² 2 4 = 1 32 × 32 + √3 × 8 × 8 2 4
= 512 + 16 × 1.732
= 512 + 27.712 = 539.712 cm²
[Note : Diagonal of a square = √2 side]
